Chapter 10: Vector Algebra

Step 1 — Choose a scale. Let a length of $$1\,\text{cm}$$ represent $$10\,\text{km}$$. Then the displacement of $$40\,\text{km}$$ is represented by a directed line segment of length $$\dfrac{40}{10}=4\,\text{cm}$$.

Step 2 — Fix the direction. Begin at a point $$O$$. The phrase '$$30^\circ$$ west of south' means: start from the south direction at $$O$$ and turn through an angle of $$30^\circ$$ towards the west.

Step 3 — Draw the vector. From $$O$$, draw a ray into the south-west region making an angle of $$30^\circ$$ with the downward (south) direction, measured towards the west. Mark a point $$P$$ on this ray so that $$OP=4\,\text{cm}$$, and place an arrowhead at $$P$$.

The directed line segment $$\overrightarrow{OP}$$ then represents the required displacement, with magnitude $$|\overrightarrow{OP}|=40\,\text{km}$$ in the direction $$30^\circ$$ west of south.

Let the positive y-axis represent North and the positive x-axis represent East.

Denote the desired displacement by $$\vec{d}$$. It has

• magnitude $$|\vec{d}| = 40\text{ km}$$,
• direction $$30^\circ$$ east of north: start from the positive y-axis (north) and rotate $$30^\circ$$ towards the positive x-axis (east).

For plotting we find its rectangular components:

$$d_x = |\vec{d}|\,\sin 30^\circ = 40\times \tfrac12 = 20\text{ km},\qquad d_y = |\vec{d}|\,\cos 30^\circ = 40\times \tfrac{\sqrt3}{2} = 20\sqrt3\text{ km}.$$

Choose a convenient scale, e.g. 1 cm = 5 km. Then

• $$d_x = 20\text{ km} \;\Rightarrow\; 4\text{ cm}$$ along the positive x-axis,
• $$d_y = 20\sqrt3\text{ km}\approx 34.6\text{ km} \;\Rightarrow\; 6.92\text{ cm}$$ along the positive y-axis.

Steps to draw the vector

1. Draw mutually perpendicular axes OX (east) and OY (north).
2. From the origin O mark 4 cm to the right (east) and 6.92 cm upward (north) to locate point P$$(20,\,20\sqrt3).$$
3. Join O to P with an arrow pointing towards P. Label the arrow $$\vec{d}\;(40\text{ km},\;30^\circ\text{ E of N}).$$

The arrow OP is the graphical representation of the required displacement.

(If an actual diagram is to be drawn, depict the axes, mark the calculated scales, and show the vector making a $$30^\circ$$ angle with the north-ward direction.)

If two vectors are equal, every corresponding component along the three coordinate axes must be equal.

The given vectors are

$$\vec a = x\hat i + 2\hat j + z\hat k$$

and

$$\vec b = 2\hat i + y\hat j + \hat k$$.

Write each vector in ordered-triplet form (coefficients of i, j, k):

$$\vec a \;\equiv\; (\,x,\;2,\;z\,)$$, $$\vec b \;\equiv\; (\,2,\;y,\;1\,)$$.

Since $$\vec a = \vec b$$, we equate the components one by one:

1. x-component: $$x = 2$$
2. y-component: $$2 = y$$
3. z-component: $$z = 1$$

Thus

$$x = 2,\; y = 2,\; z = 1$$.

Let the two vectors be

$$\vec{a}=\hat{i}+2\hat{j}, \qquad \vec{b}=2\hat{i}+\hat{j}.$$

1. Magnitude of $$\vec{a}$$
   For a two-dimensional vector $$x\hat{i}+y\hat{j}$$ the magnitude is $$\sqrt{x^{2}+y^{2}}$$.
   Here $$x=1,\;y=2$$, hence
   $$|\vec{a}| = \sqrt{1^{2}+2^{2}}=\sqrt{1+4}=\sqrt{5}.$$
2. Magnitude of $$\vec{b}$$
   For $$\vec{b}=2\hat{i}+\hat{j}$$ we have $$x=2,\;y=1$$, so
   $$|\vec{b}| = \sqrt{2^{2}+1^{2}}=\sqrt{4+1}=\sqrt{5}.$$
3. Comparison of magnitudes
   Since $$|\vec{a}|=|\vec{b}|=\sqrt{5},$$ the two vectors have equal length.
4. Are the vectors themselves equal?
   Two vectors are equal only when their corresponding components are equal.
   $$\vec{a}=\hat{i}+2\hat{j}\;(1,2), \quad \vec{b}=2\hat{i}+\hat{j}\;(2,1).$$
   The first components differ (1 ≠ 2), hence $$\vec{a}\neq\vec{b}.$$

Thus the vectors have the same magnitude, but they are not equal.

Step 1 : Write the given vector

$$\vec{a}=2\hat{i}+3\hat{j}+\hat{k}$$

Step 2 : Recall the definition of a unit vector

A unit vector $$\hat{a}$$ in the direction of a non-zero vector $$\vec{a}$$ is obtained by dividing the vector by its magnitude:

$$\hat{a}=\dfrac{\vec{a}}{|\vec{a}|}$$

Step 3 : Compute the magnitude of $$\vec{a}$$

For a vector $$\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$$,

$$|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$$

Here $$a_1=2,\;a_2=3,\;a_3=1$$, so

$$|\vec{a}|=\sqrt{2^2+3^2+1^2}$$

$$|\vec{a}|=\sqrt{4+9+1}$$

$$|\vec{a}|=\sqrt{14}$$

Step 4 : Form the unit vector

Divide each component of $$\vec{a}$$ by $$|\vec{a}|$$:

$$\hat{a}=\dfrac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}}$$

Expressing the components separately,

$$\hat{a}=\dfrac{2}{\sqrt{14}}\,\hat{i}+\dfrac{3}{\sqrt{14}}\,\hat{j}+\dfrac{1}{\sqrt{14}}\,\hat{k}$$

Verification

The magnitude of $$\hat{a}$$ should be $$1$$:

$$|\hat{a}|=\sqrt{\left(\dfrac{2}{\sqrt{14}}\right)^2+\left(\dfrac{3}{\sqrt{14}}\right)^2+\left(\dfrac{1}{\sqrt{14}}\right)^2}=\sqrt{\dfrac{4+9+1}{14}}=\sqrt{\dfrac{14}{14}}=1$$

This confirms that $$\hat{a}$$ is indeed a unit vector.

Step 1 | Write the given vector
The vector whose direction we have to follow is $$\vec a = \hat i - 2\hat j$$.

Step 2 | Find its magnitude
$$|\vec a| = \sqrt{1^{2} + (-2)^{2}} = \sqrt{1 + 4} = \sqrt 5$$

Step 3 | Unit vector in this direction
The unit vector along $$\vec a$$ is
$$\hat u = \frac{\vec a}{|\vec a|} = \frac{1}{\sqrt 5}(\hat i - 2\hat j)$$

Step 4 | Scale to the required magnitude 7
Let $$\vec v$$ be the required vector. Set
$$\vec v = 7\, \hat u = 7\Bigl(\frac{1}{\sqrt 5}(\hat i - 2\hat j)\Bigr) = \frac{7}{\sqrt 5}(\hat i - 2\hat j).$$
Thus
$$\boxed{\vec v = \frac{7}{\sqrt 5}\,\hat i - \frac{14}{\sqrt 5}\,\hat j}$$

Step 5 | Verification (optional)
$$|\vec v| = \sqrt{\left(\frac{7}{\sqrt 5}\right)^{2} + \left(-\frac{14}{\sqrt 5}\right)^{2}} = \sqrt{\frac{49 + 196}{5}} = \sqrt{\frac{245}{5}} = \sqrt{49} = 7,$$ which matches the required magnitude.

We are asked to obtain a unit vector in the direction of the sum of two given vectors.

Given

$$\vec a = 2\hat i + 2\hat j - 5\hat k, \qquad \vec b = 2\hat i + \hat j + 3\hat k$$

First, write down the sum:

$$\vec a + \vec b = (2\hat i + 2\hat j - 5\hat k) + (2\hat i + \hat j + 3\hat k)$$

Combine the corresponding components:

• $$\hat i$$-components: $$2 + 2 = 4$$
• $$\hat j$$-components: $$2 + 1 = 3$$
• $$\hat k$$-components: $$-5 + 3 = -2$$

Hence the sum is

$$\vec{c} = \vec a + \vec b = 4\hat i + 3\hat j - 2\hat k.$$ The required unit vector must be in the same direction as $$\vec c$$, so we now compute the magnitude $$|\vec c|$$:

$$|\vec c| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}.$$

Finally, divide $$\vec c$$ by its magnitude to obtain the unit vector:

$$\hat u = \frac{\vec c}{|\vec c|} = \frac{4\hat i + 3\hat j - 2\hat k}{\sqrt{29}}.$$

Therefore, the required unit vector is $$\displaystyle \hat u = \frac{4}{\sqrt{29}}\,\hat i + \frac{3}{\sqrt{29}}\,\hat j - \frac{2}{\sqrt{29}}\,\hat k.$$

For the vector $$\vec a = \hat i + \hat j - 2\hat k$$ we read the coefficients of $$\hat i,\;\hat j,\;\hat k$$ directly.

1. Direction ratios (d.r.’s)

• The coefficient of $$\hat i$$ is $$1$$.
• The coefficient of $$\hat j$$ is $$1$$.
• The coefficient of $$\hat k$$ is $$-2$$.

Hence the three direction ratios are $$1,\;1,\;-2$$.

2. Direction cosines (d.c.’s)

Let the required direction cosines be $$l,\;m,\;n$$. They are obtained by dividing each direction ratio by the magnitude of the vector.

Magnitude of $$\vec a$$:

$$|\vec a| = \sqrt{1^{2}+1^{2}+(-2)^{2}} = \sqrt{1+1+4} = \sqrt{6}$$

Therefore

$$l = \dfrac{1}{\sqrt 6},\;\; m = \dfrac{1}{\sqrt 6},\;\; n = \dfrac{-2}{\sqrt 6}$$

We can quickly verify the condition $$l^{2}+m^{2}+n^{2}=1$$: $$\left(\frac{1}{\sqrt6}\right)^{2}+\left(\frac{1}{\sqrt6}\right)^{2}+\left(\frac{-2}{\sqrt6}\right)^{2}=\frac1{6}+\frac1{6}+\frac4{6}=1$$, so the values are correct.

The coordinates of the initial point are $$P(2,3,0)$$ and those of the terminal point are $$Q(-1,-2,-4)$$.

To obtain the vector directed from $$P$$ to $$Q$$, subtract the coordinates of $$P$$ from the corresponding coordinates of $$Q$$:

$$\vec{PQ}=\bigl((-1)-(2)\bigr)\,\hat\imath + \bigl((-2)-(3)\bigr)\,\hat\jmath + \bigl((-4)-(0)\bigr)\,\hat k$$

Simplifying each component, we get

$$\vec{PQ}=-3\,\hat\imath-5\,\hat\jmath-4\,\hat k$$

Let the position vectors of the three given points be read directly from the question:

$$\vec{OA}=2\hat{i}-\hat{j}+\hat{k},\;\vec{OB}=\hat{i}-3\hat{j}-5\hat{k},\;\vec{OC}=3\hat{i}-4\hat{j}-4\hat{k}. $$

To verify that $$\triangle ABC$$ is right-angled, we examine the dot product of the vectors representing its sides. For any two sides to be perpendicular, the dot product of their corresponding vectors must be zero.

Step 1 – Form the side vectors.

• Vector $$\vec{AB}=\vec{OB}-\vec{OA}=(\,1-2,\,-3-(-1),\,-5-1\,)=(\,-1,-2,-6\,).$$
• Vector $$\vec{BC}=\vec{OC}-\vec{OB}=(\,3-1,\,-4-(-3),\,-4-(-5)\,)=(\,2,-1,1\,).$$
• Vector $$\vec{AC}=\vec{OC}-\vec{OA}=(\,3-2,\,-4-(-1),\,-4-1\,)=(\,1,-3,-5\,).$$

Step 2 – Compute the three possible dot products.

• $$\vec{AB}\!\cdot\!\vec{BC}=(-1)(2)+(-2)(-1)+(-6)(1)=-2+2-6=-6\neq0.$$
• $$\vec{AB}\!\cdot\!\vec{AC}=(-1)(1)+(-2)(-3)+(-6)(-5)=-1+6+30=35\neq0.$$
• $$\vec{BC}\!\cdot\!\vec{AC}=(2)(1)+(-1)(-3)+(1)(-5)=2+3-5=0.$$

Step 3 – Interpretation.

The dot product $$\vec{BC}\!\cdot\!\vec{AC}=0$$ shows that $$\vec{BC}$$ is perpendicular to $$\vec{AC}$$. Both of these vectors have their common terminal point at $$C$$ (they emanate from $$B$$ and $$A$$ respectively and end at $$C$$). Hence $$\angle BCA$$ is a right angle.

Conclusion. Since one interior angle of $$\triangle ABC$$ is $$90^{\circ}$$, the given points are the vertices of a right-angled triangle, right-angled at $$C$$.

For any vector $$\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$$ its magnitude (or length) is

$$|\vec{v}|=\sqrt{v_x^2+v_y^2+v_z^2}.$$

We apply the same rule to each given vector.

1. Magnitude of $$\vec{a}=\hat{i}+\hat{j}+\hat{k}$$

Identify the components: $$a_x=1,\;a_y=1,\;a_z=1.$$

$$|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{1+1+1}=\sqrt{3}.$$

2. Magnitude of $$\vec{b}=2\hat{i}-7\hat{j}-3\hat{k}$$

Components: $$b_x=2,\;b_y=-7,\;b_z=-3.$$

$$|\vec{b}|=\sqrt{2^2+(-7)^2+(-3)^2}=\sqrt{4+49+9}=\sqrt{62}.$$

3. Magnitude of $$\vec{c}=\dfrac{1}{\sqrt{3}}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j}-\dfrac{1}{\sqrt{3}}\hat{k}$$

Components: $$c_x=\dfrac{1}{\sqrt{3}},\;c_y=\dfrac{1}{\sqrt{3}},\;c_z=-\dfrac{1}{\sqrt{3}}.$$

Square each component:

$$\left(\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{1}{3}, \quad \left(\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{1}{3}, \quad \left(-\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{1}{3}.$$

Add and take the square root:

$$|\vec{c}|=\sqrt{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}=\sqrt{1}=1.$$

Hence the magnitudes are:

$$|\vec{a}|=\sqrt{3}, \qquad |\vec{b}|=\sqrt{62}, \qquad |\vec{c}|=1.$$

Concept used: Two vectors are said to have the same magnitude if the numerical values of the lengths of their directed line segments are equal, i.e. if their moduli are equal.

Let us construct two different vectors and show that their moduli coincide.

First vector
Choose $$\vec{a}=2\hat{i}+3\hat{j}+6\hat{k}$$.

Its magnitude is obtained from the Euclidean norm:
$$|\vec{a}|=\sqrt{2^{2}+3^{2}+6^{2}}=\sqrt{4+9+36}=\sqrt{49}=7.$$

Second vector
Choose another vector with different direction but with the same squared sum of components; for example

$$\vec{b}=-2\hat{i}+3\hat{j}-6\hat{k}.$$

Compute its magnitude:
$$|\vec{b}|=\sqrt{(-2)^{2}+3^{2}+(-6)^{2}}=\sqrt{4+9+36}=\sqrt{49}=7.$$

Thus $$|\vec{a}|=|\vec{b}|=7$$ while $$\vec{a}\neq \vec{b}$$, so the two vectors are different yet possess the same magnitude.

Any such pair (including simpler ones like $$\hat{i}$$ and $$\hat{j}$$, both of length 1) would satisfy the requirement, but $$\vec{a}$$ and $$\vec{b}$$ exhibited above explicitly demonstrate the idea.

Two vectors have the same direction precisely when one is a positive scalar multiple of the other.

Choose any non-zero vector; for example

$$\vec a = 2\hat i + 3\hat j + \hat k$$

Multiply it by a positive scalar, say $$2$$, to obtain a different vector:

$$\vec b = 2\vec a = 4\hat i + 6\hat j + 2\hat k$$

The vectors $$\vec a$$ and $$\vec b$$ are distinct in magnitude yet point in exactly the same direction because $$\vec b = 2\vec a$$. Hence they satisfy the requirement.

The two vectors are

$$\vec a = 2\hat i + 3\hat j$$
$$\vec b = x\hat i + y\hat j$$

For vectors to be equal, the coefficients of the corresponding unit vectors must match:

Along $$\hat i$$:  $$x = 2$$
Along $$\hat j$$:  $$y = 3$$

Therefore, the required values are $$x = 2$$ and $$y = 3$$.

Let the initial point be $$A(2,1)$$ and the terminal point be $$B(-5,7)$$.

The vector $$\overrightarrow{AB}$$ is obtained by subtracting the coordinates of the initial point from those of the terminal point:

$$\overrightarrow{AB}=\bigl(x_B-x_A\,,\;y_B-y_A\bigr)=(-5-2,\;7-1)=(-7,\;6).$$

Scalar components
The scalar components (also called the rectangular components) are simply the differences along each axis:

$$x\text{-component}=-7,\qquad y\text{-component}=6.$$

Vector components
Expressing these along the unit vectors $$\hat\imath$$ and $$\hat\jmath$$ gives the vector components:

$$-7\,\hat\imath\qquad\text{and}\qquad 6\,\hat\jmath.$$

Hence

$$\overrightarrow{AB}=-7\,\hat\imath+6\,\hat\jmath.$$

Given vectors :

$$\vec a = \hat i - 2\hat j + \hat k$$
$$\vec b = -2\hat i + 4\hat j + 5\hat k$$
$$\vec c = \hat i - 6\hat j - 7\hat k$$

We have to find the resultant vector $$\vec r$$ such that

$$\vec r = \vec a + \vec b + \vec c$$

Write each vector in ordered–triple (component) form:

$$\vec a = (1,\,-2,\,1)$$
$$\vec b = (-2,\,4,\,5)$$
$$\vec c = (1,\,-6,\,-7)$$

Add the corresponding components one by one.

1. x–components
   $$1 + (-2) + 1 = 0$$
2. y–components
   $$(-2) + 4 + (-6) = -4$$
3. z–components
   $$1 + 5 + (-7) = -1$$

Hence the resultant ordered triple is $$\,(0,\,-4,\,-1)$$.

