Chapter 11: Three Dimensional Geometry

The direction cosines $$l, m, n$$ of a line are the cosines of the angles the line makes with the positive directions of the $$x$$, $$y$$ and $$z$$-axes respectively.

Here the line makes angles $$90^\circ$$, $$60^\circ$$ and $$30^\circ$$ with the $$x$$, $$y$$, $$z$$-axes, so

$$l = \cos 90^\circ = 0$$

$$m = \cos 60^\circ = \dfrac{1}{2}$$

$$n = \cos 30^\circ = \dfrac{\sqrt{3}}{2}$$

Hence the direction cosines of the line are $$0,\ \dfrac{1}{2},\ \dfrac{\sqrt{3}}{2}$$.

If $$a, b, c$$ are the direction ratios of a line, its direction cosines are $$\dfrac{a}{\sqrt{a^2+b^2+c^2}},\ \dfrac{b}{\sqrt{a^2+b^2+c^2}},\ \dfrac{c}{\sqrt{a^2+b^2+c^2}}$$.

Here $$a = 2,\ b = -1,\ c = -2$$, so

$$\sqrt{a^2+b^2+c^2} = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$$

Therefore the direction cosines are

$$\dfrac{2}{3},\ \dfrac{-1}{3},\ \dfrac{-2}{3}$$

The direction ratios of the line through two points $$P(x_1,y_1,z_1)$$ and $$Q(x_2,y_2,z_2)$$ are $$x_2-x_1,\ y_2-y_1,\ z_2-z_1$$.

Here $$P(-2,4,-5)$$ and $$Q(1,2,3)$$, so the direction ratios are

$$1-(-2),\ 2-4,\ 3-(-5)\ \Rightarrow\ 3,\ -2,\ 8$$

The magnitude is

$$\sqrt{3^2 + (-2)^2 + 8^2} = \sqrt{9+4+64} = \sqrt{77}$$

Hence the direction cosines are

$$\dfrac{3}{\sqrt{77}},\ \dfrac{-2}{\sqrt{77}},\ \dfrac{8}{\sqrt{77}}$$

The $$x$$-axis makes angles $$0^\circ$$, $$90^\circ$$, $$90^\circ$$ with the $$x$$, $$y$$, $$z$$-axes respectively. So its direction cosines are

$$\cos 0^\circ,\ \cos 90^\circ,\ \cos 90^\circ \ \Rightarrow\ 1,\ 0,\ 0$$

The $$y$$-axis makes angles $$90^\circ$$, $$0^\circ$$, $$90^\circ$$ with the axes, so its direction cosines are

$$\cos 90^\circ,\ \cos 0^\circ,\ \cos 90^\circ \ \Rightarrow\ 0,\ 1,\ 0$$

The $$z$$-axis makes angles $$90^\circ$$, $$90^\circ$$, $$0^\circ$$ with the axes, so its direction cosines are

$$\cos 90^\circ,\ \cos 90^\circ,\ \cos 0^\circ \ \Rightarrow\ 0,\ 0,\ 1$$

Three points are collinear if the line through any two of them is parallel to the line through another pair sharing a common point.

Direction ratios of $$AB$$ (from $$A(2,3,-4)$$ to $$B(1,-2,3)$$):

$$1-2,\ -2-3,\ 3-(-4)\ \Rightarrow\ -1,\ -5,\ 7$$

Direction ratios of $$BC$$ (from $$B(1,-2,3)$$ to $$C(3,8,-11)$$):

$$3-1,\ 8-(-2),\ -11-3\ \Rightarrow\ 2,\ 10,\ -14$$

We observe that

$$(2,\ 10,\ -14) = -2(-1,\ -5,\ 7)$$

So the direction ratios of $$BC$$ are proportional to those of $$AB$$; hence $$AB$$ is parallel to $$BC$$. Since $$B$$ is a common point, the points $$A$$, $$B$$, $$C$$ are collinear.

Given data

Point on the line: $$(5,2,-4)$$     Direction vector: $$3\hat i + 2\hat j - 8\hat k$$.

Step 1 – Position vector of the given point

The position vector is $$\vec a = 5\hat i + 2\hat j - 4\hat k$$.

Step 2 – Direction vector of the line

Since the line is parallel to $$3\hat i + 2\hat j - 8\hat k$$, take $$\vec b = 3\hat i + 2\hat j - 8\hat k$$ as its direction vector.

Step 3 – Vector equation of the line

The general vector form is $$\vec r = \vec a + \lambda \vec b,\; \lambda \in \mathbb R$$. Substituting $$\vec a$$ and $$\vec b$$:

$$\vec r = 5\hat i + 2\hat j - 4\hat k + \lambda(3\hat i + 2\hat j - 8\hat k),\; \lambda \in \mathbb R$$.

Step 4 – Component form

• $$x = 5 + 3\lambda$$
• $$y = 2 + 2\lambda$$
• $$z = -4 - 8\lambda$$

Step 5 – Cartesian equation

Eliminating $$\lambda$$:

$$\dfrac{x-5}{3} = \dfrac{y-2}{2} = \dfrac{z+4}{-8}$$.

Hence both required forms are obtained.

Step 1 – Identify direction vectors

The constant multiples of the parameters give the direction vectors of the two lines:

• For Line 1: $$\vec d_1 = \hat i + 2\hat j + 2\hat k$$
• For Line 2: $$\vec d_2 = 3\hat i + 2\hat j + 6\hat k$$

Step 2 – Compute the dot product

$$\vec d_1 \cdot \vec d_2 = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$$

Step 3 – Find the magnitudes of the two vectors

Magnitude of $$\vec d_1$$:
$$|\vec d_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt9 = 3$$

Magnitude of $$\vec d_2$$:
$$|\vec d_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$

Step 4 – Apply the cosine formula

If $$\theta$$ is the angle between the two lines, then

$$\cos \theta = \frac{\vec d_1 \cdot \vec d_2}{|\vec d_1|\,|\vec d_2|} = \frac{19}{3 \times 7} = \frac{19}{21}$$

Step 5 – Express the angle

$$\theta = \cos^{-1}\left(\frac{19}{21}\right)$$

For a numerical value, $$\theta \approx 24.6^{\circ}$$ (rounded to one decimal place).