Convert back to unit–vector notation:

$$\vec r = 0\hat i - 4\hat j - \hat k$$

Therefore, the sum of the given vectors is

$$\boxed{\;\vec r = -4\hat j - \hat k\;}$$

We are given the vector $$\vec{a}=\hat{i}+\hat{j}+2\hat{k}$$ and wish to find the corresponding unit vector, i.e. a vector that has magnitude 1 and points in the same direction as $$\vec{a}$$.

Step 1 – Calculate the magnitude of $$\vec{a}$$.

The magnitude (or length) of a vector $$\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$$ is given by

$$|\vec{v}|=\sqrt{v_x^{\,2}+v_y^{\,2}+v_z^{\,2}}$$.

For $$\vec{a}=\hat{i}+\hat{j}+2\hat{k}$$ we have $$v_x=1,\,v_y=1,\,v_z=2$$, so

$$|\vec{a}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt{6}$$.

Step 2 – Divide the vector by its magnitude.

The unit vector $$\hat{u}$$ in the direction of $$\vec{a}$$ is

$$\hat{u}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}.$$

Step 3 – Write the components explicitly.

Dividing each component by $$\sqrt{6}$$ gives

$$\hat{u}=\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}.$$

This vector has magnitude 1 (you can verify by squaring each component, adding, and checking it sums to 1) and points in the same direction as $$\vec{a}$$.

Step 1 : Write the position vectors

For point P$$(1,2,3)$$ the position vector is $$\overrightarrow{OP}=1\hat{i}+2\hat{j}+3\hat{k}$$.
For point Q$$(4,5,6)$$ the position vector is $$\overrightarrow{OQ}=4\hat{i}+5\hat{j}+6\hat{k}$$.

Step 2 : Find $$\overrightarrow{PQ}$$

$$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$$

$$=\left(4-1\right)\hat{i}+\left(5-2\right)\hat{j}+\left(6-3\right)\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}$$

Step 3 : Magnitude of $$\overrightarrow{PQ}$$

$$|\overrightarrow{PQ}|=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{27}=3\sqrt{3}$$

Step 4 : Unit vector in the direction of $$\overrightarrow{PQ}$$

The required unit vector is $$\dfrac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\dfrac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}=\left(\dfrac{1}{\sqrt{3}}\right)\hat{i}+\left(\dfrac{1}{\sqrt{3}}\right)\hat{j}+\left(\dfrac{1}{\sqrt{3}}\right)\hat{k}$$.

Thus, the unit vector is $$\left(\dfrac{1}{\sqrt{3}},\,\dfrac{1}{\sqrt{3}},\,\dfrac{1}{\sqrt{3}}\right)$$.

Given
$$\vec a = 2\hat i-\hat j+2\hat k \quad\text{and}\quad \vec b=-\hat i+\hat j-\hat k.$$

1. Form the required vector.
$$\vec a+\vec b=(2\hat i-\hat j+2\hat k)+(-\hat i+\hat j-\hat k).$$

• Combine the \(\hat i\) components: $$2+(-1)=1$$
• Combine the \(\hat j\) components: $$-1+1=0$$
• Combine the \(\hat k\) components: $$2+(-1)=1$$

Hence
$$\vec a+\vec b = 1\hat i+0\hat j+1\hat k = \hat i+\hat k.$$

2. Find the magnitude of \(\vec a+\vec b\).
$$|\vec a+\vec b|=\sqrt{1^{2}+0^{2}+1^{2}}=\sqrt{2}.$$

3. Obtain the unit vector.
A unit vector $$\hat u$$ in the direction of any non-zero vector $$\vec v$$ is $$\hat u=\dfrac{\vec v}{|\vec v|}.$$
Therefore $$\text{unit vector} = \frac{\hat i+\hat k}{\sqrt{2}} = \frac{1}{\sqrt 2}\,\hat i+0\,\hat j+\frac{1}{\sqrt 2}\,\hat k.$$

Given vector
$$\vec{a}=5\hat{i}-\hat{j}+2\hat{k}$$

Step 1: Magnitude of $$\vec{a}$$
$$|\vec{a}|=\sqrt{5^{2}+(-1)^{2}+2^{2}}=\sqrt{25+1+4}=\sqrt{30}$$

Step 2: Unit vector in the direction of $$\vec{a}$$
$$\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}$$

Step 3: Required vector of magnitude 8 units
Let $$\vec{v}$$ be the required vector. Since $$\vec{v}=8\,\hat{a}$$,
$$\vec{v}=8\left(\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}\right)=\frac{8}{\sqrt{30}}\bigl(5\hat{i}-\hat{j}+2\hat{k}\bigr)$$

Verification
$$|\vec{v}|=\frac{8}{\sqrt{30}}\sqrt{5^{2}+(-1)^{2}+2^{2}}=\frac{8}{\sqrt{30}}\times\sqrt{30}=8$$ — as required.

Let

$$\mathbf{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$$
$$\mathbf{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}$$

To test collinearity, find a real number $$\lambda$$ such that $$\mathbf{b} = \lambda\mathbf{a}$$.

Equate the corresponding components:

• Along $$\hat{i}:\; -4 = 2\lambda \implies \lambda = -2$$
• Along $$\hat{j}:\; 6 = -3\lambda \implies \lambda = -2$$
• Along $$\hat{k}:\; -8 = 4\lambda \implies \lambda = -2$$

The same value $$\lambda = -2$$ satisfies all three conditions, hence $$\mathbf{b} = -2\mathbf{a}$$.

Because one vector is a scalar multiple of the other, the two vectors are collinear.

Let the given vector be denoted by $$\vec{a}$$.

$$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$

Write its components along the coordinate axes:

• $$a_x = 1$$ (coefficient of $$\hat{i}$$)
• $$a_y = 2$$ (coefficient of $$\hat{j}$$)
• $$a_z = 3$$ (coefficient of $$\hat{k}$$)

The magnitude (length) of $$\vec{a}$$ is

$$|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}.$$

If $$\alpha,\,\beta,\,\gamma$$ are the angles that $$\vec{a}$$ makes with the positive $$x,\,y,\,z$$ axes respectively, then the direction cosines are defined as

$$\cos\alpha = \dfrac{a_x}{|\vec{a}|},\qquad \cos\beta = \dfrac{a_y}{|\vec{a}|},\qquad \cos\gamma = \dfrac{a_z}{|\vec{a}|}.$$

Substituting the values just obtained:

$$\cos\alpha = \dfrac{1}{\sqrt{14}},\qquad \cos\beta = \dfrac{2}{\sqrt{14}},\qquad \cos\gamma = \dfrac{3}{\sqrt{14}}.$$

Thus the required direction cosines of the vector $$\hat{i}+2\hat{j}+3\hat{k}$$ are $$\left(\dfrac{1}{\sqrt{14}},\;\dfrac{2}{\sqrt{14}},\;\dfrac{3}{\sqrt{14}}\right).$$

Step 1 : Write the position vectors of the two points.

The position vector of $$A(1,2,-3)$$ is $$\vec{OA}=1\,\hat\imath+2\,\hat\jmath-3\,\hat k$$.
The position vector of $$B(-1,-2,1)$$ is $$\vec{OB}=-1\,\hat\imath-2\,\hat\jmath+1\,\hat k$$.

Step 2 : Form the vector $$\vec{AB}$$ directed from $$A$$ to $$B$$.

For any two points, $$\vec{AB}=\vec{OB}-\vec{OA}$$. Hence

$$\vec{AB}=\bigl(-1\,\hat\imath-2\,\hat\jmath+1\,\hat k\bigr)-\bigl(1\,\hat\imath+2\,\hat\jmath-3\,\hat k\bigr)$$

$$\phantom{\vec{AB}}=(-1-1)\,\hat\imath+(-2-2)\,\hat\jmath+(1-(-3))\,\hat k$$

$$\boxed{\vec{AB}=-2\,\hat\imath-4\,\hat\jmath+4\,\hat k}$$

Step 3 : Find the magnitude of $$\vec{AB}$$.

$$|\vec{AB}|=\sqrt{(-2)^2+(-4)^2+(4)^2}=\sqrt{4+16+16}=\sqrt{36}=6$$

Step 4 : Compute the direction cosines.

If $$\vec{v}=a\,\hat\imath+b\,\hat\jmath+c\,\hat k$$ and its magnitude is $$|\vec v|$$, the direction cosines $$(l,m,n)$$ are

$$l=\dfrac{a}{|\vec v|},\;m=\dfrac{b}{|\vec v|},\;n=\dfrac{c}{|\vec v|}$$

Thus for $$\vec{AB}=-2\,\hat\imath-4\,\hat\jmath+4\,\hat k$$:

$$l=\dfrac{-2}{6}=-\dfrac13,\quad m=\dfrac{-4}{6}=-\dfrac23,\quad n=\dfrac{4}{6}=\dfrac23$$

Therefore, the direction cosines of the vector $$\vec{AB}$$ are $$\bigl(-\dfrac13,-\dfrac23,\dfrac23\bigr)$$.

Let the given vector be

$$\vec a = \hat i + \hat j + \hat k .$$

1. Magnitude of the vector

$$|\vec a| = \sqrt{1^{2}+1^{2}+1^{2}} = \sqrt3.$$

2. Direction cosines

The direction cosines l, m, n of a vector are obtained by dividing each component by the magnitude:

$$l = \dfrac{1}{\sqrt3},\; m = \dfrac{1}{\sqrt3},\; n = \dfrac{1}{\sqrt3}.$$

These are, respectively, $$\cos\alpha,\;\cos\beta,\;\cos\gamma$$, where $$\alpha,\beta,\gamma$$ are the angles that $$\vec a$$ makes with the positive OX, OY and OZ axes.

3. Equality of the angles

Since

$$\cos\alpha=\cos\beta=\cos\gamma=\dfrac{1}{\sqrt3},$$

and the direction angles satisfy $$0\le \alpha,\beta,\gamma\le \pi$$, on which interval the cosine function is one-to-one (strictly decreasing), we conclude

$$\alpha = \beta = \gamma.$$

4. Conclusion

The vector $$\hat i + \hat j + \hat k$$ makes equal angles with each of the three coordinate axes. Therefore it is equally inclined to OX, OY and OZ.

Step 1 : Write the position vectors of the given points.

The position vector of the point $$P(2,3,4)$$ is

$$\vec{OP}=2\,\hat i+3\,\hat j+4\,\hat k$$

and the position vector of the point $$Q(4,1,-2)$$ is

$$\vec{OQ}=4\,\hat i+1\,\hat j-2\,\hat k$$

Step 2 : Use the midpoint formula in vector form.

If $$M$$ is the midpoint of the line segment $$PQ$$, then its position vector is the average of $$\vec{OP}$$ and $$\vec{OQ}$$:

$$\vec{OM}=\dfrac{\vec{OP}+\vec{OQ}}{2}$$

Step 3 : Add the two position vectors.

$$\vec{OP}+\vec{OQ}=\bigl(2\,\hat i+3\,\hat j+4\,\hat k\bigr)+\bigl(4\,\hat i+1\,\hat j-2\,\hat k\bigr)$$

Collect the coefficients of each unit vector:

$$=\bigl(2+4\bigr)\hat i+\bigl(3+1\bigr)\hat j+\bigl(4-2\bigr)\hat k$$

$$=6\,\hat i+4\,\hat j+2\,\hat k$$

Step 4 : Divide by 2 to obtain the midpoint vector.

$$\vec{OM}=\dfrac{6\,\hat i+4\,\hat j+2\,\hat k}{2}=3\,\hat i+2\,\hat j+1\,\hat k$$

Conclusion

The position vector of the midpoint $$M$$ of the segment joining $$P(2,3,4)$$ and $$Q(4,1,-2)$$ is

$$\boxed{3\,\hat i+2\,\hat j+1\,\hat k}$$

Let the position vectors of the three given points be

$$\vec a = 3\hat i - 4\hat j - 4\hat k,$$
$$\vec b = 2\hat i - \hat j + \hat k,$$
$$\vec c = \hat i - 3\hat j - 5\hat k.$$

To decide the nature of the triangle formed by A, B, C we compare the three side vectors.

1. Side vectors

Vector $$\overrightarrow{AB}$$:

$$\overrightarrow{AB}=\vec b-\vec a=(2-3)\hat i+(-1+4)\hat j+(1+4)\hat k=-\hat i+3\hat j+5\hat k.$$

Vector $$\overrightarrow{BC}$$:

$$\overrightarrow{BC}=\vec c-\vec b=(1-2)\hat i+(-3+1)\hat j+(-5-1)\hat k=-\hat i-2\hat j-6\hat k.$$

Vector $$\overrightarrow{CA}$$:

$$\overrightarrow{CA}=\vec a-\vec c=(3-1)\hat i+(-4+3)\hat j+(-4+5)\hat k=2\hat i-\hat j+\hat k.$$

2. Checking perpendicularity

The dot product of two vectors is zero when they are perpendicular.

$$\overrightarrow{AB}\,\cdot\,\overrightarrow{CA}=(-1)(2)+3(-1)+5(1)=-2-3+5=0.$$

Since $$\overrightarrow{AB}\perp\overrightarrow{CA},$$ the angle between AB and AC is a right angle; hence the triangle is right-angled at A.

3. Cross-verification with Pythagoras

Magnitudes squared:

$$|\overrightarrow{AB}|^{2}=(-1)^2+3^2+5^2=35,$$
$$|\overrightarrow{CA}|^{2}=2^2+(-1)^2+1^2=6,$$
$$|\overrightarrow{BC}|^{2}=(-1)^2+(-2)^2+(-6)^2=41.$$

Clearly $$|\overrightarrow{AB}|^{2}+|\overrightarrow{CA}|^{2}=35+6=41=|\overrightarrow{BC}|^{2},$$ which confirms that BC is the hypotenuse, in accord with the result above.

Therefore A, B, C are the vertices of a right-angled triangle, right-angled at A.

Step 1 : Recall the basic side-relation in a triangle
Starting from vertex A, moving along $$\overrightarrow{AB}$$ and then along $$\overrightarrow{BC}$$ reaches C, so

$$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\quad\cdots(1)$$

Taking the vector from C to A, we have $$\overrightarrow{CA}=-\overrightarrow{AC}$$ and from C to B, $$\overrightarrow{CB}=-\overrightarrow{BC}$$.

Step 2 : Test each option

1. (A) $$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}$$
   Replace $$\overrightarrow{CA}$$ by $$-\overrightarrow{AC}$$ and use (1):

   $$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA} =\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC} =(\overrightarrow{AC})-\overrightarrow{AC}=\vec{0}.$$
   Hence (A) is true.

2. (B) $$\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}$$
   Using (1) directly:

   $$\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC} =(\overrightarrow{AC})-\overrightarrow{AC}=\vec{0}.$$
   Hence (B) is true.

3. (C) $$\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}$$
   First change $$-\overrightarrow{CA}$$ into $$-(-\overrightarrow{AC})=\overrightarrow{AC}$$, then substitute from (1):

   $$\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA} =\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{AC} =(\overrightarrow{AC})+\overrightarrow{AC}=2\,\overrightarrow{AC}\neq\vec{0}$$ (unless the triangle collapses).
   Therefore (C) is not true.

4. (D) $$\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}$$
   Replace $$-\overrightarrow{CB}$$ by $$\overrightarrow{BC}$$ and simplify:

   $$\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA} =\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA} =\vec{0}$$ from part (A).
   Hence (D) is true.

Conclusion: Options (A), (B) and (D) are correct; option (C) is the only one that does not hold for a non-degenerate triangle.

Two vectors $$\vec a$$ and $$\vec b$$ are collinear (parallel) when one is a scalar multiple of the other. Hence there exists a scalar $$\lambda$$ such that

$$\vec b = \lambda\,\vec a.$$

We test each option against this defining property.

(A) $$\vec b = \lambda\vec a$$ for some scalar $$\lambda.$$
This is precisely the definition of collinearity, so (A) is correct.

(B) $$\vec a = \pm\vec b.$$
From $$\vec b = \lambda\vec a$$ we get $$\vec a = (1/\lambda)\vec b,$$ so $$\vec a = \pm\vec b$$ would force $$\lambda = \pm 1.$$ But $$\lambda$$ can be any non-zero real scalar — e.g. taking $$\lambda = 2$$ gives $$\vec b = 2\vec a,$$ with $$|\vec b| = 2|\vec a|,$$ so $$\vec a \neq \pm\vec b.$$ Hence (B) is incorrect.

(C) The respective components of $$\vec a$$ and $$\vec b$$ are not proportional.
Writing $$\vec a = a_1\hat i + a_2\hat j + a_3\hat k$$ and $$\vec b = b_1\hat i + b_2\hat j + b_3\hat k,$$ the relation $$\vec b = \lambda\vec a$$ gives

$$b_1 = \lambda a_1,\quad b_2 = \lambda a_2,\quad b_3 = \lambda a_3,$$

so $$\dfrac{b_1}{a_1} = \dfrac{b_2}{a_2} = \dfrac{b_3}{a_3} = \lambda$$ (whenever the $$a_i$$ are non-zero); the components are proportional. The statement claiming they are not proportional is therefore false, so (C) is incorrect.

(D) Both vectors $$\vec a$$ and $$\vec b$$ have the same direction but different magnitudes.
From $$\vec b = \lambda\vec a:$$

• if $$\lambda \gt 0,$$ they point in the same direction; if $$\lambda \lt 0,$$ they point in opposite directions — so “same direction” is not guaranteed;
• even when $$\lambda \gt 0,$$ the value $$\lambda = 1$$ gives $$|\vec b| = |\vec a|$$ — so “different magnitudes” is not guaranteed either.

Hence (D) does not hold for every pair of collinear vectors, so (D) is incorrect.

The incorrect statements are (B), (C) and (D).

Let the angle between $$\vec{a}$$ and $$\vec{b}$$ be $$\theta$$.

The scalar (dot) product of two non-zero vectors is related to the angle between them by

$$\vec{a}\,\cdot\,\vec{b}=|\vec{a}|\,|\vec{b}|\cos\theta$$

According to the question,

$$|\vec{a}| = 1, \; |\vec{b}| = 2 \;\text{and}\; \vec{a}\,\cdot\,\vec{b}=1.$$

Substituting these values in the above relation:

$$1 = (1)(2)\cos\theta$$

$$\Rightarrow \; 1 = 2\cos\theta$$

$$\Rightarrow \; \cos\theta = \dfrac{1}{2}$$

The acute angle whose cosine is $$\dfrac12$$ is $$60^{\circ}$$ (or $$\dfrac{\pi}{3}$$ radians).