The two given straight lines are expressed in symmetric form:

1. $$\dfrac{x+3}{3}=\dfrac{y-1}{5}=\dfrac{z+3}{4}$$
2. $$\dfrac{x+1}{1}=\dfrac{y-4}{1}=\dfrac{z-5}{2}$$

For a line written as $$\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$$ the numbers $$a,\,b,\,c$$ serve as a set of direction ratios (d.r.s) of that line.

Hence the required direction ratios are

• First line: $$a_1=3,\;b_1=5,\;c_1=4$$
• Second line: $$a_2=1,\;b_2=1,\;c_2=2$$

If $$\theta$$ is the acute angle between two lines having direction ratios $$(a_1,b_1,c_1)$$ and $$(a_2,b_2,c_2)$$, then

$$\cos\theta=\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}.$$

Compute the numerator (dot product):
$$a_1a_2+b_1b_2+c_1c_2=3\times1+5\times1+4\times2=3+5+8=16.$$

Compute the magnitudes of the two direction-ratio vectors:

$$\begin{aligned}\sqrt{a_1^2+b_1^2+c_1^2}&=\sqrt{3^2+5^2+4^2}=\sqrt{9+25+16}=\sqrt{50}=5\sqrt2,\\[4pt]\sqrt{a_2^2+b_2^2+c_2^2}&=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt6.\end{aligned}$$

Therefore

$$\begin{aligned}\cos\theta&=\dfrac{16}{(5\sqrt2)(\sqrt6)}\\&=\dfrac{16}{\sqrt{50\times6}}=\dfrac{16}{\sqrt{300}}\\&=\dfrac{16}{10\sqrt3}=\dfrac{8}{5\sqrt3}.\end{aligned}$$

As the cosine value is positive, $$\theta$$ itself is acute:

$$\theta=\cos^{-1}\!\left(\dfrac{8}{5\sqrt3}\right).$$

Numerically, $$\theta\approx22.8^{\circ}$$ (correct to one decimal place).

Step 1 · Identify fixed points and direction vectors

For the first line $$l_1$$

$$\vec r = \hat i + \hat j + \lambda\,(2\hat i - \hat j + \hat k)$$

• Take $$\lambda = 0$$ ⇒ fixed point $$A(1,1,0)$$
• Direction vector $$\vec a = 2\hat i - \hat j + \hat k \;(\,2,-1,1\,)$$

For the second line $$l_2$$

$$\vec r = 2\hat i + \hat j - \hat k + \mu\,(3\hat i - 5\hat j + 2\hat k)$$

• Take $$\mu = 0$$ ⇒ fixed point $$B(2,1,-1)$$
• Direction vector $$\vec b = 3\hat i - 5\hat j + 2\hat k \;(\,3,-5,2\,)$$

Step 2 · Form the vector joining the chosen points

$$\vec{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2-1)\hat i + (1-1)\hat j + (-1-0)\hat k = \hat i - \hat k \;(\,1,0,-1\,)$$

Step 3 · Compute the cross product of the direction vectors

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -1 & 1\\ 3 & -5 & 2 \end{vmatrix} = 3\hat i - \hat j - 7\hat k \;(\,3,-1,-7\,)$$

Step 4 · Magnitude of the cross product

$$|\vec a \times \vec b| = \sqrt{3^{2}+(-1)^{2}+(-7)^{2}} = \sqrt{9+1+49} = \sqrt{59}$$

Step 5 · Scalar triple product

$$\vec{AB}\,\cdot\,(\vec a \times \vec b) = (1,0,-1)\cdot(3,-1,-7) = 1\times3 + 0\times(-1) + (-1)\times(-7) = 3+7 = 10$$

Step 6 · Shortest distance formula

The shortest (perpendicular) distance $$d$$ between two skew lines is

$$d = \frac{|\vec{AB}\,\cdot\,(\vec a \times \vec b)|}{|\vec a \times \vec b|} = \frac{|10|}{\sqrt{59}} = \frac{10}{\sqrt{59}}$$

Hence the required shortest distance is $$\dfrac{10}{\sqrt{59}}$$ units.

For two lines written in vector form as

$$\vec r = \vec a_1 + \lambda\,\vec b \quad\text{and}\quad \vec r = \vec a_2 + \mu\,\vec b,$$

if they share the same direction vector $$\vec b$$ (hence are parallel), the shortest distance between them is

$$d = \dfrac{\lVert (\vec a_2-\vec a_1) \times \vec b \rVert}{\lVert\vec b\rVert}.$$

Identify the fixed (position) vectors and the common direction vector from the given equations:

• First line: $$\vec a_1 = \hat i + 2\hat j - 4\hat k = \langle 1,2,-4 \rangle$$
• Second line: $$\vec a_2 = 3\hat i + 3\hat j - 5\hat k = \langle 3,3,-5 \rangle$$
• Common direction: $$\vec b = 2\hat i + 3\hat j + 6\hat k = \langle 2,3,6 \rangle$$

Step 1 – Vector joining points on the two lines:

$$\vec{a_2}-\vec{a_1}=\langle 3-1,\,3-2,\,-5-(-4)\rangle=\langle 2,1,-1\rangle.$$

Step 2 – Cross product with the direction vector:

$$\begin{aligned} (\vec{a_2}-\vec{a_1}) \times \vec b &= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & -1\\ 2 & 3 & 6 \end{vmatrix} \\ &= \hat i\,(1\cdot6-(-1)\cdot3) - \hat j\,(2\cdot6-(-1)\cdot2) + \hat k\,(2\cdot3-1\cdot2) \\ &= \hat i\,9 - \hat j\,14 + \hat k\,4 \\ &= \langle 9,-14,4 \rangle. \end{aligned}$$

Step 3 – Magnitudes:

$$\lVert (\vec{a_2}-\vec{a_1}) \times \vec b \rVert = \sqrt{9^2+(-14)^2+4^2}=\sqrt{81+196+16}=\sqrt{293}.$$

$$\lVert\vec b\rVert = \sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7.$$

Step 4 – Distance:

$$d = \dfrac{\sqrt{293}}{7}.$$

Hence the perpendicular distance between the given parallel lines is $$\dfrac{\sqrt{293}}{7}$$ units.