Hence,

$$\theta = 60^{\circ}\; ( = \dfrac{\pi}{3}\,\text{rad}).$$

We want the angle $$\theta$$ between the vectors

$$\vec a = \hat i + \hat j - \hat k$$ and $$\vec b = \hat i - \hat j + \hat k$$.

Step 1 – Dot product

$$\vec a \cdot \vec b = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$$

Step 2 – Magnitudes of the vectors

$$|\vec a| = \sqrt{1^{2} + 1^{2} + (-1)^{2}} = \sqrt{3}$$
$$|\vec b| = \sqrt{1^{2} + (-1)^{2} + 1^{2}} = \sqrt{3}$$

Step 3 – Using the cosine formula

The angle $$\theta$$ between two vectors is given by

$$\cos\theta = \dfrac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|}$$

Substitute the values:

$$\cos\theta = \dfrac{-1}{(\sqrt{3})(\sqrt{3})} = -\dfrac{1}{3}$$

Step 4 – Angle

$$\theta = \cos^{-1}\left(-\dfrac{1}{3}\right)$$

In degrees, this is approximately $$109^{\circ}\,28'$$ (or about $$1.911\text{ rad}$$).

Given vectors

$$\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$$
$$\vec{b} = \hat{i} + 3\hat{j} - 5\hat{k}$$

First find $$\vec{a} + \vec{b}$$ component-wise:

$$\vec{a} + \vec{b} = (5 + 1)\hat{i} + (-1 + 3)\hat{j} + (-3 - 5)\hat{k}$$
$$\qquad\; = 6\hat{i} + 2\hat{j} - 8\hat{k}$$

Next find $$\vec{a} - \vec{b}$$:

$$\vec{a} - \vec{b} = (5 - 1)\hat{i} + (-1 - 3)\hat{j} + (-3 + 5)\hat{k}$$
$$\qquad\; = 4\hat{i} - 4\hat{j} + 2\hat{k}$$

Compute the dot product of the two results:

$$ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = (6)(4) + (2)(-4) + (-8)(2) $$
$$ = 24 - 8 - 16 = 0 $$

Since the dot product is zero, the angle between the vectors is $$90^\circ$$. Therefore $$\vec{a} + \vec{b}$$ and $$\vec{a} - \vec{b}$$ are perpendicular.

We are asked to find the projection of $$\vec a = 2\hat{i} + 3\hat{j} + 2\hat{k}$$ on $$\vec b = \hat{i} + 2\hat{j} + \hat{k}.$$

By the NCERT definition, the (scalar) projection of $$\vec a$$ on $$\vec b$$ is

$$\text{Projection of }\vec a\text{ on }\vec b \;=\; \dfrac{\vec a \cdot \vec b}{|\vec b|}.$$

Step 1 — Dot product $$\vec a \cdot \vec b$$

$$\vec a \cdot \vec b = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10.$$

Step 2 — Magnitude $$|\vec b|$$

$$|\vec b| = \sqrt{1^{2} + 2^{2} + 1^{2}} = \sqrt{1+4+1} = \sqrt{6}.$$

Step 3 — Substitute

$$\dfrac{\vec a \cdot \vec b}{|\vec b|} \;=\; \dfrac{10}{\sqrt{6}}.$$

Rationalising the denominator:

$$\dfrac{10}{\sqrt{6}} \;=\; \dfrac{10}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} \;=\; \dfrac{10\sqrt{6}}{6} \;=\; \dfrac{5\sqrt{6}}{3}.$$

Hence the required projection is

$$\dfrac{5\sqrt{6}}{3}.$$

We have two vectors $$\vec a$$ and $$\vec b$$ for which

• $$|\vec a| = 2$$
• $$|\vec b| = 3$$
• $$\vec a \cdot \vec b = 4$$

We wish to find $$|\vec a - \vec b|$$.

Step 1 : Express the required magnitude through a dot-product.

By definition, the square of the magnitude of a vector equals the dot-product of the vector with itself, hence

$$|\vec a - \vec b|^{2} = (\vec a - \vec b) \cdot (\vec a - \vec b).$$

Step 2 : Expand the dot-product.

Using the distributive property of the dot product,

$$\begin{aligned}(\vec a - \vec b) \cdot (\vec a - \vec b) &= \vec a \cdot \vec a \, - \, \vec a \cdot \vec b \, - \, \vec b \cdot \vec a \, + \, \vec b \cdot \vec b\\ &= |\vec a|^{2} - 2(\vec a \cdot \vec b) + |\vec b|^{2}.\end{aligned}$$

Step 3 : Substitute the given numerical values.

Substituting $$|\vec a| = 2$$, $$|\vec b| = 3$$ and $$\vec a \cdot \vec b = 4$$, we get

$$|\vec a - \vec b|^{2} = (2)^{2} - 2(4) + (3)^{2}.$$

Simplifying,

$$|\vec a - \vec b|^{2} = 4 + 9 - 8 = 5.$$

Step 4 : Take the positive square root.

Since magnitude is always non-negative,

$$|\vec a - \vec b| = \sqrt{5}.$$

This completes the calculation.

Let $$\vec{a}$$ be a given unit vector, so that $$|\vec{a}| = 1$$.

The condition in the question is

$$(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8.$$

Expand the left-hand side using the distributive and commutative properties of the dot product:

$$(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = \vec{x}\cdot\vec{x} + \vec{x}\cdot\vec{a} - \vec{a}\cdot\vec{x} - \vec{a}\cdot\vec{a}.$$

Because the dot product is commutative (i.e. $$\vec{x}\cdot\vec{a} = \vec{a}\cdot\vec{x}$$), the middle two terms cancel:

$$\vec{x}\cdot\vec{a} - \vec{a}\cdot\vec{x} = 0.$$

Hence

$$(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = \vec{x}\cdot\vec{x} - \vec{a}\cdot\vec{a}.$$

Recognising dot products of a vector with itself, we write

$$\vec{x}\cdot\vec{x} = |\vec{x}|^2, \qquad \vec{a}\cdot\vec{a} = |\vec{a}|^2.$$

Therefore

$$|\vec{x}|^2 - |\vec{a}|^2 = 8.$$

But $$|\vec{a}| = 1$$ (unit vector), so $$|\vec{a}|^2 = 1$$. Substitute:

$$|\vec{x}|^2 - 1 = 8 \;\;\Longrightarrow\;\; |\vec{x}|^2 = 9.$$

Taking the (non-negative) square root gives

$$|\vec{x}| = 3.$$

Thus the magnitude of $$\vec{x}$$ is $$3$$.

Let $$\vec a$$ and $$\vec b$$ be any two vectors in three–dimensional space. For a real parameter $$\lambda$$ define $$\vec c = \vec a - \lambda\,\vec b.$$ Since the square of the magnitude of a vector is always non–negative, $$|\vec c|^{2}= (\vec a-\lambda\vec b)\cdot(\vec a-\lambda\vec b) \ge 0.$$

Expand the dot product: $$|\vec c|^{2}= \vec a\cdot\vec a -2\lambda\,\vec a\cdot\vec b + \lambda^{2}\,\vec b\cdot\vec b = |\vec a|^{2}-2\lambda\,(\vec a\cdot\vec b)+\lambda^{2}|\vec b|^{2}\ge 0.$$

Regard the right side as a quadratic expression in $$\lambda$$: $$f(\lambda)=|\vec b|^{2}\,\lambda^{2}-2(\vec a\cdot\vec b)\,\lambda+|\vec a|^{2}\ge 0 \,\,\text{for all real }\lambda.$$

A real quadratic is non-negative for every $$\lambda$$ only when its discriminant is non-positive: $$\Delta=[-2(\vec a\cdot\vec b)]^{2}-4|\vec b|^{2}|\vec a|^{2}\le 0.$$

Simplifying, $$4(\vec a\cdot\vec b)^{2}-4|\vec a|^{2}|\vec b|^{2}\le 0 \quad\Rightarrow\quad (\vec a\cdot\vec b)^{2}\le |\vec a|^{2}|\vec b|^{2}.$$

Both sides are non-negative, so taking square roots gives the required Cauchy–Schwarz inequality: $$|\vec a\cdot\vec b| \le |\vec a|\,|\vec b|.$$

Equality condition: The discriminant becomes zero precisely when $$\vec c=\vec a-\lambda\vec b=\vec 0,$$ i.e. when $$\vec a$$ and $$\vec b$$ are parallel (or one of them is the zero vector).

Given : any two vectors $$\vec a$$ and $$\vec b$$ in three–dimensional space.

We must prove the triangle inequality

$$|\vec a+\vec b| \le |\vec a|+|\vec b|.$$

Step 1 : Write the square of the magnitude of the sum.

$$|\vec a+\vec b|^{2}=(\vec a+\vec b)\cdot(\vec a+\vec b).$$

Step 2 : Expand the dot product.

$$(\vec a+\vec b)\cdot(\vec a+\vec b)=\vec a\cdot\vec a+2\vec a\cdot\vec b+\vec b\cdot\vec b.$$

But $$\vec a\cdot\vec a=|\vec a|^{2}\;\;\text{and}\;\;\vec b\cdot\vec b=|\vec b|^{2}.$$ Hence

$$|\vec a+\vec b|^{2}=|\vec a|^{2}+2\,\vec a\cdot\vec b+|\vec b|^{2}.$$

Step 3 : Replace the mixed dot product by its trigonometric form.

Let $$\theta$$ be the angle between $$\vec a$$ and $$\vec b$$. Then $$\vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta$$, so

$$|\vec a+\vec b|^{2}=|\vec a|^{2}+2|\vec a|\,|\vec b|\cos\theta+|\vec b|^{2}.$$

Step 4 : Use the bound on $$\cos\theta$$.

Because $$-1\le\cos\theta\le1$$, the maximum possible value of the middle term is obtained when $$\cos\theta=1$$. Therefore

$$|\vec a+\vec b|^{2}\le |\vec a|^{2}+2|\vec a|\,|\vec b|+|\vec b|^{2}=(|\vec a|+|\vec b|)^{2}.$$

Step 5 : Take the non-negative square root.

The magnitudes are non-negative, so the square-root function preserves the inequality:

$$|\vec a+\vec b|\le |\vec a|+|\vec b|.$$

Equality condition : Equality holds when $$\cos\theta=1$$, i.e. when $$\theta=0^{\circ}$$ and the two vectors point in exactly the same direction.

Thus the triangle inequality is proved for all vectors $$\vec a$$ and $$\vec b$$.

Given position vectors

A : $$\vec a=-2\hat{i}+3\hat{j}+5\hat{k}$$
B : $$\vec b=\;\;\hat{i}+2\hat{j}+3\hat{k}$$
C : $$\vec c=\;7\hat{i}\;\;-\;\hat{k}$$

Step 1 – Form the direction vectors

For any two points, the vector joining them is obtained by subtracting their position vectors.

$$\overrightarrow{AB}=\vec b-\vec a=(1-(-2))\hat{i}+(2-3)\hat{j}+(3-5)\hat{k}$$

$$\Rightarrow\;\overrightarrow{AB}=3\hat{i}-\hat{j}-2\hat{k}$$

Similarly, $$\overrightarrow{AC}=\vec c-\vec a=(7-(-2))\hat{i}+(0-3)\hat{j}+(-1-5)\hat{k}$$

$$\Rightarrow\;\overrightarrow{AC}=9\hat{i}-3\hat{j}-6\hat{k}$$

Step 2 – Check if the direction vectors are parallel

Write $$\overrightarrow{AC}$$ in terms of $$\overrightarrow{AB}$$:

$$\overrightarrow{AC}=9\hat{i}-3\hat{j}-6\hat{k}=3\,(3\hat{i}-\hat{j}-2\hat{k})=3\,\overrightarrow{AB}$$

Since $$\overrightarrow{AC}=3\,\overrightarrow{AB}$$, the two vectors are scalar multiples, i.e. they are parallel.

Step 3 – Conclude collinearity

If two vectors drawn from the same initial point are parallel, the three points concerned lie on the same straight line. Therefore A, B and C are collinear.

Let $$\theta$$ be the angle between the two vectors $$\vec a$$ and $$\vec b$$.

By the definition of the dot product,

$$\vec a \cdot \vec b = |\vec a|\,|\vec b|\cos\theta.$$

The data given are

• $$\vec a \cdot \vec b = \sqrt{6},$$
• $$|\vec a| = \sqrt{3},$$
• $$|\vec b| = 2.$$

Substituting these values,

$$\sqrt{6} = (\sqrt{3})(2)\cos\theta.$$

Multiply the magnitudes:

$$ (\sqrt{3})(2) = 2\sqrt{3}, $$

so

$$ \sqrt{6} = 2\sqrt{3}\cos\theta. $$

Solve for $$\cos\theta$$:

$$ \cos\theta = \frac{\sqrt{6}}{2\sqrt{3}}. $$

Simplify the right–hand side:

$$ \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}\,\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}. $$

Thus

$$ \cos\theta = \frac{1}{\sqrt{2}}. $$

Taking the inverse cosine,

$$ \theta = \cos^{-1}\!\left(\frac{1}{\sqrt{2}}\right). $$

Since $$\cos 45^{\circ} = \dfrac{1}{\sqrt{2}},$$

$$ \theta = 45^{\circ} \; (\text{or } \pi/4 \text{ radians}). $$

Therefore, the angle between the two vectors is $$45^{\circ}$$.

Given vectors

$$\vec{a}=\hat{i}-2\hat{j}+3\hat{k}, \qquad \vec{b}=3\hat{i}-2\hat{j}+\hat{k}$$

The angle $$\theta$$ between them satisfies

$$\vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\,\cos\theta.$$

1. Dot product

$$\vec{a}\cdot\vec{b}=1\cdot 3+(-2)\cdot(-2)+3\cdot 1=3+4+3=10.$$

2. Magnitudes

$$|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14},$$

$$|\vec{b}|=\sqrt{3^{2}+(-2)^{2}+1^{2}}=\sqrt{9+4+1}=\sqrt{14}.$$

3. Cosine of the angle

$$\cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}=\dfrac{10}{\sqrt{14}\,\sqrt{14}}=\dfrac{10}{14}=\dfrac{5}{7}.$$

4. Required angle

$$\theta=\cos^{-1}\!\left(\dfrac{5}{7}\right).$$

Let the vectors be $$\vec a = \hat i - \hat j$$ and $$\vec b = \hat i + \hat j$$.

The (scalar) projection of $$\vec a$$ on $$\vec b$$ is given by

$$\mathrm{comp}_{\vec b}\vec a = \frac{\vec a \cdot \vec b}{|\vec b|}.$$

Compute the dot product:

$$\vec a \cdot \vec b = (1)(1) + (-1)(1) = 1 - 1 = 0.$$

Compute the magnitude of $$\vec b$$:

$$|\vec b| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}.$$

Hence,

$$\mathrm{comp}_{\vec b}\vec a = \frac{0}{\sqrt{2}} = 0.$$

Therefore, the projection of $$\hat i - \hat j$$ on $$\hat i + \hat j$$ is $$0$$ (the zero vector if the vector projection is considered).

Let $$\vec{a}=\hat{i}+3\hat{j}+7\hat{k}$$ and $$\vec{b}=7\hat{i}-\hat{j}+8\hat{k}$$.

The (scalar) projection of $$\vec{a}$$ on $$\vec{b}$$ is defined as
$$\operatorname{proj}_{\vec{b}}\vec{a}=\dfrac{\vec{a}\,\cdot\,\vec{b}}{|\vec{b}|}.$$

Step 1: Compute the dot product $$\vec{a}\cdot\vec{b}$$.
$$\vec{a}\cdot\vec{b}=1\times7+3\times(-1)+7\times8=7-3+56=60.$$

Step 2: Find the magnitude of $$\vec{b}$$.
$$|\vec{b}|=\sqrt{7^{2}+(-1)^{2}+8^{2}}=\sqrt{49+1+64}=\sqrt{114}.$$

Step 3: Substitute in the formula.
$$\operatorname{proj}_{\vec{b}}\vec{a}=\dfrac{60}{\sqrt{114}}.$$

Thus, the required projection of $$\vec{a}$$ on $$\vec{b}$$ is $$\dfrac{60}{\sqrt{114}}$$.

Let us denote the three vectors by

• $$\vec a = \dfrac{1}{7}\,(2\hat i+3\hat j+6\hat k)=\left(\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}\right)$$
• $$\vec b = \dfrac{1}{7}\,(3\hat i-6\hat j+2\hat k)=\left(\dfrac{3}{7},-\dfrac{6}{7},\dfrac{2}{7}\right)$$
• $$\vec c = \dfrac{1}{7}\,(6\hat i+2\hat j-3\hat k)=\left(\dfrac{6}{7},\dfrac{2}{7},-\dfrac{3}{7}\right)$$

1. Show that each vector is a unit vector

Magnitude of $$\vec a$$:

$$|\vec a| = \sqrt{\left(\dfrac{2}{7}\right)^2 + \left(\dfrac{3}{7}\right)^2 + \left(\dfrac{6}{7}\right)^2} = \sqrt{\dfrac{4+9+36}{49}} = \sqrt{\dfrac{49}{49}} = 1$$

Magnitude of $$\vec b$$:

$$|\vec b| = \sqrt{\left(\dfrac{3}{7}\right)^2 + \left(-\dfrac{6}{7}\right)^2 + \left(\dfrac{2}{7}\right)^2} = \sqrt{\dfrac{9+36+4}{49}} = \sqrt{\dfrac{49}{49}} = 1$$

Magnitude of $$\vec c$$:

$$|\vec c| = \sqrt{\left(\dfrac{6}{7}\right)^2 + \left(\dfrac{2}{7}\right)^2 + \left(-\dfrac{3}{7}\right)^2} = \sqrt{\dfrac{36+4+9}{49}} = \sqrt{\dfrac{49}{49}} = 1$$

Hence $$\vec a,\,\vec b,\,\vec c$$ are all unit vectors.