Let the angles made by the line with the positive coordinate axes be

$$\alpha = 90^{\circ},\; \beta = 135^{\circ},\; \gamma = 45^{\circ}.$$

By definition, the direction cosines $$l, m, n$$ of the line are the cosines of these angles.

• $$l = \cos \alpha = \cos 90^{\circ} = 0$$
• $$m = \cos \beta = \cos 135^{\circ} = -\dfrac{1}{\sqrt 2}$$
• $$n = \cos \gamma = \cos 45^{\circ} = \dfrac{1}{\sqrt 2}$$

Verification of the fundamental relation:

$$l^{2}+m^{2}+n^{2}=0^{2}+\left(-\dfrac{1}{\sqrt 2}\right)^{2}+\left(\dfrac{1}{\sqrt 2}\right)^{2}=0+\dfrac12+\dfrac12=1,$$

so the obtained values are consistent.

Therefore, the direction cosines of the given line are
$$\boxed{\bigl(\,l,\,m,\,n\bigr)=\left(0,-\dfrac{1}{\sqrt 2},\dfrac{1}{\sqrt 2}\right)}.$$

Let the given line make angles $$\alpha, \beta, \gamma$$ with the $$x$$-, $$y$$- and $$z$$-axes respectively. By definition, its direction cosines are

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

Because the line is said to make equal angles with the three coordinate axes, we have

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

Hence

$$l = m = n.$$

Direction cosines are always constrained by

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

Substituting $$l = m = n$$:

$$l^{2}+l^{2}+l^{2}=1 \;\;\Longrightarrow\;\; 3l^{2}=1.$$

Solving for $$l$$ gives

$$l = \pm\frac{1}{\sqrt{3}}.$$

Therefore

$$l = m = n = \frac{1}{\sqrt{3}}$$ for one sense (direction) of the line, and

$$l = m = n = -\frac{1}{\sqrt{3}}$$ for the opposite sense.

Thus the required direction cosines are

$$\bigl(\tfrac{1}{\sqrt{3}},\;\tfrac{1}{\sqrt{3}},\;\tfrac{1}{\sqrt{3}}\bigr)$$ (or all three negative together).

Step 1 · Recall the relation between direction ratios and direction cosines
If a line has direction ratios (d.r.) $$a, b, c$$, then its direction cosines (d.c.) $$l,m,n$$ are obtained by dividing each ratio by the magnitude of the vector $$\vec{d}=a\hat i+b\hat j+c\hat k$$.
Thus $$l = \dfrac{a}{\|\vec d\|}, \; m = \dfrac{b}{\|\vec d\|}, \; n = \dfrac{c}{\|\vec d\|}$$ where $$\|\vec d\| = \sqrt{a^{2}+b^{2}+c^{2}}$$.

Step 2 · Identify the given direction ratios
For the line we are given
$$a = -18, \; b = 12, \; c = -4$$.

Step 3 · Compute the magnitude of the direction-ratio vector
$$\|\vec d\| = \sqrt{(-18)^{2}+12^{2}+(-4)^{2}} = \sqrt{324+144+16} = \sqrt{484} = 22.$$

Step 4 · Divide each direction ratio by the magnitude

• $$l = \dfrac{-18}{22} = -\dfrac{9}{11}$$
• $$m = \dfrac{12}{22} = \dfrac{6}{11}$$
• $$n = \dfrac{-4}{22} = -\dfrac{2}{11}$$

Step 5 · Verify the result (optional check)
$$l^{2}+m^{2}+n^{2} = \left(-\dfrac{9}{11}\right)^{2}+\left(\dfrac{6}{11}\right)^{2}+\left(-\dfrac{2}{11}\right)^{2} = \dfrac{81+36+4}{121}=1,$$ confirming they are valid direction cosines.

Hence, the direction cosines of the line are $$l = -\dfrac{9}{11},\; m = \dfrac{6}{11},\; n = -\dfrac{2}{11}$$.

Let the points be $$A(2,3,4)$$, $$B(-1,-2,1)$$ and $$C(5,8,7)$$.

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

$$\vec{OA}=2\hat i+3\hat j+4\hat k, \; \vec{OB}=-\hat i-2\hat j+\hat k, \; \vec{OC}=5\hat i+8\hat j+7\hat k.$$

Find the vector $$\vec{AB}$$:

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

Find the vector $$\vec{AC}$$:

$$\vec{AC}=\vec{OC}-\vec{OA}=(5-2)\hat i+(8-3)\hat j+(7-4)\hat k=3\hat i+5\hat j+3\hat k.$$

Observe that

$$\vec{AC} = -1\times \vec{AB}.$$

Because $$\vec{AB}$$ and $$\vec{AC}$$ are non-zero scalar multiples of one another, they are parallel. Hence the points A, B and C all lie on the same straight line; that is, they are collinear.

Thus, the points $$(2,3,4)$$, $$(-1,-2,1)$$ and $$(5,8,7)$$ are collinear.