2. Show that they are mutually perpendicular

Compute their dot products:

$$\vec a\,\cdot\,\vec b = \dfrac{2}{7}\cdot\dfrac{3}{7} + \dfrac{3}{7}\cdot\left(-\dfrac{6}{7}\right) + \dfrac{6}{7}\cdot\dfrac{2}{7} = \dfrac{6 - 18 + 12}{49}=0$$

$$\vec a\,\cdot\,\vec c = \dfrac{2}{7}\cdot\dfrac{6}{7} + \dfrac{3}{7}\cdot\dfrac{2}{7} + \dfrac{6}{7}\cdot\left(-\dfrac{3}{7}\right) = \dfrac{12 + 6 - 18}{49}=0$$

$$\vec b\,\cdot\,\vec c = \dfrac{3}{7}\cdot\dfrac{6}{7} + \left(-\dfrac{6}{7}\right)\cdot\dfrac{2}{7} + \dfrac{2}{7}\cdot\left(-\dfrac{3}{7}\right) = \dfrac{18 - 12 - 6}{49}=0$$

Since the dot product of every pair is zero, the vectors are mutually perpendicular.

Thus, each of the given vectors is a unit vector and they are pairwise perpendicular.

We have two pieces of information:

• $$(\vec a+\vec b)\cdot(\vec a-\vec b)=8$$
• $$|\vec a| = 8\,|\vec b|$$

Step 1: Expand the given scalar product.

Recall that for any vectors $$\vec p,\vec q$$ we have $$\vec p\cdot\vec q = \vec q\cdot\vec p$$ and $$\vec p\cdot\vec p = |\vec p|^2$$. Using distributive and commutative properties,

$$ (\vec a+\vec b)\cdot(\vec a-\vec b) = \vec a\cdot\vec a\; -\; \vec a\cdot\vec b\; +\; \vec b\cdot\vec a\; -\; \vec b\cdot\vec b. $$

Because $$\vec a\cdot\vec b = \vec b\cdot\vec a$$, the middle two terms cancel:

$$ (\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2. $$

Hence the first condition becomes

$$ |\vec a|^2 - |\vec b|^2 = 8. \qquad(1) $$

Step 2: Introduce a single unknown.

Let $$|\vec b| = t\;(t>0).$$ Then by the second given condition,

$$ |\vec a| = 8t. \qquad(2) $$

Step 3: Substitute into equation (1).

Using (2) in (1):

$$ (8t)^2 - t^2 = 8 \;\Longrightarrow\; 64t^2 - t^2 = 8 \;\Longrightarrow\; 63t^2 = 8. $$

So

$$ t^2 = \frac{8}{63}\;\Longrightarrow\; t = \sqrt{\frac{8}{63}} = \frac{\sqrt8}{\sqrt{63}} = \frac{2\sqrt2}{3\sqrt7}. $$

Rationalising the denominator:

$$ t = \frac{2\sqrt2}{3\sqrt7}\times\frac{\sqrt7}{\sqrt7}=\frac{2\sqrt{14}}{21}. $$

Therefore,

$$ |\vec b| = \frac{2\sqrt{14}}{21}. \qquad(3) $$

Step 4: Find $$|\vec a|$$.

Using (2) and (3):

$$ |\vec a| = 8t = 8\left(\frac{2\sqrt{14}}{21}\right)=\frac{16\sqrt{14}}{21}. $$

Conclusion.

$$ |\vec a| = \dfrac{16\sqrt{14}}{21}, \qquad |\vec b| = \dfrac{2\sqrt{14}}{21}. $$

The dot (scalar) product is distributive over vector addition and compatible with scalar multiplication:

$$ (k_1\vec u+k_2\vec v)\cdot\vec w=k_1(\vec u\cdot\vec w)+k_2(\vec v\cdot\vec w) \quad\text{and}\quad (k\vec u)\cdot\vec v=k(\vec u\cdot\vec v). $$

Using these properties on $$(3\vec a-5\vec b)\cdot(2\vec a+7\vec b)$$:

First expand by the distributive law:

$$ (3\vec a-5\vec b)\cdot(2\vec a+7\vec b)=3\vec a\cdot(2\vec a+7\vec b)-5\vec b\cdot(2\vec a+7\vec b). $$

Now distribute each dot product:

$$ 3\vec a\cdot(2\vec a)+3\vec a\cdot(7\vec b)-5\vec b\cdot(2\vec a)-5\vec b\cdot(7\vec b). $$

Take the numerical coefficients out of each dot product:

$$ 3\times2(\vec a\cdot\vec a)+3\times7(\vec a\cdot\vec b)-5\times2(\vec b\cdot\vec a)-5\times7(\vec b\cdot\vec b). $$

Because the dot product is commutative, $$\vec a\cdot\vec b=\vec b\cdot\vec a,$$ so combine identical terms:

$$ 6(\vec a\cdot\vec a)+21(\vec a\cdot\vec b)-10(\vec a\cdot\vec b)-35(\vec b\cdot\vec b). $$

Collect the middle coefficients, $$21-10=11$$:

$$ 6(\vec a\cdot\vec a)+11(\vec a\cdot\vec b)-35(\vec b\cdot\vec b). $$

Remember that $$\vec a\cdot\vec a=|\vec a|^{2}$$ and $$\vec b\cdot\vec b=|\vec b|^{2},$$ so the product can also be written as

$$ 6|\vec a|^{2}+11(\vec a\cdot\vec b)-35|\vec b|^{2}. $$

This is the required expansion of $$(3\vec a-5\vec b)\cdot(2\vec a+7\vec b).$$

Let the common magnitude of the two vectors be $$m$$; that is, $$|\vec{a}| = |\vec{b}| = m$$.

The angle between them is $$60^\circ$$, so by the definition of scalar (dot) product,

$$\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}| \cos 60^\circ = m \times m \times \frac{1}{2} = \frac{m^2}{2}.$$

The scalar product is given to be $$\dfrac{1}{2}$$. Hence

$$\frac{m^2}{2} = \frac{1}{2} \;\Longrightarrow\; m^2 = 1.$$

Because a magnitude is always non-negative, $$m = 1$$.

Therefore,

$$|\vec{a}| = |\vec{b}| = 1.$$

Given a unit vector $$\vec a$$ (so $$|\vec a| = 1$$) and the relation

$$ (\vec x - \vec a) \cdot (\vec x + \vec a) = 12 $$

expand the dot product using distributivity:

$$ (\vec x - \vec a) \cdot (\vec x + \vec a) = \vec x \cdot \vec x + \vec x \cdot \vec a - \vec a \cdot \vec x - \vec a \cdot \vec a $$

Because the dot product is commutative, $$\vec x \cdot \vec a = \vec a \cdot \vec x$$, so the middle terms cancel:

$$ \vec x \cdot \vec a - \vec a \cdot \vec x = 0 $$

Hence

$$ (\vec x - \vec a) \cdot (\vec x + \vec a) = \vec x \cdot \vec x - \vec a \cdot \vec a. $$

Recognising dot products with themselves as squared magnitudes:

$$ \vec x \cdot \vec x = |\vec x|^2, \qquad \vec a \cdot \vec a = |\vec a|^2 = 1.$$

Substitute these into the equation:

$$ |\vec x|^2 - 1 = 12. $$

Solve for $$|\vec x|^2$$:

$$ |\vec x|^2 = 12 + 1 = 13. $$

Taking the non-negative square root (since a magnitude cannot be negative):

$$ |\vec x| = \sqrt{13}. $$

To make the vector $$\vec{a}+\lambda\vec{b}$$ perpendicular to $$\vec{c}$$, their dot-product must be zero.

1. Write each vector in component form:
   $$\vec{a} = 2\hat{i}+2\hat{j}+3\hat{k},\qquad \vec{b} = -\hat{i}+2\hat{j}+\hat{k},\qquad \vec{c}=3\hat{i}+\hat{j}+0\hat{k}.$$
2. Form the required vector:
   $$\vec{a}+\lambda\vec{b} = (2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}.$$
3. Impose perpendicularity using the dot-product:
   $$(\vec{a}+\lambda\vec{b})\cdot\vec{c}=0.$$
4. Compute the dot-product with components:
   $$[(2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}]\cdot[3\hat{i}+\hat{j}+0\hat{k}]=0.$$
   This gives
   $$3(2-\lambda) + 1\,(2+2\lambda) + 0\,(3+\lambda)=0.$$
5. Simplify step by step:
   $$6-3\lambda + 2+2\lambda = 0 \;\Rightarrow\; 8-\lambda = 0.$$
6. Solve for $$\lambda$$:
   $$\lambda = 8.$$

Hence the required value is $$\lambda = 8$$.

Let $$\vec u = |\vec a|\,\vec b + |\vec b|\,\vec a$$ and $$\vec v = |\vec a|\,\vec b - |\vec b|\,\vec a$$, where $$\vec a$$ and $$\vec b$$ are any two non-zero vectors.

To show that $$\vec u$$ and $$\vec v$$ are perpendicular, we prove that their scalar (dot) product is zero.

Compute $$\vec u\,\cdot\,\vec v$$:

$$\vec u\,\cdot\,\vec v = (|\vec a|\,\vec b + |\vec b|\,\vec a) \cdot (|\vec a|\,\vec b - |\vec b|\,\vec a)$$

Distribute the dot product term by term:

$$\begin{aligned} \vec u\,\cdot\,\vec v & = |\vec a|^2(\vec b\,\cdot\,\vec b) \; - \; |\vec a||\vec b|\,(\vec b\,\cdot\,\vec a) \\[-2pt] & \quad + |\vec a||\vec b|\,(\vec a\,\cdot\,\vec b) \; - \; |\vec b|^2(\vec a\,\cdot\,\vec a). \end{aligned}$$

Notice that $$\vec b\,\cdot\,\vec a = \vec a\,\cdot\,\vec b$$, so the mixed terms cancel:

$$- |\vec a||\vec b|\,(\vec a\,\cdot\,\vec b) + |\vec a||\vec b|\,(\vec a\,\cdot\,\vec b) = 0.$$

The remaining two terms involve the magnitudes of $$\vec a$$ and $$\vec b$$:

$$\begin{aligned} \vec u\,\cdot\,\vec v & = |\vec a|^2(\vec b\,\cdot\,\vec b) - |\vec b|^2(\vec a\,\cdot\,\vec a) \\[2pt] & = |\vec a|^2 \;|\vec b|^2 - |\vec b|^2 \;|\vec a|^2 \\[2pt] & = 0. \end{aligned}$$

Conclusion: Since $$\vec u\,\cdot\,\vec v = 0$$, the vectors $$|\vec a|\vec b + |\vec b|\vec a$$ and $$|\vec a|\vec b - |\vec b|\vec a$$ are perpendicular to each other.

Given: $$\vec a \cdot \vec a = 0\quad\text{and}\quad \vec a \cdot \vec b = 0$$.

Step 1. Interpret $$\vec a \cdot \vec a$$
For any vector $$\vec a$$, the dot-product with itself equals the squared magnitude:

$$\vec a \cdot \vec a = |\vec a|^{2}.$$

Step 2. Use the given value
The statement $$\vec a \cdot \vec a = 0$$ implies

$$|\vec a|^{2}=0 \;\Longrightarrow\; |\vec a| = 0.$$

A vector has zero magnitude only when it is the zero vector, so

$$\vec a = \vec 0.$$

Step 3. Examine $$\vec a \cdot \vec b$$ with $$\vec a = \vec 0$$
Substituting $$\vec a = \vec 0$$ into the second given condition:

$$\vec a \cdot \vec b = \vec 0 \cdot \vec b = 0,$$

which is an identity (always true) for every vector $$\vec b$$ because the zero vector is orthogonal to all vectors.

Conclusion
The second condition places no restriction on $$\vec b$$ once we know $$\vec a = \vec 0$$. Hence $$\vec b$$ can be any vector in the space.

We are given that $$\vec a,\vec b,\vec c$$ are unit vectors and that

$$\vec a+\vec b+\vec c=\vec 0.$$

Our goal is to evaluate $$\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a.$$

Step 1: Square the given vector equation.

Take the dot product of $$\vec a+\vec b+\vec c$$ with itself:

$$ (\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)=\vec 0\cdot\vec 0=0. $$

Step 2: Expand the left-hand side.

Using distributivity of the dot product,

$$ \vec a\cdot\vec a+\vec b\cdot\vec b+\vec c\cdot\vec c \, + \, 2\,(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0. $$

Step 3: Substitute the magnitudes.

Because $$\vec a,\vec b,\vec c$$ are unit vectors, $$\vec a\cdot\vec a=\vec b\cdot\vec b=\vec c\cdot\vec c=1.$$ Hence

$$ 1+1+1+2\,(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0. $$

Step 4: Solve for the required sum.

$$ 3+2\,(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0 \;\;\Rightarrow\;\; \vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=-\tfrac{3}{2}. $$

We have to discuss both directions of the statement.

1. If either vector is the zero vector, the dot product is certainly zero.

Write the dot product in component form for three‑dimensional vectors (the proof is identical in two dimensions):

$$\vec a \cdot \vec b=(a_1\mathbf i+a_2\mathbf j+a_3\mathbf k)\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k) =a_1b_1+a_2b_2+a_3b_3.$$

• If $$\vec a=\vec0$$, then $$a_1=a_2=a_3=0$$, so every term on the right is 0 and hence $$\vec a\cdot\vec b=0$$.
• If $$\vec b=\vec0$$, then $$b_1=b_2=b_3=0$$ and again $$\vec a\cdot\vec b=0$$.

Thus $$\boxed{\text{If }\vec a=\vec0 \text{ or } \vec b=\vec0,\;\;\vec a\cdot\vec b=0.}$$

2. The converse is not always true.

The converse would read: “If $$\vec a\cdot\vec b=0$$, then either $$\vec a=\vec0$$ or $$\vec b=\vec0$$.” To show that this is false we need one counter-example where the dot product is 0 even though both vectors are non-zero.

Take, in the plane,

$$\vec a = \mathbf i = (1,0),\qquad \vec b = \mathbf j = (0,1).$$

Both are clearly non-zero vectors. Their dot product is

$$\vec a\cdot\vec b = 1\cdot0 + 0\cdot1 = 0.$$

Another viewpoint uses the geometric definition of the dot product:

$$\vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta.$$

When $$\theta=90^\circ$$ (vectors are perpendicular), $$\cos\theta=0$$, so the dot product vanishes even though both magnitudes $$|\vec a|,|\vec b|$$ are non-zero. Hence orthogonal (perpendicular) non-zero vectors always give a zero dot product.

Therefore the converse is disproved: a zero dot product does not compel either vector to be the zero vector.

The required angle is placed at vertex B, so we need the vectors whose initial point is B and whose terminal points are A and C, respectively.

1. Form the two vectors

Vector $$\overrightarrow{BA}$$: subtract the coordinates of B from those of A.

$$\overrightarrow{BA} = A - B = (1 - (-1),\; 2 - 0,\; 3 - 0) = (2,\,2,\,3)$$

Vector $$\overrightarrow{BC}$$: subtract the coordinates of B from those of C.

$$\overrightarrow{BC} = C - B = (0 - (-1),\; 1 - 0,\; 2 - 0) = (1,\,1,\,2)$$

2. Dot product of the two vectors

$$\overrightarrow{BA}\,\cdot\,\overrightarrow{BC} = 2\cdot1 + 2\cdot1 + 3\cdot2 = 2 + 2 + 6 = 10$$

3. Magnitudes of the vectors

$$|\overrightarrow{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$$

$$|\overrightarrow{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$

4. Cosine of the angle using the dot-product formula

$$\cos \angle ABC = \dfrac{\overrightarrow{BA}\,\cdot\,\overrightarrow{BC}}{|\overrightarrow{BA}|\,|\overrightarrow{BC}|} = \dfrac{10}{\sqrt{17}\,\sqrt{6}} = \dfrac{10}{\sqrt{102}}$$

5. Angle at B

$$\angle ABC = \cos^{-1}\left(\dfrac{10}{\sqrt{102}}\right) \approx 8.17^{\circ}$$

Let $$O$$ be the origin. The position vectors of the given points are

$$\vec a = 1\hat i + 2\hat j + 7\hat k, \;\; \vec b = 2\hat i + 6\hat j + 3\hat k, \;\; \vec c = 3\hat i + 10\hat j - 1\hat k.$$

Step 1: Find $$\vec{AB}$$.

$$\vec{AB} = \vec b - \vec a = (2-1)\hat i + (6-2)\hat j + (3-7)\hat k = 1\hat i + 4\hat j - 4\hat k.$$

Step 2: Find $$\vec{AC}$$.

$$\vec{AC} = \vec c - \vec a = (3-1)\hat i + (10-2)\hat j + (-1-7)\hat k = 2\hat i + 8\hat j - 8\hat k.$$

Step 3: Compare $$\vec{AB}$$ and $$\vec{AC}$$.

Multiply $$\vec{AB}$$ by $$2$$:

$$2\vec{AB} = 2(1\hat i + 4\hat j - 4\hat k) = 2\hat i + 8\hat j - 8\hat k = \vec{AC}.$$

Thus $$\vec{AC}$$ is a scalar multiple of $$\vec{AB}$$, so the two vectors are parallel.

Step 4: Conclude collinearity.

Since the direction vectors from $$A$$ to $$B$$ and from $$A$$ to $$C$$ are parallel, the three points $$A(1,2,7)$$, $$B(2,6,3)$$ and $$C(3,10,-1)$$ lie on the same straight line; hence they are collinear.

Let the given vectors be the position vectors of the three points

• $$\vec a = 2\hat i - \hat j + \hat k$$
• $$\vec b = \hat i - 3\hat j - 5\hat k$$
• $$\vec c = 3\hat i - 4\hat j - 4\hat k$$

Thus the triangle has vertices

$$A(2,-1,1), \; B(1,-3,-5), \; C(3,-4,-4).$$

Step 1  Find the vectors representing the sides.

$$\vec{AB}=\vec b-\vec a=(1-2)\hat i+(-3+1)\hat j+(-5-1)\hat k=-\hat i-2\hat j-6\hat k$$

$$\vec{BC}=\vec c-\vec b=(3-1)\hat i+(-4+3)\hat j+(-4+5)\hat k=2\hat i-\hat j+\hat k$$

$$\vec{CA}=\vec a-\vec c=(2-3)\hat i+(-1+4)\hat j+(1+4)\hat k=-\hat i+3\hat j+5\hat k$$

Step 2  Examine the dot products.

$$\vec{AB}\cdot\vec{BC}=(-1)(2)+(-2)(-1)+(-6)(1)=-2+2-6=-6\neq0$$

$$\vec{BC}\cdot\vec{CA}=(2)(-1)+(-1)(3)+(1)(5)=-2-3+5=0$$

$$\vec{CA}\cdot\vec{AB}=(-1)(-1)+(3)(-2)+(5)(-6)=1-6-30=-35\neq0$$

Since $$\vec{BC}\cdot\vec{CA}=0$$, the two sides $$BC$$ and $$CA$$ are perpendicular.