Let the vertices of the triangle be denoted by
$$A(3,5,-4), \; B(-1,1,2), \; C(-5,-5,-2).$$

To obtain the direction cosines of any side we proceed in two steps:

1. Form the vector representing that side (direction ratios).
2. Divide each direction ratio by the magnitude of the vector to get the required direction cosines \((l,m,n)\).

Throughout, recall that for a vector with direction ratios $$(a,b,c)$$ its magnitude is $$|\vec r|=\sqrt{a^{2}+b^{2}+c^{2}}$$ and its direction cosines are $$l=\dfrac a{|\vec r|},\; m=\dfrac b{|\vec r|},\; n=\dfrac c{|\vec r|}.$$

1. Side \(\overrightarrow{AB}\)

Direction ratios:

$$\overrightarrow{AB}=B-A=(-1-3,\;1-5,\;2-(-4))=(-4,-4,6).$$

Magnitude:

$$|\overrightarrow{AB}|=\sqrt{(-4)^{2}+(-4)^{2}+6^{2}}=\sqrt{16+16+36}=\sqrt{68}=2\sqrt{17}.$$

Direction cosines:

$$l_{AB}=\dfrac{-4}{2\sqrt{17}}=-\dfrac{2}{\sqrt{17}},\qquad m_{AB}=\dfrac{-4}{2\sqrt{17}}=-\dfrac{2}{\sqrt{17}},\qquad n_{AB}=\dfrac{6}{2\sqrt{17}}=\dfrac{3}{\sqrt{17}}.$$

2. Side \(\overrightarrow{BC}\)

Direction ratios:

$$\overrightarrow{BC}=C-B=(-5-(-1),\;-5-1,\;-2-2)=(-4,-6,-4).$$

Magnitude:

$$|\overrightarrow{BC}|=\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}=\sqrt{16+36+16}=\sqrt{68}=2\sqrt{17}.$$

Direction cosines:

$$l_{BC}=\dfrac{-4}{2\sqrt{17}}=-\dfrac{2}{\sqrt{17}},\qquad m_{BC}=\dfrac{-6}{2\sqrt{17}}=-\dfrac{3}{\sqrt{17}},\qquad n_{BC}=\dfrac{-4}{2\sqrt{17}}=-\dfrac{2}{\sqrt{17}}.$$

3. Side \(\overrightarrow{CA}\)

Direction ratios:

$$\overrightarrow{CA}=A-C=(3-(-5),\;5-(-5),\;-4-(-2))=(8,10,-2).$$

Magnitude:

$$|\overrightarrow{CA}|=\sqrt{8^{2}+10^{2}+(-2)^{2}}=\sqrt{64+100+4}=\sqrt{168}=2\sqrt{42}.$$

Direction cosines:

$$l_{CA}=\dfrac{8}{2\sqrt{42}}=\dfrac{4}{\sqrt{42}},\qquad m_{CA}=\dfrac{10}{2\sqrt{42}}=\dfrac{5}{\sqrt{42}},\qquad n_{CA}=\dfrac{-2}{2\sqrt{42}}=-\dfrac{1}{\sqrt{42}}.$$

Hence the direction cosines of the three sides of the triangle are:

• Side AB : $$\bigl(-\dfrac{2}{\sqrt{17}},\; -\dfrac{2}{\sqrt{17}},\; \dfrac{3}{\sqrt{17}}\bigr)$$
• Side BC : $$\bigl(-\dfrac{2}{\sqrt{17}},\; -\dfrac{3}{\sqrt{17}},\; -\dfrac{2}{\sqrt{17}}\bigr)$$
• Side CA : $$\bigl(\dfrac{4}{\sqrt{42}},\; \dfrac{5}{\sqrt{42}},\; -\dfrac{1}{\sqrt{42}}\bigr)$$

These comply with the relation $$l^{2}+m^{2}+n^{2}=1$$ for each side, establishing their correctness.

Let the three given lines have direction–cosine triplets

• $$l_1 = \dfrac{12}{13},\; m_1 = \dfrac{-3}{13},\; n_1 = \dfrac{-4}{13}$$
• $$l_2 = \dfrac{4}{13},\; m_2 = \dfrac{12}{13},\; n_2 = \dfrac{3}{13}$$
• $$l_3 = \dfrac{3}{13},\; m_3 = \dfrac{-4}{13},\; n_3 = \dfrac{12}{13}$$

Each triple is a set of direction cosines, so every one of them represents a unit direction vector.

Two lines are perpendicular ↔ the scalar (dot) product of their direction vectors is zero. We therefore evaluate the three pair-wise dot products.

(i) First and second lines

$$l_1l_2 + m_1m_2 + n_1n_2 = \frac{12}{13}\cdot\frac{4}{13} + \frac{-3}{13}\cdot\frac{12}{13} + \frac{-4}{13}\cdot\frac{3}{13}$$ $$= \frac{48 - 36 - 12}{169} = \frac{0}{169} = 0$$

(ii) Second and third lines

$$l_2l_3 + m_2m_3 + n_2n_3 = \frac{4}{13}\cdot\frac{3}{13} + \frac{12}{13}\cdot\frac{-4}{13} + \frac{3}{13}\cdot\frac{12}{13}$$ $$= \frac{12 - 48 + 36}{169} = \frac{0}{169} = 0$$

(iii) Third and first lines

$$l_3l_1 + m_3m_1 + n_3n_1 = \frac{3}{13}\cdot\frac{12}{13} + \frac{-4}{13}\cdot\frac{-3}{13} + \frac{12}{13}\cdot\frac{-4}{13}$$ $$= \frac{36 + 12 - 48}{169} = \frac{0}{169} = 0$$

All three scalar products are zero. Hence every pair of direction vectors is orthogonal, and therefore the three given lines are mutually perpendicular.

Given points

• Line 1 passes through $$A(1,-1,2)$$ and $$B(3,4,-2)$$
• Line 2 passes through $$C(0,3,2)$$ and $$D(3,5,6)$$

Step 1 : Direction ratios of line 1

The direction ratios are the coordinate differences of $$B$$ and $$A$$:

$$l_1 = 3-1 = 2, \; m_1 = 4-(-1) = 5, \; n_1 = -2-2 = -4$$

Thus a direction vector of line 1 is $$\vec v_1 = 2\hat i + 5\hat j - 4\hat k$$.

Step 2 : Direction ratios of line 2

Using points $$D$$ and $$C$$:

$$l_2 = 3-0 = 3, \; m_2 = 5-3 = 2, \; n_2 = 6-2 = 4$$

Hence a direction vector of line 2 is $$\vec v_2 = 3\hat i + 2\hat j + 4\hat k$$.

Step 3 : Dot product of the two direction vectors

$$\vec v_1 \cdot \vec v_2 = (2)(3) + (5)(2) + (-4)(4) = 6 + 10 - 16 = 0$$

Step 4 : Perpendicularity test

Since the dot product of the direction vectors is zero, the vectors (and therefore the two lines) are perpendicular.

Result

The line through $$(1,-1,2)$$, $$(3,4,-2)$$ is perpendicular to the line through $$(0,3,2)$$, $$(3,5,6)$$.

Step 1 · Direction ratios of the first line

Let the points be $$A(4,7,8)$$ and $$B(2,3,4)$$.

The vector $$\overrightarrow{AB}$$ is obtained by subtracting the coordinates of $$A$$ from those of $$B$$:

$$\overrightarrow{AB} = (2-4,\;3-7,\;4-8) = (-2,-4,-4).$$

Thus the direction ratios (d.r.’s) of the line through $$A$$ and $$B$$ are $$-2,-4,-4$$.

Step 2 · Direction ratios of the second line

Let the points be $$C(-1,-2,1)$$ and $$D(1,2,5)$$.

The vector $$\overrightarrow{CD}$$ is

$$\overrightarrow{CD} = (1-(-1),\;2-(-2),\;5-1) = (2,4,4).$$

Hence the direction ratios of the line through $$C$$ and $$D$$ are $$2,4,4$$.

Step 3 · Comparing direction ratios

Compare the two sets of d.r.’s:

$$\dfrac{-2}{2} = \dfrac{-4}{4} = \dfrac{-4}{4} = -1.$$

All three ratios are equal (each equals $$-1$$). Therefore one set of direction ratios is a constant multiple of the other.

Conclusion

Because the direction ratios of the two lines are proportional, the two lines are parallel.

Given data

• Point through which the line must pass: $$P(1, 2, 3)$$.
• Vector parallel to the required line: $$\vec{d}=3\hat{i}+2\hat{j}-2\hat{k}$$.

Step 1 — Write the direction ratios

The components of the parallel vector give the direction ratios of the line:

$$a = 3, \; b = 2, \; c = -2$$

Step 2 — Vector equation of the line

A line that passes through a fixed position vector $$\vec{r_0}$$ and is parallel to $$\vec{d}$$ is written as

$$\vec{r}=\vec{r_0}+\lambda\,\vec{d},\;\;\lambda\in\mathbb{R}.$$

Here $$\vec{r_0}=\hat{i}+2\hat{j}+3\hat{k}$$ (because the point is (1,2,3)). Hence

$$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(3\hat{i}+2\hat{j}-2\hat{k}).$$

Step 3 — Parametric (component) form

Write each component separately:

• $$x = 1 + 3\lambda$$
• $$y = 2 + 2\lambda$$
• $$z = 3 - 2\lambda$$

These three equations together give the parametric form of the required line.

Step 4 — Symmetric form

Eliminate $$\lambda$$ from the parametric equations. From each component, solve for $$\lambda$$:

• From $$x = 1 + 3\lambda$$, $$\lambda = \dfrac{x-1}{3}$$
• From $$y = 2 + 2\lambda$$, $$\lambda = \dfrac{y-2}{2}$$
• From $$z = 3 - 2\lambda$$, $$\lambda = -\dfrac{z-3}{2}$$

Equating these three expressions for $$\lambda$$ gives the symmetric form:

$$\dfrac{x-1}{3}=\dfrac{y-2}{2}=\dfrac{z-3}{-2}.$$

(The negative sign in the third fraction is often written by moving the minus to the numerator as $$\dfrac{3-z}{2}$$; both are correct.)

Conclusion

The line through $$(1,2,3)$$ parallel to $$3\hat{i}+2\hat{j}-2\hat{k}$$ can be expressed in any of the following equivalent ways:

• Vector form: $$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(3\hat{i}+2\hat{j}-2\hat{k}).$$
• Parametric form: $$x = 1 + 3\lambda,\; y = 2 + 2\lambda,\; z = 3 - 2\lambda.$$
• Symmetric form: $$\dfrac{x-1}{3}=\dfrac{y-2}{2}=\dfrac{z-3}{-2}.$$

Given data

• Position vector of the given point (say A): $$\vec a = 2\hat i-\hat j+4\hat k$$
• Direction vector of the required line: $$\vec b = \hat i+2\hat j-\hat k$$

1. Vector (parametric) equation of the line

For a line through a fixed point having position vector $$\vec a$$ and parallel to the vector $$\vec b$$, the general vector equation is $$\vec r = \vec a + \lambda\, \vec b$$, where $$\lambda \in \mathbb R$$.