Step 3  Conclusion.

Sides $$BC$$ and $$CA$$ meet at $$C$$, so the angle at vertex $$C$$ is a right angle. Hence the triangle whose vertices are given by the three vectors is right angled at $$C$$.

(For completeness, $$|BC|^{2}=6,\;|CA|^{2}=35,\;|AB|^{2}=41$$ and $$6+35=41$$, satisfying the Pythagoras theorem.)

Therefore, the given vectors form the vertices of a right angled triangle.

The given vector $$\vec a$$ has magnitude $$a\;\bigl(a\gt 0\bigr)$$ and $$\lambda\;(\neq 0)$$ is a scalar.
To decide when $$\lambda\vec a$$ is a unit vector, recall the rule for the magnitude of a scalar–multiple:

$$|\lambda\vec a| = |\lambda|\,|\vec a| = |\lambda|\,a.$$

A unit vector has magnitude 1, hence we must have

$$|\lambda|\,a = 1.$$

Solving for $$a$$ gives

$$a = \frac{1}{|\lambda|}.$$

This matches option (D).

Given vectors
$$\vec a = 2\hat i + \hat j + 3\hat k, \qquad \vec b = 3\hat i + 5\hat j - 2\hat k$$

The magnitude $$|\vec a \times \vec b|$$ can be obtained in two stages:

1. First compute the cross-product $$\vec a \times \vec b$$.
2. Then take its magnitude.

1. Computing $$\vec a \times \vec b$$

Write the determinant involving the unit vectors:

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & 3\\ 3 & 5 & -2 \end{vmatrix}$$

Expand along the first row:

$$\vec a \times \vec b = \hat i\,(1\,(-2) - 3\,5) - \hat j\,(2\,(-2) - 3\,3) + \hat k\,(2\,5 - 1\,3)$$

Simplify each bracket:

• For $$\hat i$$: $$1(-2) - 3\cdot5 = -2 - 15 = -17$$
• For $$\hat j$$: $$2(-2) - 3\cdot3 = -4 - 9 = -13$$ (note the minus sign in the expansion)
• For $$\hat k$$: $$2\cdot5 - 1\cdot3 = 10 - 3 = 7$$

So,

$$\vec a \times \vec b = -17\hat i + 13\hat j + 7\hat k$$

2. Magnitude of the cross-product

The magnitude of a vector $$p\hat i + q\hat j + r\hat k$$ is $$\sqrt{p^{2}+q^{2}+r^{2}}$$. Here $$p=-17,\,q=13,\,r=7$$, hence

$$|\vec a \times \vec b| = \sqrt{(-17)^{2} + 13^{2} + 7^{2}} = \sqrt{289 + 169 + 49} = \sqrt{507}$$

Factorise the radicand:

$$507 = 169 \times 3 = (13)^{2} \times 3 \;\;\Rightarrow\;\; \sqrt{507} = 13\sqrt{3}$$

Therefore, $$|\vec a \times \vec b| = 13\sqrt{3}$$.

Step 1 : Form the two given combinations

$$\vec p = (\vec a + \vec b) = (\hat i + \hat j + \hat k) + (\hat i + 2\hat j + 3\hat k) = 2\hat i + 3\hat j + 4\hat k$$
$$\vec q = (\vec a - \vec b) = (\hat i + \hat j + \hat k) - (\hat i + 2\hat j + 3\hat k) = 0\hat i - \hat j - 2\hat k = -\hat j - 2\hat k$$

Step 2 : Take their cross-product
A vector perpendicular to each of $$\vec p$$ and $$\vec q$$ is their cross product $$\vec p \times \vec q$$.

Write the determinant for the cross product:

$$\vec p \times \vec q = \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 3 & 4\\ 0 & -1 & -2 \end{vmatrix}$$

Expand the determinant:

$$\vec p \times \vec q = \hat i\,(3\cdot(-2) - 4\cdot(-1)) - \hat j\,(2\cdot(-2) - 4\cdot 0) + \hat k\,(2\cdot(-1) - 3\cdot 0)$$

Simplify term by term:

• For $$\hat i$$: $$3(-2) - 4(-1) = -6 + 4 = -2$$
• For $$\hat j$$: $$2(-2) - 0 = -4$$
• For $$\hat k$$: $$2(-1) - 0 = -2$$

Hence

$$\vec p \times \vec q = -2\hat i - (-4)\hat j - 2\hat k = -2\hat i + 4\hat j - 2\hat k$$

Step 3 : Find its magnitude

$$\left|\vec p \times \vec q\right| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt 6$$

Step 4 : Convert to a unit vector

Divide by its magnitude:

$$\hat n = \dfrac{\vec p \times \vec q}{\lvert\vec p \times \vec q\rvert} = \dfrac{-2\hat i + 4\hat j - 2\hat k}{2\sqrt 6} = -\dfrac{1}{\sqrt 6}\,\hat i + \dfrac{2}{\sqrt 6}\,\hat j - \dfrac{1}{\sqrt 6}\,\hat k$$

Either direction of this vector is acceptable, so another valid answer is $$\dfrac{1}{\sqrt 6}\,\hat i - \dfrac{2}{\sqrt 6}\,\hat j + \dfrac{1}{\sqrt 6}\,\hat k$$.

Conclusion
A unit vector perpendicular to both $$\vec a + \vec b$$ and $$\vec a - \vec b$$ is

$$\boxed{\displaystyle \hat n = -\dfrac{1}{\sqrt 6}\,\hat i + \dfrac{2}{\sqrt 6}\,\hat j - \dfrac{1}{\sqrt 6}\,\hat k}$$

Let the position vectors of the three given vertices be

$$\vec a = \hat i + \hat j + \hat k, \qquad \vec b = \hat i + 2\hat j + 3\hat k, \qquad \vec c = 2\hat i + 3\hat j + \hat k.$$

To find the area of the triangle, we first determine two side vectors.

Step 1 — Find $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$.

$$\overrightarrow{AB} = \vec b - \vec a = (1-1)\hat i + (2-1)\hat j + (3-1)\hat k = \hat j + 2\hat k.$$

$$\overrightarrow{AC} = \vec c - \vec a = (2-1)\hat i + (3-1)\hat j + (1-1)\hat k = \hat i + 2\hat j.$$

Step 2 — Compute the cross-product $$\overrightarrow{AB}\times\overrightarrow{AC}$$.

$$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & 1 & 2\\ 1 & 2 & 0 \end{vmatrix}.$$

Expanding along the first row,

$$\overrightarrow{AB}\times\overrightarrow{AC} = \hat i\,(1\cdot 0 - 2\cdot 2) - \hat j\,(0\cdot 0 - 2\cdot 1) + \hat k\,(0\cdot 2 - 1\cdot 1).$$

$$\overrightarrow{AB}\times\overrightarrow{AC} = -4\hat i + 2\hat j - \hat k.$$

Step 3 — Magnitude of the cross-product.

$$|\overrightarrow{AB}\times\overrightarrow{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}.$$

Step 4 — Area of the triangle.

The area of a triangle formed by two side vectors is half the magnitude of their cross-product:

$$\text{Area} = \tfrac{1}{2}\,|\overrightarrow{AB}\times\overrightarrow{AC}| = \tfrac{1}{2}\sqrt{21}.$$

Hence the required area is $$\dfrac{\sqrt{21}}{2}$$ square units.

Given vectors

$$\vec a = 3\hat i + \hat j + 4\hat k, \qquad \vec b = \hat i - \hat j + \hat k.$$

Concept. For two adjacent sides $$\vec a$$ and $$\vec b$$ of a parallelogram, its area $$A$$ equals the magnitude of their cross-product:

$$A = |\vec a \times \vec b|.$$

Step 1 — Form the determinant for $$\vec a \times \vec b$$

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}.$$

Step 2 — Expand the determinant

• $$\hat i$$-component: $$1\cdot 1 - 4\cdot(-1) = 1 + 4 = 5.$$
• $$\hat j$$-component: $$-(3\cdot 1 - 4\cdot 1) = -(3 - 4) = 1.$$
• $$\hat k$$-component: $$3\cdot(-1) - 1\cdot 1 = -3 - 1 = -4.$$

Hence $$\vec a \times \vec b = 5\hat i + \hat j - 4\hat k.$$

Step 3 — Magnitude of the cross-product

$$|\vec a \times \vec b| = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42}.$$

Result. The area of the parallelogram is $$\sqrt{42}$$ square units.

We are given

$$\vec{a}=\hat{i}-7\hat{j}+7\hat{k}, \qquad \vec{b}=3\hat{i}-2\hat{j}+2\hat{k}.$$

The magnitude $$|\vec{a}\times\vec{b}|$$ is obtained in two steps:

1. Find the cross product

Using the determinant for the vector product,

$$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-7&7\\3&-2&2\end{vmatrix}.$$

Expand along the first row:

$$\vec{a}\times\vec{b}=\hat{i}\bigl((-7)(2)-(7)(-2)\bigr)-\hat{j}\bigl((1)(2)-(7)(3)\bigr)+\hat{k}\bigl((1)(-2)-(-7)(3)\bigr).$$

• Along $$\hat{i}$$: $$(-7)(2)-(7)(-2)=-14-(-14)=0$$
• Along $$\hat{j}$$: $$(1)(2)-(7)(3)=2-21=-19$$, then the factor $$-\hat{j}$$ makes it $$+19\hat{j}$$
• Along $$\hat{k}$$: $$(1)(-2)-(-7)(3)=-2-(-21)=-2+21=19$$

Hence

$$\vec{a}\times\vec{b}=0\hat{i}+19\hat{j}+19\hat{k}=19\hat{j}+19\hat{k}.$$

2. Compute the magnitude

$$|\vec{a}\times\vec{b}|=\sqrt{0^{2}+19^{2}+19^{2}}=\sqrt{361+361}=\sqrt{722}=19\sqrt{2}.$$

Therefore, $$|\vec{a}\times\vec{b}|=19\sqrt{2}$$.

Step 1: Compute $$\vec a+\vec b$$ and $$\vec a-\vec b$$

Given

$$\vec a = 3\hat i + 2\hat j + 2\hat k, \qquad \vec b = \hat i + 2\hat j - 2\hat k$$

1. Sum:

$$\vec a + \vec b = (3+1)\hat i + (2+2)\hat j + (2-2)\hat k = 4\hat i + 4\hat j + 0\hat k$$

2. Difference:

$$\vec a - \vec b = (3-1)\hat i + (2-2)\hat j + \bigl(2 - (-2)\bigr)\hat k = 2\hat i + 0\hat j + 4\hat k$$

Step 2: Find a vector perpendicular to both

The cross product of two vectors is perpendicular to each of them, so set

$$\vec r = (\vec a + \vec b) \times (\vec a - \vec b)$$

Using the determinant rule,

$$\vec r = \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}$$

$$\vec r = \hat i(4\cdot4 - 0\cdot0) - \hat j(4\cdot4 - 0\cdot2) + \hat k(4\cdot0 - 4\cdot2)$$

$$\vec r = 16\hat i - 16\hat j - 8\hat k$$

Simplify by dividing all components by 8:

$$\vec r = 2\hat i - 2\hat j - \hat k$$

Step 3: Convert $$\vec r$$ to a unit vector

Magnitude of $$\vec r$$:

$$|\vec r| = \sqrt{2^{2} + (-2)^{2} + (-1)^{2}} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$

Hence the required unit vector is

$$\hat u = \frac{\vec r}{|\vec r|} = \frac{1}{3}\,(2\hat i - 2\hat j - \hat k)$$

This $$\hat u$$ is perpendicular to both $$\vec a + \vec b$$ and $$\vec a - \vec b$$. (Its negative would also be acceptable.)

Let the required unit vector be $$\vec a = l\,\hat{i} + m\,\hat{j} + n\,\hat{k}$$ where $$l,m,n$$ are its direction-cosines.

By definition of direction-cosines we have

$$l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma,$$

where $$\alpha,\beta,\gamma$$ are the angles made by $$\vec a$$ with $$\hat{i},\hat{j},\hat{k}$$ respectively.

The question gives

$$\alpha = \dfrac{\pi}{3},\qquad \beta = \dfrac{\pi}{4},\qquad \gamma = \theta\;(\theta \text{ acute}).$$

Therefore

$$l = \cos\dfrac{\pi}{3} = \dfrac{1}{2},\qquad m = \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2},\qquad n = \cos\theta.$$

Because $$\vec a$$ is a unit vector, the direction-cosines satisfy

$$l^{2}+m^{2}+n^{2}=1.$$

Substituting the known values,

$$\left(\dfrac{1}{2}\right)^{2}+\left(\dfrac{\sqrt{2}}{2}\right)^{2}+n^{2}=1.$$

Simplify the numerical terms:

$$\dfrac{1}{4}+\dfrac{1}{2}+n^{2}=1 \;\;\Longrightarrow\;\; \dfrac{3}{4}+n^{2}=1.$$

Hence

$$n^{2}=1-\dfrac{3}{4}=\dfrac{1}{4} \;\;\Longrightarrow\;\; n=\pm\dfrac{1}{2}.$$

Since $$\theta$$ is acute, $$\cos\theta>0,$$ so we take the positive sign:

$$n = \dfrac{1}{2}.$$

Thus $$\cos\theta = \dfrac{1}{2} \;\;\Longrightarrow\;\; \theta = \dfrac{\pi}{3},$$ which lies in $$\left(0,\dfrac{\pi}{2}\right)$$ as required.

Finally, the components of $$\vec a$$ are

$$\vec a = \dfrac{1}{2}\,\hat{i}+\dfrac{\sqrt{2}}{2}\,\hat{j}+\dfrac{1}{2}\,\hat{k}.$$

Given: Prove that $$(\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = 2(\vec{a}\times\vec{b})$$.

1. Start with the left-hand side (LHS)

   $$ (\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) $$

2. Apply distributive property of the cross product
   For any vectors $$\vec{p},\vec{q},\vec{r}$$ we have $$\vec{p}\times(\vec{q}+\vec{r}) = \vec{p}\times\vec{q}+\vec{p}\times\vec{r}$$. Hence

   $$ (\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = (\vec{a}-\vec{b})\times\vec{a} \\ \;\;\;\;+ (\vec{a}-\vec{b})\times\vec{b}. $$

3. Distribute once more inside each term

   First term: $$(\vec{a}-\vec{b})\times\vec{a} = \vec{a}\times\vec{a} - \vec{b}\times\vec{a}.$$ Second term: $$(\vec{a}-\vec{b})\times\vec{b} = \vec{a}\times\vec{b} - \vec{b}\times\vec{b}.$$ Combine:

   $$ (\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = \vec{a}\times\vec{a} - \vec{b}\times\vec{a} + \vec{a}\times\vec{b} - \vec{b}\times\vec{b}. $$

4. Simplify the individual cross-products

   • For any vector $$\vec{v}$$, $$\vec{v}\times\vec{v}=\vec{0}$$ (cross product of a vector with itself is the zero vector).
   • For any two vectors $$\vec{p},\vec{q}$$, $$\vec{p}\times\vec{q}= -\,\vec{q}\times\vec{p}$$ (anti-commutativity).

   Thus:

   $$ \vec{a}\times\vec{a}=\vec{0}, \qquad \vec{b}\times\vec{b}=\vec{0}, \qquad -\vec{b}\times\vec{a}= -\bigl(-\,\vec{a}\times\vec{b}\bigr)=\vec{a}\times\vec{b}. $$

5. Substitute these results

   $$ (\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = \vec{0}+\vec{a}\times\vec{b}+\vec{a}\times\vec{b}+\vec{0}. $$

   $$ (\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = 2\,(\vec{a}\times\vec{b}). $$

6. The right-hand side (RHS) is exactly $$2(\vec{a}\times\vec{b})$$, so

   $$\boxed{\,(\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = 2\,(\vec{a}\times\vec{b})\,}$$

   Hence proved.

Let
$$\vec{a}=2\hat{i}+6\hat{j}+27\hat{k},\qquad \vec{b}=\hat{i}+\lambda\hat{j}+\mu\hat{k}.$$
We are told that
$$\vec{a}\times\vec{b}=\vec{0}.$$

The cross-product of two vectors in component form is obtained from the determinant

$$\vec{a}\times\vec{b}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\[2pt] 2&6&27\\[2pt] 1&\lambda&\mu \end{vmatrix}.$$

Expanding this determinant,

$$\vec{a}\times\vec{b}=\hat{i}(6\mu-27\lambda) -\hat{j}(2\mu-27) +\hat{k}(2\lambda-6).$$

For the cross-product to be the zero vector, each of the three components must be zero:

• $$6\mu-27\lambda=0$$
• $$2\mu-27=0$$
• $$2\lambda-6=0$$

Solve the equations one by one.

1. From $$2\lambda-6=0$$ we get $$\lambda=3.$$
2. From $$2\mu-27=0$$ we get $$\mu=\dfrac{27}{2}.$$
3. Substitute these values into $$6\mu-27\lambda$$ to verify consistency:
     $$6\left(\dfrac{27}{2}\right)-27(3)=81-81=0,$$ which indeed satisfies the first equation.

Hence the required values are

$$\boxed{\lambda=3,\; \mu=\dfrac{27}{2}}.$$

We are given two simultaneous conditions on the same pair of vectors $$\vec a$$ and $$\vec b$$:

• $$\vec a \cdot \vec b = 0$$  … the dot–product is zero.
• $$\vec a \times \vec b = \vec 0$$  … the cross–product is the zero vector.