Substituting $$\vec a$$ and $$\vec b$$:

$$\vec r = (2\hat i-\hat j+4\hat k)+\lambda(\hat i+2\hat j-\hat k), \; \lambda\in\mathbb R.$$

Simplifying the component form:

$$\vec r = (2+\lambda)\hat i + (-1+2\lambda)\hat j + (4-\lambda)\hat k.$$

2. Cartesian equation of the line

Let $$\vec r = x\hat i + y\hat j + z\hat k$$ be a variable point on the line. Equating its components with those of the vector equation derived above, we obtain the parametric relations

$$x = 2+\lambda, \quad y = -1+2\lambda, \quad z = 4-\lambda.$$

Eliminate the parameter $$\lambda$$ to get the Cartesian form:

• From $$x = 2+\lambda$$, we have $$\lambda = x-2.$$
• From $$y = -1+2\lambda$$, we have $$\lambda = \dfrac{y+1}{2}.$$
• From $$z = 4-\lambda$$, we have $$\lambda = 4-z.$$

Equating these three expressions for $$\lambda$$ gives

$$\dfrac{x-2}{1} = \dfrac{y+1}{2} = \dfrac{z-4}{-1}.$$

Thus, the required line is completely determined in both vector and Cartesian forms.

Step 1 : Identify the direction–ratios of the given line

The given line is
$$\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$$

Comparing with the symmetric (cartesian) form $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$$ we read the direction–ratios (d.r.’s) as
$$a=3,\;b=5,\;c=6.$$

Step 2 : Use the same direction–ratios for the required parallel line

If two lines are parallel, their direction–ratios are proportional. Hence the required line will also have d.r.’s $$3,5,6$$.

Step 3 : Insert the given point in the symmetric form

The required line passes through the point $$(-2,4,-5)$$. Substituting $$x_1=-2,\;y_1=4,\;z_1=-5$$ and the d.r.’s just obtained, the symmetric (cartesian) equation becomes

$$\dfrac{x-(-2)}{3}=\dfrac{y-4}{5}=\dfrac{z-(-5)}{6}.$$

Simplifying the numerators, we get

$$\boxed{\dfrac{x+2}{3}=\dfrac{y-4}{5}=\dfrac{z+5}{6}}.$$

This is the cartesian equation of the required line.

The given cartesian equation is $$\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$$.

Let the common value be $$t$$. Then

• $$x-5 = 3t \;\Rightarrow\; x = 5+3t$$
• $$y+4 = 7t \;\Rightarrow\; y = -4+7t$$
• $$z-6 = 2t \;\Rightarrow\; z = 6+2t$$

The position vector of any point $$P(x,y,z)$$ on the line is $$\vec r = x\hat i + y\hat j + z\hat k$$.

Substituting the above expressions,

$$\vec r = (5+3t)\hat i + (-4+7t)\hat j + (6+2t)\hat k$$

Group the constant and variable parts:

$$\vec r = \bigl(5\hat i -4\hat j + 6\hat k\bigr) + t\,\bigl(3\hat i + 7\hat j + 2\hat k\bigr)$$

Therefore the vector form of the line is

$$\boxed{\vec r = (5\hat i -4\hat j + 6\hat k) + t(3\hat i + 7\hat j + 2\hat k)}, \; t\in\mathbb R$$.

Here $$\vec a = 5\hat i -4\hat j + 6\hat k$$ is the position vector of the fixed point $$A(5,-4,6)$$ on the line, and $$\vec b = 3\hat i + 7\hat j + 2\hat k$$ gives the direction ratios of the line.

For two straight lines to be perpendicular, the dot–product of any pair of their direction vectors must be zero.

1. Direction ratios of the first line

Given $$\dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2}$$.
Put this common value equal to a parameter $$\lambda$$.

• $$\dfrac{1-x}{3}=\lambda \;\Rightarrow\; x = 1-3\lambda$$
• $$\dfrac{7y-14}{2p}=\lambda \;\Rightarrow\; 7y-14 = 2p\lambda \;\Rightarrow\; y = \dfrac{2p\lambda}{7}+2$$
• $$\dfrac{z-3}{2}=\lambda \;\Rightarrow\; z = 3+2\lambda$$

Thus the direction ratios (increment per unit change in $$\lambda$$) are

$$(-3,\;\dfrac{2p}{7},\;2).$$

Multiplying by 7 to clear the fraction, choose the direction ratios of the first line as

$$\vec d_1 = (-21,\;2p,\;14).$$

2. Direction ratios of the second line

Given $$\dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5}$$.
Set this common value equal to a parameter $$\mu$$.

• $$\dfrac{7-7x}{3p}=\mu \;\Rightarrow\; 7-7x = 3p\mu \;\Rightarrow\; x = 1-\dfrac{3p\mu}{7}$$
• $$y-5 = \mu \;\Rightarrow\; y = 5+\mu$$
• $$\dfrac{6-z}{5}=\mu \;\Rightarrow\; 6-z = 5\mu \;\Rightarrow\; z = 6-5\mu$$

Therefore the direction ratios are

$$\left(-\dfrac{3p}{7},\;1,\;-5\right).$$

Multiplying by 7, choose

$$\vec d_2 = (-3p,\;7,\;-35).$$

3. Perpendicularity condition

The lines are perpendicular when $$\vec d_1\cdot\vec d_2 = 0$$.

Compute the dot product:

$$\begin{aligned} \vec d_1\cdot\vec d_2 &= (-21)(-3p) + (2p)(7) + (14)(-35) \\ &= 63p + 14p - 490 \\ &= 77p - 490. \end{aligned}$$

Set it equal to zero:

$$77p - 490 = 0 \;\Rightarrow\; 77p = 490 \;\Rightarrow\; p = \dfrac{490}{77} = \dfrac{70}{11}.$$

4. Required value

The given lines are at right angles when

$$p = \dfrac{70}{11}.$$

For two lines in space to be perpendicular, the angle $$\theta$$ between their direction vectors must satisfy $$\cos\theta = 0$$, that is, the dot-product of the two direction vectors must be zero.