Recall the geometric interpretations for any two vectors $$\vec u,\vec v$$ in three–dimensional space:

1. Dot product: $$\vec u \cdot \vec v = 0$$ means that $$\vec u$$ is perpendicular to $$\vec v$$ unless at least one of them is the zero vector.
2. Cross product: $$\vec u \times \vec v = \vec 0$$ means that $$\vec u$$ is parallel to $$\vec v$$ unless at least one of them is the zero vector.

Apply these facts to our pair $$\vec a,\vec b$$:

• From $$\vec a \times \vec b = \vec 0$$ we conclude that either the vectors are parallel or at least one of $$\vec a,\vec b$$ is the zero vector.
• From $$\vec a \cdot \vec b = 0$$ we conclude that either the vectors are perpendicular or at least one of $$\vec a,\vec b$$ is the zero vector.

If both $$\vec a$$ and $$\vec b$$ were non-zero simultaneously, the two statements would read:

• “$$\vec a$$ is parallel to $$\vec b$$” (from the cross product), and
• “$$\vec a$$ is perpendicular to $$\vec b$$” (from the dot product).

A single non-zero vector cannot be both parallel and perpendicular to another non-zero vector at the same time. Hence the assumption that both vectors are non-zero leads to a contradiction.

Therefore, the only possibility that satisfies both conditions is that at least one of the vectors is the zero vector.

Conclusion: $$\vec a = \vec 0$$ or $$\vec b = \vec 0$$ (or both).

Let

$$\vec a = a_1\hat i + a_2\hat j + a_3\hat k, \;\;\; \vec b = b_1\hat i + b_2\hat j + b_3\hat k, \;\;\; \vec c = c_1\hat i + c_2\hat j + c_3\hat k.$$

First form the sum $$\vec b + \vec c$$:

$$\vec b + \vec c = (b_1+c_1)\hat i + (b_2+c_2)\hat j + (b_3+c_3)\hat k.$$

Now compute the cross–product $$\vec a \times (\vec b + \vec c)$$ with the determinant formula:

$$\vec a \times (\vec b + \vec c) =\begin{vmatrix}\hat i & \hat j & \hat k\\ a_1 & a_2 & a_3\\ (b_1+c_1) & (b_2+c_2) & (b_3+c_3)\end{vmatrix}.$$

Expanding the determinant,

$$\begin{aligned}\vec a \times (\vec b + \vec c)=\;&\hat i\bigl[a_2(b_3+c_3)-a_3(b_2+c_2)\bigr]\\[2pt]&-\hat j\bigl[a_1(b_3+c_3)-a_3(b_1+c_1)\bigr]\\[2pt]&+\hat k\bigl[a_1(b_2+c_2)-a_2(b_1+c_1)\bigr].\end{aligned}$$

Distribute each bracket:

$$\begin{aligned}\vec a \times (\vec b + \vec c)=\;&\hat i\bigl[a_2b_3-a_3b_2 + a_2c_3-a_3c_2\bigr]\\[2pt]&-\hat j\bigl[a_1b_3-a_3b_1 + a_1c_3-a_3c_1\bigr]\\[2pt]&+\hat k\bigl[a_1b_2-a_2b_1 + a_1c_2-a_2c_1\bigr].\end{aligned}$$

Group the b–terms and the c–terms separately:

$$\vec a \times (\vec b + \vec c)= \Bigl[\hat i(a_2b_3-a_3b_2)-\hat j(a_1b_3-a_3b_1)+\hat k(a_1b_2-a_2b_1)\Bigr]$$
$$\quad+\Bigl[\hat i(a_2c_3-a_3c_2)-\hat j(a_1c_3-a_3c_1)+\hat k(a_1c_2-a_2c_1)\Bigr].$$

The first bracket is exactly $$\vec a \times \vec b$$ and the second bracket is $$\vec a \times \vec c$$. Hence

$$\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c.$$

Thus the cross product is distributive over vector addition, as required.

We know that the cross-product of two vectors is defined by

$$\vec a \times \vec b = |\vec a|\,|\vec b|\sin\theta\;\hat n,$$

where $$\theta$$ is the angle between $$\vec a$$ and $$\vec b$$ and $$\hat n$$ is the unit vector perpendicular to the plane containing them.

Given proposition: If either $$\vec a = \vec 0$$ or $$\vec b = \vec 0$$, then clearly at least one of the magnitudes $$|\vec a|, |\vec b|$$ is zero, so $$|\vec a|\,|\vec b|\sin\theta = 0$$ and therefore $$\vec a \times \vec b = \vec 0$$. The statement is correct.

Converse statement: “If $$\vec a \times \vec b = \vec 0$$, then either $$\vec a = \vec 0$$ or $$\vec b = \vec 0$$.”

From the definition, $$\vec a \times \vec b = \vec 0$$ when at least one of the following is true:

• $$|\vec a|=0$$ (i.e. $$\vec a=\vec 0$$),
• $$|\vec b|=0$$ (i.e. $$\vec b=\vec 0$$),
• $$\sin\theta = 0$$, which happens when $$\theta = 0^{\circ}$$ or $$\theta = 180^{\circ}$$, i.e. the two non-zero vectors are parallel (collinear).

Thus the converse need not be true: $$\vec a \times \vec b$$ can vanish even when neither vector is the zero vector, provided the vectors are parallel.

Example disproving the converse

Choose $$\vec a = \hat i \quad\text{and}\quad \vec b = 2\hat i.$$

Both vectors are non-zero and clearly parallel, so $$\theta = 0^{\circ}$$ and $$\sin\theta = 0$$.

Therefore $$\vec a \times \vec b = |\vec a|\,|\vec b|\sin0^{\circ}\;\hat n = (1)(2)(0)\;\hat n = 0,$$ i.e. $$\vec a \times \vec b = \vec 0,$$ even though neither $$\vec a$$ nor $$\vec b$$ is the zero vector.

Hence the converse is false.

The area of a triangle in space can be found from the magnitudes of the vectors along two of its sides.

Taking vertex $$A$$ as the common point, form the side vectors:

$$\vec{AB}=B-A=(2-1)\hat{i}+(3-1)\hat{j}+(5-2)\hat{k}=1\hat{i}+2\hat{j}+3\hat{k}$$

$$\vec{AC}=C-A=(1-1)\hat{i}+(5-1)\hat{j}+(5-2)\hat{k}=0\hat{i}+4\hat{j}+3\hat{k}$$

The area is half the magnitude of the cross product $$\vec{AB}\times\vec{AC}$$, so compute that cross product:

$$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\0&4&3\end{vmatrix}$$

Expanding,

$$\vec{AB}\times\vec{AC}=\hat{i}(2\cdot3-3\cdot4)-\hat{j}(1\cdot3-3\cdot0)+\hat{k}(1\cdot4-2\cdot0)=-6\hat{i}-3\hat{j}+4\hat{k}$$

Its magnitude is

$$|\vec{AB}\times\vec{AC}|=\sqrt{(-6)^2+(-3)^2+4^2}=\sqrt{36+9+16}=\sqrt{61}$$

Therefore,

$$\text{Area}=\frac{1}{2}|\vec{AB}\times\vec{AC}|=\frac{\sqrt{61}}{2}\;\text{square units}$$

Given vectors:

$$\vec{a}= \hat{i}-\hat{j}+3\hat{k},\qquad \vec{b}=2\hat{i}-7\hat{j}+\hat{k}$$

The area $$\Delta$$ of the parallelogram formed by $$\vec{a}$$ and $$\vec{b}$$ equals the magnitude of their cross product:

$$\Delta = |\vec{a}\times\vec{b}|$$

Step 1: Compute the cross product

$$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1\end{vmatrix}$$

Expanding,

$$\vec{a}\times\vec{b}= \hat{i}\bigl((-1)(1)-3(-7)\bigr) -\hat{j}\bigl(1\cdot1-3\cdot2\bigr)+\hat{k}\bigl(1(-7)-(-1)\cdot2\bigr)$$

• $$\hat{i}$$-component: $$(-1)(1)-3(-7)= -1+21=20$$
• $$\hat{j}$$-component: $$-\left(1-6\right)=-(-5)=5$$
• $$\hat{k}$$-component: $$1(-7)-(-1)\cdot2= -7+2=-5$$

Hence, $$\vec{a}\times\vec{b}=20\hat{i}+5\hat{j}-5\hat{k}$$

Step 2: Find its magnitude

$$|\vec{a}\times\vec{b}|=\sqrt{20^{2}+5^{2}+(-5)^{2}}=\sqrt{400+25+25}=\sqrt{450}$$

$$\sqrt{450}=\sqrt{9\cdot50}=3\sqrt{50}=3\sqrt{25\cdot2}=3\cdot5\sqrt{2}=15\sqrt{2}$$

Area of the parallelogram:

$$\boxed{15\sqrt{2}\ \text{square units}}$$

For any two vectors $$\vec a$$ and $$\vec b$$ the magnitude of their cross-product is

$$|\vec a \times \vec b| = |\vec a|\,|\vec b|\sin\theta,$$

where $$\theta$$ is the (acute or obtuse) angle between the two vectors.

According to the question the cross-product itself must be a unit vector, i.e. its magnitude equals 1:

$$|\vec a \times \vec b| = 1.$$

The given magnitudes are

$$|\vec a| = 3, \quad |\vec b| = \dfrac{\sqrt2}{3}.$$

Substituting these values in the formula,

$$1 = |\vec a \times \vec b| = (3)\left(\dfrac{\sqrt2}{3}\right)\sin\theta.$$

Simplifying,

$$1 = \sqrt2\,\sin\theta \quad\Longrightarrow\quad \sin\theta = \frac1{\sqrt2} = \frac{\sqrt2}{2}.$$

The principal angles in $$[0,\pi]$$ for which $$\sin\theta = \frac{\sqrt2}{2}$$ are

$$\theta = \frac{\pi}{4}\quad\text{or}\quad \theta = \frac{3\pi}{4}.$$

Among the alternatives provided, only $$\theta = \dfrac{\pi}{4}$$ appears.

Hence the required angle between $$\vec a$$ and $$\vec b$$ is $$\boxed{\dfrac{\pi}{4}}$$, i.e. option (B).

Let the position vectors of the four vertices be written in Cartesian form.

• $$A(-1,\,\tfrac12,\,4)$$
• $$B(1,\,\tfrac12,\,4)$$
• $$C(1,\,-\tfrac12,\,4)$$
• $$D(-1,\,-\tfrac12,\,4)$$

Take the order $$A \to B \to C \to D$$ so that consecutive points are adjacent vertices of the rectangle.

Step 1 – Find two adjacent sides as vectors.

$$\vec{AB}=\vec{B}-\vec{A}=(1-(-1),\;\tfrac12-\tfrac12,\;4-4)=(2,0,0)$$

$$\vec{BC}=\vec{C}-\vec{B}=(1-1,\;-\tfrac12-\tfrac12,\;4-4)=(0,-1,0)$$

Step 2 – Check perpendicularity (optional).

$$\vec{AB}\cdot\vec{BC}=2\cdot0+0\cdot(-1)+0\cdot0=0,$$ hence the sides are perpendicular, confirming a rectangle.

Step 3 – Compute the lengths of the two adjacent sides.

$$|\vec{AB}|=\sqrt{2^{2}+0^{2}+0^{2}}=2$$

$$|\vec{BC}|=\sqrt{0^{2}+(-1)^{2}+0^{2}}=1$$

Step 4 – Area of the rectangle.

The area is the product of the lengths of two adjacent sides:

$$\text{Area}=|\vec{AB}|\,|\vec{BC}| = 2\times1 = 2.$$

Therefore the area is $$2$$, so the correct option is (C).

Let $$\vec{a}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k}$$ be any vector.

Because the vector lies in the $$XY$$-plane, its $$z$$-component is zero, so $$z=0$$ and $$\vec{a}=x\,\hat{i}+y\,\hat{j}.$$

For $$\vec{a}$$ to be a unit vector we need its magnitude to equal 1:

$$|\vec{a}|=\sqrt{x^{2}+y^{2}}=1 \;\Longrightarrow\; x^{2}+y^{2}=1.$$

The equation $$x^{2}+y^{2}=1$$ represents the unit circle in the $$XY$$-plane. A convenient way to write every ordered pair $$(x,y)$$ on this circle is

$$x=\cos\theta,\quad y=\sin\theta,\qquad 0\le\theta<2\pi.$$

Substituting these values in $$\vec{a}=x\,\hat{i}+y\,\hat{j}$$ gives the desired family of unit vectors:

$$\vec{u}(\theta)=\cos\theta\,\hat{i}+\sin\theta\,\hat{j},\qquad 0\le\theta<2\pi.$$

Every choice of $$\theta$$ in the stated interval produces one unit vector, and every unit vector in the $$XY$$-plane is produced in this way. Hence the set of all unit vectors in the $$XY$$-plane is

$$\{\,\cos\theta\,\hat{i}+\sin\theta\,\hat{j} \;\mid\; 0\le\theta<2\pi\,\}.$$

Let the position vectors of points A, B, C and D be

$$\vec{OA}=\hat{i}+\hat{j}+\hat{k},\;\; \vec{OB}=2\hat{i}+5\hat{j},\;\; \vec{OC}=3\hat{i}+2\hat{j}-3\hat{k},\;\; \vec{OD}=\hat{i}-6\hat{j}-\hat{k}.$$

Step 1 – Find $$\vec{AB}$$:

$$\vec{AB}=\vec{OB}-\vec{OA}=(2-1)\hat{i}+(5-1)\hat{j}+(0-1)\hat{k}=\hat{i}+4\hat{j}-\hat{k}.$$

Step 2 – Find $$\vec{CD}$$:

$$\vec{CD}=\vec{OD}-\vec{OC}=(1-3)\hat{i}+(-6-2)\hat{j}+(-1+3)\hat{k}=-2\hat{i}-8\hat{j}+2\hat{k}.$$

Step 3 – Angle between $$\vec{AB}$$ and $$\vec{CD}$$:

Dot product:
$$\vec{AB}\cdot\vec{CD}=1(-2)+4(-8)+(-1)(2)=-2-32-2=-36.$$

Magnitudes:
$$|\vec{AB}|=\sqrt{1^{2}+4^{2}+(-1)^{2}}=\sqrt{18}=3\sqrt{2},$$
$$|\vec{CD}|=\sqrt{(-2)^{2}+(-8)^{2}+2^{2}}=\sqrt{72}=6\sqrt{2}.$$

Therefore
$$\cos\theta=\dfrac{\vec{AB}\cdot\vec{CD}}{|\vec{AB}|\,|\vec{CD}|}=\dfrac{-36}{(3\sqrt{2})(6\sqrt{2})}=\dfrac{-36}{36}=-1.$$

Hence $$\theta=180^{\circ}.$$

Step 4 – Deduction about collinearity:

A cosine of $$-1$$ shows that $$\vec{AB}$$ and $$\vec{CD}$$ are parallel but point in opposite directions. Since each vector joins two points of the same line of action, the line through A and B is the same line through C and D. Thus $$\vec{AB}$$ and $$\vec{CD}$$ are collinear.

We are given three non-zero vectors $$\vec{a},\vec{b},\vec{c}$$ that satisfy

• $$|\vec{a}| = 3$$, $$|\vec{b}| = 4$$, $$|\vec{c}| = 5$$,
• each vector is perpendicular to the sum of the other two.

Translate the perpendicularity conditions into dot-product equations:

$$\vec{a}\cdot(\vec{b}+\vec{c}) = 0, \qquad \vec{b}\cdot(\vec{c}+\vec{a}) = 0, \qquad \vec{c}\cdot(\vec{a}+\vec{b}) = 0.$$

Introduce the unknown pairwise dot products

$$x = \vec{a}\cdot\vec{b}, \qquad y = \vec{b}\cdot\vec{c}, \qquad z = \vec{c}\cdot\vec{a}.$$

Expanding the three perpendicularity equations gives

(1)   $$x + z = 0 \;\Rightarrow\; z = -x,$$

(2)   $$y + x = 0 \;\Rightarrow\; y = -x,$$

(3)   $$z + y = 0.$$

Substituting the expressions for $$y$$ and $$z$$ from (1) and (2) into (3):

$$(-x) + (-x) = 0 \;\Longrightarrow\; -2x = 0 \;\Longrightarrow\; x = 0.$$

Hence $$x = y = z = 0,$$ which shows that every pair of vectors is perpendicular:

$$\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0.$$

Now compute the magnitude of their sum. Start with the square:

$$|\vec{a}+\vec{b}+\vec{c}|^{2} = (\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}).$$

Expand using distributivity:

$$|\vec{a}+\vec{b}+\vec{c}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} + |\vec{c}|^{2} + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}).$$

The cross terms are zero, so

$$|\vec{a}+\vec{b}+\vec{c}|^{2} = 3^{2} + 4^{2} + 5^{2} = 9 + 16 + 25 = 50.$$

Taking the positive square root gives

$$|\vec{a}+\vec{b}+\vec{c}| = \sqrt{50} = 5\sqrt{2}.$$

Therefore, the required magnitude is $$5\sqrt{2}.$$

Given data

• $$\vec a+\vec b+\vec c=\vec 0$$
• $$|\vec a|=3\,,\;|\vec b|=4\,,\;|\vec c|=2$$
• Required: $$\mu=\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$$

Step 1 · Form an equation involving $$\mu$$

Take the dot product of both sides of $$\vec a+\vec b+\vec c=\vec 0$$ with itself:

$$ (\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)=0 $$

Expand the left–hand side using distributivity:

$$ \vec a\cdot\vec a+\vec b\cdot\vec b+\vec c\cdot\vec c+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0 $$

By definition of magnitudes, $$\vec a\cdot\vec a=|\vec a|^{2}$$, etc. Hence

$$ |\vec a|^{2}+|\vec b|^{2}+|\vec c|^{2}+2\mu=0 $$

Step 2 · Insert the given magnitudes

$$ 3^{2}+4^{2}+2^{2}+2\mu=0 \quad\Longrightarrow\quad 9+16+4+2\mu=0 $$

Simplify the numerical sum:

$$ 29+2\mu=0 $$

Step 3 · Solve for $$\mu$$

$$ 2\mu=-29\;\;\Longrightarrow\;\;\mu=-\dfrac{29}{2} $$

Conclusion

The required value is $$\boxed{-\dfrac{29}{2}}$$.

Given vectors $$\vec{\alpha}=3\hat i-\hat j$$ and $$\vec{\beta}=2\hat i+\hat j-3\hat k$$.

We need to write $$\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}$$ so that

• $$\vec{\beta_1}\parallel\vec{\alpha}$$ (that is, $$\vec{\beta_1}$$ is a scalar multiple of $$\vec{\alpha}$$)
• $$\vec{\beta_2}\perp\vec{\alpha}$$ (so $$\vec{\alpha}\cdot\vec{\beta_2}=0$$).

The vector parallel to $$\vec{\alpha}$$ obtained from $$\vec{\beta}$$ is its projection on $$\vec{\alpha}$$:

$$\vec{\beta_1}=\operatorname{proj}_{\vec{\alpha}}\vec{\beta}=\frac{\vec{\beta}\cdot\vec{\alpha}}{\lvert\vec{\alpha}\rvert^{2}}\,\vec{\alpha}$$.

Step 1 – Dot product $$\vec{\beta}\cdot\vec{\alpha}$$

$$\vec{\beta}\cdot\vec{\alpha}=(2\hat i+\hat j-3\hat k)\cdot(3\hat i-\hat j)$$
$$=2\times3+1\times(-1)+(-3)\times0$$
$$=6-1+0$$
$$=5$$.

Step 2 – Magnitude squared of $$\vec{\alpha}$$

$$\lvert\vec{\alpha}\rvert^{2}=3^{2}+(-1)^{2}+0^{2}=9+1=10$$.

Step 3 – Compute $$\vec{\beta_1}$$

$$\vec{\beta_1}=\frac{5}{10}\,(3\hat i-\hat j)=\frac12\,(3\hat i-\hat j)=\frac32\hat i-\frac12\hat j$$.

Step 4 – Compute $$\vec{\beta_2}$$

$$\vec{\beta_2}=\vec{\beta}-\vec{\beta_1}$$
$$=(2\hat i+\hat j-3\hat k)-\left(\frac32\hat i-\frac12\hat j\right)$$
$$=\left(2-\frac32\right)\hat i+\left(1+\frac12\right)\hat j-3\hat k$$
$$=\frac12\hat i+\frac32\hat j-3\hat k$$.

Step 5 – Verification of perpendicularity

$$\vec{\alpha}\cdot\vec{\beta_2}=(3\hat i-\hat j)\cdot\left(\frac12\hat i+\frac32\hat j-3\hat k\right)$$
$$=3\times\frac12+(-1)\times\frac32+0\times(-3)=\frac32-\frac32=0$$, as required.

Therefore

$$\vec{\beta}=\underbrace{\left(\frac32\hat i-\frac12\hat j\right)}_{\vec{\beta_1}\parallel\vec{\alpha}}+\underbrace{\left(\frac12\hat i+\frac32\hat j-3\hat k\right)}_{\vec{\beta_2}\perp\vec{\alpha}}$$.

Let the required unit vector be denoted by $$\vec{u}=u_x\hat{i}+u_y\hat{j}$$, where $$u_x$$ and $$u_y$$ are its components along the $$x$$- and $$y$$-axes respectively.

Because the vector lies in the $$XY$$-plane and makes an angle of $$30^\circ$$ with the positive direction of the $$x$$-axis, its direction cosines are given directly by elementary trigonometry:

• Component along $$x$$-axis (adjacent side): $$u_x=\cos 30^\circ$$
• Component along $$y$$-axis (opposite side): $$u_y=\sin 30^\circ$$

Compute the exact values:

$$\cos 30^\circ = \frac{\sqrt{3}}{2},\quad \sin 30^\circ = \frac{1}{2}$$

Therefore

$$\vec{u}=\cos 30^\circ\,\hat{i}+\sin 30^\circ\,\hat{j}=\frac{\sqrt{3}}{2}\,\hat{i}+\frac{1}{2}\,\hat{j}.$$

Finally, verify that $$\vec{u}$$ is indeed a unit vector:

Magnitude $$|\vec{u}|=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{3}{4}+\frac{1}{4}}=\sqrt{1}=1.$$

Thus the required unit vector is $$\frac{\sqrt{3}}{2}\,\hat{i}+\frac{1}{2}\,\hat{j}$$.

Let $$P(x_1, y_1, z_1)$$ and $$Q(x_2, y_2, z_2)$$ be two points in three–dimensional space.

Their position vectors with respect to the origin $$O(0,0,0)$$ are

$$\vec{OP}=x_1\hat i+y_1\hat j+z_1\hat k, \qquad \vec{OQ}=x_2\hat i+y_2\hat j+z_2\hat k.$$

The vector that joins $$P$$ to $$Q$$ is obtained by subtracting these position vectors:

$$\vec{PQ}=\vec{OQ}-\vec{OP}.$$

Substituting the expressions for $$\vec{OP}$$ and $$\vec{OQ}$$ we get

$$\vec{PQ}=(x_2\hat i+y_2\hat j+z_2\hat k)-(x_1\hat i+y_1\hat j+z_1\hat k).$$

Grouping like terms gives

$$\vec{PQ}=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k.$$

Hence the scalar components of $$\vec{PQ}$$ are

• along $$\hat i$$ : $$x_2-x_1$$,
• along $$\hat j$$ : $$y_2-y_1$$,
• along $$\hat k$$ : $$z_2-z_1$$.

The magnitude (length) of the vector is

$$|\vec{PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.$$

Thus, the required scalar components and magnitude are obtained.

Choose a rectangular coordinate system with

• origin at the girl’s starting point O,
• positive $$x$$-axis towards east,
• positive $$y$$-axis towards north.

Step 1 – First walk
4 km towards west is represented by

$$\vec v_1=-4\,\hat i$$

Step 2 – Second walk
The next 3 km are at $$30^\circ$$ east of north.
Resolving,

$$v_{2x}=3\sin30^\circ=3\left(\tfrac12\right)=1.5,$$
$$v_{2y}=3\cos30^\circ=3\left(\tfrac{\sqrt3}{2}\right)=\tfrac{3\sqrt3}{2}\approx2.598.$$

Hence

$$\vec v_2=1.5\,\hat i+2.598\,\hat j.$$

Step 3 – Resultant displacement

$$\begin{aligned}\vec r&=\vec v_1+\vec v_2\\&=(-4+1.5)\,\hat i+2.598\,\hat j\\&=-2.5\,\hat i+2.598\,\hat j.\end{aligned}$$

Magnitude

$$|\vec r|=\sqrt{(-2.5)^2+(2.598)^2}=\sqrt{6.25+6.7604}=\sqrt{13.0104}\approx3.61\text{ km}.$$

Direction

With $$\theta$$ measured from north towards west,

$$\tan\theta=\frac{|x|}{y}=\frac{2.5}{2.598}\approx0.961\;\Rightarrow\;\theta\approx43^\circ.$$

Thus the displacement is $$3.6\text{ km}$$ making $$43^\circ$$ west of north (or $$46^\circ$$ north of west).

Result: $$\boxed{3.6\text{ km at }43^\circ\text{ west of north}}$$

We are given $$\vec a = \vec b + \vec c.$$ Take the dot product of both sides with itself to obtain the magnitude of $$\vec a:$$

$$|\vec a|^{2} = \vec a \cdot \vec a = (\vec b + \vec c)\cdot(\vec b + \vec c).$$

Expanding (and using $$\vec b\cdot\vec c = \vec c\cdot\vec b$$),

$$|\vec a|^{2} = |\vec b|^{2} + |\vec c|^{2} + 2\,\vec b\cdot\vec c.$$

If $$\theta$$ is the angle between $$\vec b$$ and $$\vec c,$$ then $$\vec b\cdot\vec c = |\vec b|\,|\vec c|\cos\theta,$$ so

$$|\vec a|^{2} = |\vec b|^{2} + |\vec c|^{2} + 2|\vec b|\,|\vec c|\cos\theta. \quad (1)$$

On the other hand,

$$\bigl(|\vec b| + |\vec c|\bigr)^{2} = |\vec b|^{2} + |\vec c|^{2} + 2|\vec b|\,|\vec c|. \quad (2)$$

Comparing $$(1)$$ with $$(2),$$ the equality $$|\vec a| = |\vec b| + |\vec c|$$ holds if and only if $$\cos\theta = 1,$$ i.e. $$\theta = 0^{\circ}$$ — meaning $$\vec b$$ and $$\vec c$$ are parallel and point in the same direction.

For any other angle $$\theta$$ in the range $$0^{\circ} \lt \theta \le 180^{\circ},$$ we have $$\cos\theta \lt 1,$$ and hence

$$|\vec a| \lt |\vec b| + |\vec c|.$$

In general, therefore, $$|\vec a| \le |\vec b| + |\vec c|$$ — the well-known triangle inequality.

Counter-example. Let $$\vec b = \hat\imath$$ and $$\vec c = \hat\jmath,$$ so that $$\vec a = \hat\imath + \hat\jmath.$$ Then

• $$|\vec b| + |\vec c| = 1 + 1 = 2,$$
• $$|\vec a| = \sqrt{1^{2}+1^{2}} = \sqrt{2}.$$

Since $$\sqrt{2} \neq 2,$$ the equality fails here.

Conclusion. The statement $$|\vec a| = |\vec b| + |\vec c|$$ is not true in general; it holds only when $$\vec b$$ and $$\vec c$$ are parallel and point in the same direction.

Let $$\vec{a}=x(\hat{i}+\hat{j}+\hat{k})$$.

The magnitude of $$\vec{a}$$ is

$$|\vec{a}|=|x|\,|\hat{i}+\hat{j}+\hat{k}|$$

because multiplying a vector by a scalar changes its length by the absolute value of that scalar.

Compute $$|\hat{i}+\hat{j}+\hat{k}|$$:

$$|\hat{i}+\hat{j}+\hat{k}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$$.

Thus

$$|\vec{a}|=|x|\sqrt{3}$$.

For $$\vec{a}$$ to be a unit vector, we need $$|\vec{a}|=1$$.

Hence

$$|x|\sqrt{3}=1\;\Longrightarrow\;|x|=\dfrac{1}{\sqrt{3}}$$.

Therefore, the permissible values of $$x$$ are

$$x=\pm\dfrac{1}{\sqrt{3}}$$.

Step 1 — Find the resultant of $$\vec a$$ and $$\vec b$$

$$\vec a = 2\hat{i}+3\hat{j}-\hat{k}, \;\; \vec b = \hat{i}-2\hat{j}+\hat{k}$$

Resultant:

$$\vec r = \vec a + \vec b$$

Component–wise addition:

• $$i\text{-component}:\; 2+1 = 3$$
• $$j\text{-component}:\; 3-2 = 1$$
• $$k\text{-component}:\; -1+1 = 0$$

Hence $$\vec r = 3\hat{i}+\hat{j}$$.

Step 2 — Determine the magnitude of $$\vec r$$

$$|\vec r| = \sqrt{3^{2}+1^{2}+0^{2}} = \sqrt{9+1}=\sqrt{10}$$

Step 3 — Unit vector parallel to $$\vec r$$

$$\hat{r}=\dfrac{\vec r}{|\vec r|}=\dfrac{3\hat{i}+\hat{j}}{\sqrt{10}}=\frac{3}{\sqrt{10}}\hat{i}+\frac{1}{\sqrt{10}}\hat{j}$$

Step 4 — Vector of magnitude 5 units parallel to $$\vec r$$

Multiply the unit vector by 5:

$$\vec v = 5\hat{r}=5\left(\frac{3}{\sqrt{10}}\hat{i}+\frac{1}{\sqrt{10}}\hat{j}\right)=\frac{15}{\sqrt{10}}\hat{i}+\frac{5}{\sqrt{10}}\hat{j}$$

Rationalising (optional): $$\dfrac{15}{\sqrt{10}}=\dfrac{15\sqrt{10}}{10}=\dfrac{3\sqrt{10}}{2}$$ and $$\dfrac{5}{\sqrt{10}}=\dfrac{\sqrt{10}}{2}$$.

Therefore,

$$\vec v = \left(\dfrac{3\sqrt{10}}{2}\right)\hat{i}+\left(\dfrac{\sqrt{10}}{2}\right)\hat{j}$$

which indeed has magnitude 5 and is parallel to $$\vec a+\vec b$$.

We are asked to obtain a unit vector that is parallel to the vector $$2\vec a-\vec b+3\vec c$$, where

$$\vec a=\hat i+\hat j+\hat k, \qquad \vec b=2\hat i-\hat j+3\hat k, \qquad \vec c=\hat i-2\hat j+\hat k.$$

Step 1 : Compute the required vector

Twice \(\vec a\): $$2\vec a=2(\hat i+\hat j+\hat k)=2\hat i+2\hat j+2\hat k.$$

Three times \(\vec c\): $$3\vec c=3(\hat i-2\hat j+\hat k)=3\hat i-6\hat j+3\hat k.$$

Now form $$2\vec a-\vec b+3\vec c$$ component-wise:

$$\begin{aligned} 2\vec a-\vec b+3\vec c&=(2\hat i+2\hat j+2\hat k) - (2\hat i-\hat j+3\hat k) + (3\hat i-6\hat j+3\hat k)\\ &=\underbrace{(2-2+3)}_{=3}\hat i \, +\, \underbrace{(2-(-1)-6)}_{=-3}\hat j \, +\, \underbrace{(2-3+3)}_{=2}\hat k\\ &=3\hat i-3\hat j+2\hat k. \end{aligned}$$

Denote this vector by $$\vec r=3\hat i-3\hat j+2\hat k.$$

Step 2 : Find its magnitude

$$|\vec r|=\sqrt{3^{2}+(-3)^{2}+2^{2}}=\sqrt{9+9+4}=\sqrt{22}.$$

Step 3 : Form the unit vector

Divide \(\vec r\) by its magnitude:

$$\hat u=\dfrac{\vec r}{|\vec r|}=\frac{3\hat i-3\hat j+2\hat k}{\sqrt{22}} =\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}}\hat j+\frac{2}{\sqrt{22}}\hat k.$$

Hence, $$\boxed{\displaystyle \hat u = \frac{3}{\sqrt{22}}\hat i - \frac{3}{\sqrt{22}}\hat j + \frac{2}{\sqrt{22}}\hat k}.$$

Step 1 : Form the vectors $$\vec{AB}$$ and $$\vec{AC}$$

$$\vec{AB}=\bigl(5-1\,,\;0-(-2)\,,\;-2-(-8)\bigr)=(4,2,6)$$
$$\vec{AC}=\bigl(11-1\,,\;3-(-2)\,,\;7-(-8)\bigr)=(10,5,15)$$

Step 2 : Test for collinearity

Observe that $$\vec{AC}=\dfrac{5}{2}\,\vec{AB}$$ because
$$\dfrac{5}{2}\,(4,2,6)=(10,5,15)=\vec{AC}.$$
Since $$\vec{AB}$$ and $$\vec{AC}$$ are non-zero scalar multiples, they are parallel; hence A, B and C are collinear.

Step 3 : Find the ratio in which B divides $$\overline{AC}$$

Because $$\vec{AB}+\vec{BC}=\vec{AC},$$ we obtain
$$\vec{BC}=\vec{AC}-\vec{AB}=(10,5,15)-(4,2,6)=(6,3,9).$$

Compare corresponding components:
$$\dfrac{AB}{BC}=\dfrac{4}{6}=\dfrac{2}{3}.$$

Therefore $$B$$ divides the segment $$AC$$ internally in the ratio $$2:3$$ (i.e. $$AB:BC=2:3$$).

Given data

• Position vector of P: $$\vec p = 2\vec a + \vec b$$
• Position vector of Q: $$\vec q = \vec a - 3\vec b$$

Point R divides the line joining P and Q externally in the ratio 1 : 2, i.e. $$PR:RQ = 1:2$$.

1. Position vector of R (external section formula)

If a point divides the line joining $$P(\vec p)$$ and $$Q(\vec q)$$ externally in the ratio $$m:n$$ ( $$PR:RQ = m:n$$ ), its position vector is

$$\vec r = \frac{m\,\vec q - n\,\vec p}{m - n}.$$

Here $$m = 1$$ and $$n = 2$$, so

$$\vec r = \frac{1\,\vec q - 2\,\vec p}{1 - 2} = - \left( \vec q - 2\,\vec p \right).$$

Substituting $$\vec p$$ and $$\vec q$$:

$$\vec r = - \Bigl[\,(\vec a - 3\vec b) - 2\,(2\vec a + \vec b)\,\Bigr]$$

$$\;\;\;\;= - \Bigl[\,\vec a - 3\vec b - 4\vec a - 2\vec b\Bigr]$$

$$\;\;\;\;= - \Bigl[\,-3\vec a - 5\vec b\Bigr]$$

$$\boxed{\vec r = 3\vec a + 5\vec b}$$

2. Verifying that P is the midpoint of RQ

The midpoint M of the segment RQ has position vector

$$\vec m = \frac{\vec r + \vec q}{2}.$$

Compute the numerator:

$$\vec r + \vec q = (3\vec a + 5\vec b) + (\vec a - 3\vec b) = 4\vec a + 2\vec b.$$

Factor 2:

$$\vec r + \vec q = 2\,(2\vec a + \vec b) = 2\vec p.$$

Hence

$$\vec m = \frac{2\vec p}{2} = \vec p.$$

Thus $$\vec p$$ is exactly the midpoint vector, so point P is the midpoint of the segment RQ.

Result proved.

Let the adjacent sides of the parallelogram be denoted by

$$\vec a = 2\hat i - 4\hat j + 5\hat k, \qquad \vec b = \hat i - 2\hat j - 3\hat k.$$

1. Unit vector parallel to a diagonal

One diagonal of a parallelogram is given by the vector sum of its adjacent sides:

$$\vec d = \vec a + \vec b.$$

Hence

$$\vec d = (2+1)\hat i + (-4-2)\hat j + (5-3)\hat k = 3\hat i - 6\hat j + 2\hat k.$$

The magnitude of this diagonal is

$$|\vec d| = \sqrt{3^{2} + (-6)^{2} + 2^{2}} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7.$$

Therefore, the required unit vector is

$$\hat u = \dfrac{\vec d}{|\vec d|} = \dfrac{3\hat i - 6\hat j + 2\hat k}{7}.$$

2. Area of the parallelogram

The area is the magnitude of the cross-product of the two adjacent side vectors:

$$\text{Area} = |\vec a \times \vec b|.$$

Compute the cross product:

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix}$$

$$= \hat i\bigl((-4)(-3) - (5)(-2)\bigr) \, - \, \hat j\bigl(2(-3) - 5(1)\bigr) \, + \, \hat k\bigl(2(-2) - (-4)(1)\bigr).$$

Simplifying each component:

• $$\hat i:\; 12 - (-10) = 22$$
• $$-\hat j:\; -6 - 5 = -11 \;\Rightarrow\; +11\hat j$$
• $$\hat k:\; -4 - (-4) = 0$$

Thus

$$\vec a \times \vec b = 22\hat i + 11\hat j + 0\hat k.$$

Its magnitude is

$$|\vec a \times \vec b| = \sqrt{22^{2} + 11^{2}} = \sqrt{484 + 121} = \sqrt{605}.$$

Because $$605 = 121\times5,$$ we obtain

$$\sqrt{605} = 11\sqrt{5}.$$

Hence, the area of the parallelogram is $$11\sqrt{5}$$ square units.