Step 1 · Identify direction ratios of each line

The given symmetric equation of the first line is

$$\dfrac{x-5}{7} = \dfrac{y+2}{-5} = \dfrac{z}{1}$$

Hence, a set of direction ratios (d.r.’s) for this line is

$$a_1 = 7,\; b_1 = -5,\; c_1 = 1$$

The given symmetric equation of the second line is

$$\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$$

Thus a set of direction ratios for this line is

$$a_2 = 1,\; b_2 = 2,\; c_2 = 3$$

Step 2 · Form the direction vectors

Corresponding direction vectors are therefore

$$\vec{d_1} = 7\hat\imath - 5\hat\jmath + 1\hat k$$
$$\vec{d_2} = 1\hat\imath + 2\hat\jmath + 3\hat k$$

Step 3 · Compute the dot product

$$\vec{d_1} \cdot \vec{d_2} = (7)(1) + (-5)(2) + (1)(3)$$

$$\vec{d_1} \cdot \vec{d_2} = 7 - 10 + 3$$

$$\vec{d_1} \cdot \vec{d_2} = 0$$

Step 4 · Interpretation

Because the dot product of the direction vectors is zero, the angle between them is $$90^\circ$$. Therefore the two lines are perpendicular to each other.

Step 1 : Identify a point and a direction vector for each line

• Line 1  $$\vec r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)$$
    • Point $A(1,2,1)$    • Direction vector $$\vec a=(1,-1,1)$$
• Line 2  $$\vec r=2\hat i-\hat j-\hat k+\mu(2\hat i+\hat j+2\hat k)$$
    • Point $B(2,-1,-1)$    • Direction vector $$\vec b=(2,1,2)$$

Step 2 : Vector joining the given points

$$\vec{AB}=\overrightarrow{B}-\overrightarrow{A}=(2-1,\,-1-2,\,-1-1)=(1,-3,-2)$$

Step 3 : Cross product of the direction vectors

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

$$\vec a\times\vec b=\hat i((-1)(2)-1\cdot1)-\hat j(1\cdot2-1\cdot2)+\hat k(1\cdot1-(-1)\cdot2)=(-3,0,3)$$

Magnitude of the cross product:

$$|\vec a\times\vec b|=\sqrt{(-3)^2+0^2+3^2}=\sqrt{18}=3\sqrt2$$

Step 4 : Scalar triple product

$$\vec{AB}\cdot(\vec a\times\vec b)=(1,-3,-2)\cdot(-3,0,3)=-3+0-6=-9$$

Absolute value: $$|\vec{AB}\cdot(\vec a\times\vec b)|=9$$

Step 5 : Shortest distance between the skew lines

$$d=\dfrac{|\vec{AB}\cdot(\vec a\times\vec b)|}{|\vec a\times\vec b|}=\dfrac{9}{3\sqrt2}=\dfrac{3}{\sqrt2}=\dfrac{3\sqrt2}{2}$$

The two lines are skew (they neither intersect nor are parallel), so this distance is the required shortest distance.

Step 1 – Write the two lines in parametric form

For the first line L₁:

$$\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}=t \;\;(t\in\mathbb R)$$

$$x=-1+7t, \; y=-1-6t, \; z=-1+t$$

Hence a point on L₁ is $$A(-1,-1,-1)$$ and its direction vector is $$\vec a=7\hat i-6\hat j+\hat k$$.

For the second line L₂:

$$\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}=s \;\;(s\in\mathbb R)$$

$$x=3+s, \; y=5-2s, \; z=7+s$$

Hence a point on L₂ is $$B(3,5,7)$$ and its direction vector is $$\vec b=\hat i-2\hat j+\hat k$$.

Step 2 – Check that the lines are not parallel and do not intersect

• $$\vec a=(7,-6,1),\; \vec b=(1,-2,1)$$ are not proportional, so the lines are not parallel.
• To see that they do not intersect, equate the coordinates:
  $$-1+7t=3+s,$$ $$-1-6t=5-2s,$$ $$-1+t=7+s.$$ Solving the first two gives $$t=\frac74$$, while the first and third give $$t=-\frac23$$. The contradiction shows the lines do not meet. Therefore they are skew.

Step 3 – Formula for the shortest distance between two skew lines

For skew lines with direction vectors $$\vec a,\vec b$$ and any vector $$\overrightarrow{AB}$$ joining one point on each line, the shortest distance $$d$$ is

$$d=\dfrac{|\overrightarrow{AB}\,\cdot\,(\vec a\times\vec b)|}{|\vec a\times\vec b|}.$$

Step 4 – Compute the required vectors

$$\overrightarrow{AB}=B-A=(3-(-1),\,5-(-1),\,7-(-1))=(4,6,8).$$

The cross-product

$$\vec a\times\vec b=\begin{vmatrix}\hat i & \hat j & \hat k\\7 & -6 & 1\\1 & -2 & 1\end{vmatrix}$$

$$=\hat i\big((-6)(1)-(1)(-2)\big)-\hat j\big(7\cdot1-1\cdot1\big)+\hat k\big(7(-2)-(-6)(1)\big)$$

$$=(-4,-6,-8).$$

Magnitude:

$$|\vec a\times\vec b|=\sqrt{(-4)^2+(-6)^2+(-8)^2}=\sqrt{16+36+64}=\sqrt{116}=2\sqrt{29}.$$

The scalar triple product

$$\overrightarrow{AB}\,\cdot\,(\vec a\times\vec b)=(4,6,8)\cdot(-4,-6,-8)=-16-36-64=-116.$$

Step 5 – Shortest distance

$$d=\dfrac{| -116 |}{2\sqrt{29}}=\dfrac{116}{2\sqrt{29}}=\dfrac{58}{\sqrt{29}}=\dfrac{58\sqrt{29}}{29}=2\sqrt{29}.$$

Hence, the shortest distance between the given lines is $$2\sqrt{29}$$ units.