Let the vector $$\vec v$$ make equal angles with the coordinate axes OX, OY, OZ. Denote the common angle by $$\theta$$.

By definition the direction-cosines are

$$l = \cos\theta,\qquad m = \cos\theta,\qquad n = \cos\theta,$$

so that

$$l = m = n = \cos\theta.$$

For every vector the direction-cosines satisfy the fundamental relation

$$l^{2}+m^{2}+n^{2}=1.$$

Substituting $$l=m=n=\cos\theta$$:

$$3\cos^{2}\theta = 1$$

$$\Longrightarrow\;\;\cos^{2}\theta = \dfrac{1}{3}$$

$$\Longrightarrow\;\;\cos\theta = \pm\dfrac{1}{\sqrt{3}}.$$

Hence

$$l = m = n = \pm\dfrac{1}{\sqrt{3}}.$$

Therefore the direction cosines of a vector that is equally inclined to the three coordinate axes are

$$\left(\pm\dfrac{1}{\sqrt{3}},\;\pm\dfrac{1}{\sqrt{3}},\;\pm\dfrac{1}{\sqrt{3}}\right)$$

(all three signs identical — positive for the vector pointing towards the first octant, negative for the opposite direction).

We are given

$$\vec a = \hat i + 4\hat j + 2\hat k, \;\; \vec b = 3\hat i - 2\hat j + 7\hat k, \;\; \vec c = 2\hat i - \hat j + 4\hat k.$$

We need a vector $$\vec d$$ such that

• $$\vec d$$ is perpendicular to both $$\vec a$$ and $$\vec b$$, i.e. $$\vec d \perp \vec a$$ and $$\vec d \perp \vec b$$;
• $$\vec c \cdot \vec d = 15.$$

Step 1 Find a direction perpendicular to both $$\vec a$$ and $$\vec b$$.

The cross product $$\vec a \times \vec b$$ is perpendicular to both $$\vec a$$ and $$\vec b$$, so compute it.

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix}$$

$$= \hat i\,(4\cdot7-2\cdot(-2)) - \hat j\,(1\cdot7-2\cdot3) + \hat k\,(1\cdot(-2)-4\cdot3)$$

$$= \hat i\,(28+4) - \hat j\,(7-6) + \hat k\,(-2-12)$$

$$= 32\hat i - \hat j - 14\hat k.$$

Thus any vector perpendicular to both $$\vec a$$ and $$\vec b$$ is a scalar multiple of this cross product. Write

$$\vec d = t\,(32\hat i - \hat j - 14\hat k), \; t \in \mathbb R.$$

Step 2 Use the dot-product condition $$\vec c \cdot \vec d = 15$$.

Compute $$\vec c \cdot \vec d$$:

$$\vec c \cdot \vec d = (2\hat i - \hat j + 4\hat k) \cdot t\,(32\hat i - \hat j - 14\hat k)$$

$$= t\,[2\cdot32 + (-1)\cdot(-1) + 4\cdot(-14)]$$

$$= t\,[64 + 1 - 56] = t\,(9).$$

Given $$\vec c \cdot \vec d = 15$$, we have $$9t = 15 \;\Rightarrow\; t = \dfrac{15}{9} = \dfrac{5}{3}.$$

Step 3 Write the required vector $$\vec d$$.

Substitute $$t = \dfrac{5}{3}$$ into $$\vec d = t\,(32\hat i - \hat j - 14\hat k)$$:

$$\vec d = \frac{5}{3}\,(32\hat i - \hat j - 14\hat k)$$

$$= \frac{160}{3}\hat i - \frac{5}{3}\hat j - \frac{70}{3}\hat k.$$

Hence, one vector satisfying all the given conditions is

$$\boxed{\displaystyle \vec d = \frac{160}{3}\,\hat i - \frac{5}{3}\,\hat j - \frac{70}{3}\,\hat k}.$$

Let

$$\vec{a}=\hat{i}+\hat{j}+\hat{k},\qquad \vec{b}=2\hat{i}+4\hat{j}-5\hat{k},\qquad \vec{c}=\lambda\hat{i}+2\hat{j}+3\hat{k}.$$

The vector along whose direction we need the unit vector is the sum

$$\vec{s}=\vec{b}+\vec{c}=(2+\lambda)\hat{i}+(4+2)\hat{j}+(-5+3)\hat{k}=(\lambda+2)\hat{i}+6\hat{j}-2\hat{k}.$$

Its magnitude is

$$|\vec{s}|=\sqrt{(\lambda+2)^2+6^2+(-2)^2}=\sqrt{(\lambda+2)^2+40}.$$

Therefore the required unit vector is

$$\hat{u}=\dfrac{\vec{s}}{|\vec{s}|}=\dfrac{(\lambda+2)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(\lambda+2)^2+40}}.$$

The condition given in the question is that the scalar (dot) product of $$\vec{a}$$ with $$\hat{u}$$ equals 1:

$$\vec{a}\cdot\hat{u}=1.$$

Compute the dot product:

$$\vec{a}\cdot\hat{u}=\dfrac{\vec{a}\cdot\vec{s}}{|\vec{s}|}.$$

Now

$$\vec{a}\cdot\vec{s}=(1)(\lambda+2)+(1)(6)+(1)(-2)=\lambda+2+6-2=\lambda+6.$$

Hence

$$\dfrac{\lambda+6}{\sqrt{(\lambda+2)^2+40}}=1.$$

Cross–multiply and square:

$$\lambda+6=\sqrt{(\lambda+2)^2+40}\;\;\Longrightarrow\;\;(\lambda+6)^2=(\lambda+2)^2+40.$$

Expand both sides:

$$\lambda^2+12\lambda+36=\lambda^2+4\lambda+4+40.$$

Simplify:

$$\lambda^2+12\lambda+36=\lambda^2+4\lambda+44\;\;\Longrightarrow\;\;12\lambda+36=4\lambda+44.$$

Bring like terms together:

$$12\lambda-4\lambda=44-36\;\;\Longrightarrow\;\;8\lambda=8.$$

Therefore

$$\lambda=1.$$

Since $$\lambda+6=7>0$$, the original (unsquared) condition is satisfied. Hence the unique value of $$\lambda$$ is 1.

Let the magnitudes of the three given vectors be equal to a common value, say $$m$$; that is

$$|\vec a| = |\vec b| = |\vec c| = m.$$

Because the vectors are mutually perpendicular, their pair-wise scalar products vanish:

$$\vec a\cdot\vec b = \vec b\cdot\vec c = \vec c\cdot\vec a = 0.$$

Form the vector

$$\vec r = \vec a + \vec b + \vec c.$$

Our task is to compare the angles that $$\vec r$$ makes with $$\vec a,\;\vec b$$ and $$\vec c$$. For a generic vector $$\vec p$$ the cosine of the angle $$\theta_p$$ between $$\vec r$$ and $$\vec p$$ is given by

$$\cos\theta_p = \dfrac{\vec r\cdot\vec p}{|\vec r|\,|\vec p|}.$$

1. Scalar products with the three given vectors

(i) With $$\vec a$$:

$$\vec r\cdot\vec a = (\vec a + \vec b + \vec c)\cdot\vec a = \vec a\cdot\vec a + \vec b\cdot\vec a + \vec c\cdot\vec a = m^{2} + 0 + 0 = m^{2}.$$

(ii) With $$\vec b$$:

$$\vec r\cdot\vec b = (\vec a + \vec b + \vec c)\cdot\vec b = \vec a\cdot\vec b + \vec b\cdot\vec b + \vec c\cdot\vec b = 0 + m^{2} + 0 = m^{2}.$$

(iii) With $$\vec c$$:

$$\vec r\cdot\vec c = (\vec a + \vec b + \vec c)\cdot\vec c = \vec a\cdot\vec c + \vec b\cdot\vec c + \vec c\cdot\vec c = 0 + 0 + m^{2} = m^{2}.$$

Thus each dot-product numerator is the same value $$m^{2}.$$

2. Magnitude of $$\vec r$$

$$|\vec r|^{2} = (\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c)$$

$$= \vec a\cdot\vec a + \vec b\cdot\vec b + \vec c\cdot\vec c + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a)$$

$$= m^{2} + m^{2} + m^{2} + 2(0 + 0 + 0) = 3m^{2}.$$

Hence $$|\vec r| = \sqrt{3}\,m.$$

3. Cosines of the three angles

(a) With $$\vec a$$:

$$\cos\theta_a = \dfrac{m^{2}}{|\vec r|\,m} = \dfrac{m^{2}}{(\sqrt{3}\,m)\,m} = \dfrac{1}{\sqrt{3}}.$$

(b) With $$\vec b$$:

$$\cos\theta_b = \dfrac{m^{2}}{|\vec r|\,m} = \dfrac{1}{\sqrt{3}}.$$

(c) With $$\vec c$$:

$$\cos\theta_c = \dfrac{m^{2}}{|\vec r|\,m} = \dfrac{1}{\sqrt{3}}.$$

4. Conclusion

The cosines of the angles that $$\vec r = \vec a + \vec b + \vec c$$ makes with $$\vec a,\;\vec b$$ and $$\vec c$$ are identical. Therefore the three angles themselves are equal. Hence, the vector $$\vec a + \vec b + \vec c$$ is equally inclined to each of the mutually perpendicular, equal-magnitude vectors $$\vec a,\;\vec b$$ and $$\vec c$$.

In fact, each angle equals $$\cos^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right) \approx 54.7^{\circ}.$$

Given non-zero vectors $$\vec a,\,\vec b\neq\vec 0$$.

Recall the following facts for any vectors $$\vec u,\vec v$$:

• The dot product is commutative: $$\vec u\cdot\vec v = \vec v\cdot\vec u.$$
• Magnitude (length) is defined by $$|\vec u|=\sqrt{\vec u\cdot\vec u},$$ so $$|\vec u|^{2}=\vec u\cdot\vec u.$$
• Two non-zero vectors are perpendicular (orthogonal) if and only if their dot product is zero: $$\vec u\perp\vec v \iff \vec u\cdot\vec v = 0.$$

We have to prove the bi-conditional statement

$$(\vec a + \vec b)\cdot(\vec a + \vec b)=|\vec a|^{2}+|\vec b|^{2} \iff \vec a\perp\vec b.$$

Because a bi-conditional demands both directions, we treat them one by one.

(1) If $$\vec a\perp\vec b$$, then $$(\vec a + \vec b)\cdot(\vec a + \vec b)=|\vec a|^{2}+|\vec b|^{2}.$$

When $$\vec a\perp\vec b$$, we have $$\vec a\cdot\vec b=0.$$ Now expand the left-hand side:

$$(\vec a+\vec b)\cdot(\vec a+\vec b) = \vec a\cdot\vec a+\vec a\cdot\vec b+\vec b\cdot\vec a+\vec b\cdot\vec b$$

$$= |\vec a|^{2}+0+0+|\vec b|^{2} = |\vec a|^{2}+|\vec b|^{2}.$$

This completes the first implication.

(2) If $$(\vec a + \vec b)\cdot(\vec a + \vec b)=|\vec a|^{2}+|\vec b|^{2},$$ then $$\vec a\perp\vec b.$$

Start again from the general expansion:

$$(\vec a+\vec b)\cdot(\vec a+\vec b) = \vec a\cdot\vec a+\vec a\cdot\vec b+\vec b\cdot\vec a+\vec b\cdot\vec b$$

$$= |\vec a|^{2}+2\,(\vec a\cdot\vec b)+|\vec b|^{2}.$$

By hypothesis, this equals $$|\vec a|^{2}+|\vec b|^{2}.$$ Hence

$$|\vec a|^{2}+2\,(\vec a\cdot\vec b)+|\vec b|^{2}=|\vec a|^{2}+|\vec b|^{2}.$$

Subtracting $$|\vec a|^{2}+|\vec b|^{2}$$ from both sides gives

$$2\,(\vec a\cdot\vec b)=0 \quad\Longrightarrow\quad \vec a\cdot\vec b=0.$$

Since both vectors are non-zero, $$\vec a\cdot\vec b=0$$ exactly means $$\vec a\perp\vec b.$$

Conclusion. Both directions are proved, therefore

$$(\vec a + \vec b)\cdot(\vec a + \vec b)=|\vec a|^{2}+|\vec b|^{2} \iff \vec a\perp\vec b,$$

provided $$\vec a\neq\vec 0$$ and $$\vec b\neq\vec 0.$$ The statement is thus established.

The angle $$\theta$$ between two non-zero vectors is defined so that $$0 \leq \theta \leq \pi$$.

Using the definition of the scalar (dot) product,

$$\vec a \cdot \vec b = |\vec a|\,|\vec b|\cos\theta.$$

Because magnitudes are always positive, the sign of $$\vec a \cdot \vec b$$ is determined solely by $$\cos\theta$$:

• If $$\cos\theta>0$$ then $$\vec a \cdot \vec b>0.$$
• If $$\cos\theta=0$$ then $$\vec a \cdot \vec b=0.$$
• If $$\cos\theta<0$$ then $$\vec a \cdot \vec b<0.$$

The cosine function is non-negative exactly when

$$0 \leq \theta \leq \frac{\pi}{2}.$$

Hence $$\vec a \cdot \vec b \ge 0$$ only for angles in the closed interval $$[0,\;\pi/2]$$. Among the given choices, this corresponds to option (B).

Let $$|\vec a| = |\vec b| = 1$$ and let $$\theta$$ be the angle between them, so that

$$\vec a \cdot \vec b = |\vec a|\,|\vec b|\,\cos\theta = \cos\theta.$$

The squared magnitude of $$\vec a + \vec b$$ is

$$|\vec a + \vec b|^{2} = (\vec a + \vec b) \cdot (\vec a + \vec b)$$

$$= \vec a \cdot \vec a + \vec b \cdot \vec b + 2\,\vec a \cdot \vec b$$

$$= |\vec a|^{2} + |\vec b|^{2} + 2\cos\theta$$

$$= 1 + 1 + 2\cos\theta = 2(1 + \cos\theta).$$

For $$\vec a + \vec b$$ to be a unit vector, we need $$|\vec a + \vec b| = 1,$$ hence

$$|\vec a + \vec b|^{2} = 1 \;\Rightarrow\; 2(1 + \cos\theta) = 1.$$

Simplifying,

$$1 + \cos\theta = \tfrac{1}{2} \;\Rightarrow\; \cos\theta = -\tfrac{1}{2}.$$

In the interval $$0 \le \theta \le \pi$$, the value $$\cos\theta = -\tfrac{1}{2}$$ corresponds to

$$\theta = \dfrac{2\pi}{3}.$$

Therefore $$\vec a + \vec b$$ is a unit vector only when $$\theta = \dfrac{2\pi}{3},$$ i.e. option (D).

Let us evaluate each term separately, using the standard right-handed orthonormal basis $$\hat{i},\;\hat{j},\;\hat{k}$$ for which the well-known cross-products are

• $$\hat{i}\times\hat{j}=\hat{k}$$
• $$\hat{j}\times\hat{k}=\hat{i}$$
• $$\hat{k}\times\hat{i}=\hat{j}$$

(The above immediately gives the negatives when the order of the vectors is interchanged.)

1. First term
   $$\hat{i}\cdot(\hat{j}\times\hat{k})$$
   Since $$\hat{j}\times\hat{k}=\hat{i}$$, we have
   $$\hat{i}\cdot(\hat{j}\times\hat{k})=\hat{i}\cdot\hat{i}=1.$$
2. Second term
   $$\hat{j}\cdot(\hat{i}\times\hat{k})$$
   From $$\hat{i}\times\hat{k}= -\hat{j}$$ (because $$\hat{k}\times\hat{i}=\hat{j}$$), we get
   $$\hat{j}\cdot(\hat{i}\times\hat{k})=\hat{j}\cdot(-\hat{j})=-1.$$
3. Third term
   $$\hat{k}\cdot(\hat{i}\times\hat{j})$$
   As $$\hat{i}\times\hat{j}=\hat{k}$$, it follows that
   $$\hat{k}\cdot(\hat{i}\times\hat{j})=\hat{k}\cdot\hat{k}=1.$$

Adding the three obtained values:

$$1+(-1)+1=1.$$

Hence the given expression equals $$1$$.

Therefore, the correct option is (C).

Let $$\theta$$ be the angle between the non-zero vectors $$\vec a$$ and $$\vec b$$.

• The magnitude of their dot product is $$\bigl|\vec a \cdot \vec b\bigr| = |\vec a|\,|\vec b|\,|\cos\theta|.$$
• The magnitude of their cross product is $$\bigl|\vec a \times \vec b\bigr| = |\vec a|\,|\vec b|\,|\sin\theta|.$$

The given condition is

$$|\vec a \cdot \vec b| = |\vec a \times \vec b|.$$

Substituting the above expressions,

$$|\vec a|\,|\vec b|\,|\cos\theta| = |\vec a|\,|\vec b|\,|\sin\theta|.$$

Because $$|\vec a| \ne 0$$ and $$|\vec b| \ne 0$$, we can divide by $$|\vec a|\,|\vec b|$$ to get

$$|\cos\theta| = |\sin\theta|.$$

On the interval $$0 \le \theta \le \pi$$ this equality is satisfied for

$$\theta = \dfrac{\pi}{4}\quad\text{or}\quad\theta = \dfrac{3\pi}{4}.$$

Among the given options, only $$\theta = \dfrac{\pi}{4}$$ appears.

Hence the correct choice is (B).