Given lines

Line $$L_1: \; \vec r = (\hat i + 2\hat j + 3\hat k) + \lambda(\hat i - 3\hat j + 2\hat k)$$
Line $$L_2: \; \vec r = 4\hat i + 5\hat j + 6\hat k + \mu(2\hat i + 3\hat j + \hat k)$$

Identify

• a point on $$L_1$$: $$\vec a = \hat i + 2\hat j + 3\hat k\; (1,2,3)$$
• direction vector of $$L_1$$: $$\vec b = \hat i - 3\hat j + 2\hat k\; (1,-3,2)$$
• a point on $$L_2$$: $$\vec c = 4\hat i + 5\hat j + 6\hat k\; (4,5,6)$$
• direction vector of $$L_2$$: $$\vec d = 2\hat i + 3\hat j + \hat k\; (2,3,1)$$

Step 1 : Vector joining the two reference points

$$\vec{AC}=\vec c-\vec a=(4-1)\hat i+(5-2)\hat j+(6-3)\hat k=3\hat i+3\hat j+3\hat k\;(3,3,3)$$

Step 2 : Cross product of direction vectors

$$\vec b\times\vec d =\begin{vmatrix}\hat i & \hat j & \hat k\\[4pt]1 & -3 & 2\\[4pt]2 & 3 & 1\end{vmatrix}$$

$$\quad=\hat i((-3)(1)-2\cdot3) - \hat j(1\cdot1-2\cdot2)+\hat k(1\cdot3-(-3)\cdot2)$$
$$\quad=\hat i(-3-6)-\hat j(1-4)+\hat k(3+6)$$
$$\quad=-9\hat i+3\hat j+9\hat k$$

Thus $$\vec b\times\vec d =\langle-9,3,9\rangle$$ and
$$|\vec b\times\vec d|=\sqrt{(-9)^2+3^2+9^2}=\sqrt{81+9+81}=\sqrt{171}=3\sqrt{19}$$

Step 3 : Scalar triple product

$$|\vec{AC}\,\cdot\,(\vec b\times\vec d)|=(3,3,3)\cdot(-9,3,9)$$
$$=3(-9)+3(3)+3(9)=-27+9+27=9$$

Step 4 : Shortest distance formula

For two skew lines, the shortest distance $$d$$ is
$$d=\dfrac{|\vec{AC}\,\cdot\,(\vec b\times\vec d)|}{|\vec b\times\vec d|}$$

Therefore
$$d=\dfrac{9}{3\sqrt{19}}=\dfrac{3}{\sqrt{19}}$$

Hence, the shortest distance between the given lines is $$\displaystyle \mathbf{\frac{3}{\sqrt{19}}}$$.

The given lines are

Line 1: $$\vec r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\;\;(t\in\mathbb R)$$
Line 2: $$\vec r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\;\;(s\in\mathbb R)$$

Step 1 : Direction vectors

For Line 1 the coefficient of $$t$$ is $$\vec d_1=(-1,1,-2)$$.
For Line 2 the coefficient of $$s$$ is $$\vec d_2=(1,2,-2)$$.

Step 2 : One point on each line

Put $$t=0$$ in Line 1 ⇒ $$\vec a_1=(1,-2,3)$$.
Put $$s=0$$ in Line 2 ⇒ $$\vec a_2=(1,-1,-1)$$.

Step 3 : Cross product $$\vec d_1\times\vec d_2$$

$$\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\-1&1&-2\\1&2&-2\end{vmatrix}=2\hat i-4\hat j-3\hat k=(2,-4,-3).$$

Step 4 : Magnitude of the cross product

$$|\vec d_1\times\vec d_2|=\sqrt{2^2+(-4)^2+(-3)^2}=\sqrt{29}.$$

Step 5 : Vector joining the chosen points

$$\vec a_2-\vec a_1=(1,-1,-1)-(1,-2,3)=(0,1,-4).$$

Step 6 : Scalar triple product

$$\bigl[(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\bigr]= (0,1,-4)\cdot(2,-4,-3)=0(2)+1(-4)+(-4)(-3)=-4+12=8.$$

Step 7 : Shortest distance

$$D=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}=\dfrac{8}{\sqrt{29}}.$$

Hence the shortest distance between the two given skew lines is $$\displaystyle \frac{8}{\sqrt{29}}$$ units.

Let the two lines be represented by their direction–ratio (d.r.) sets

Line 1 : $$a,\;b,\;c$$    and   Line 2 : $$(b-c),\;(c-a),\;(a-b).$$

The angle $$\theta$$ between two lines whose d.r.s are $$a_1,\,b_1,\,c_1$$ and $$a_2,\,b_2,\,c_2$$ is obtained from

$$\cos\theta = \dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2}}\,\sqrt{a_2^{2}+b_2^{2}+c_2^{2}}}.$$

Step 1: Compute the dot product (numerator)

$$\begin{aligned} a_1a_2+b_1b_2+c_1c_2 &= a(b-c)+b(c-a)+c(a-b) \\ &= \bigl(ab-ac\bigr)+\bigl(bc-ab\bigr)+\bigl(ca-bc\bigr) \\ &= ab-ac+bc-ab+ca-bc \\ &= 0. \end{aligned}$$

Thus the dot product is $$0$$.

Step 2: Compute the magnitudes (denominator)

Since at least one of $$a,b,c$$ differs from the others (otherwise the second set of d.r.s would all be zero and would not define a line), we have

$$\sqrt{a^{2}+b^{2}+c^{2}}>0$$ and $$\sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}>0.$$

Step 3: Obtain $$\cos\theta$$

$$\cos\theta = \dfrac{0}{\sqrt{a^{2}+b^{2}+c^{2}}\,\sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}=0.$$

Step 4: Interpret the result

Since $$\cos\theta=0$$, the angle between the given lines is

$$\theta = 90^{\circ}.$$

Hence the two lines are perpendicular (right–angled) provided they are properly defined (i.e. not all of $$a,b,c$$ are equal).

Given: A line that is (i) parallel to the $$x$$-axis and (ii) passes through the origin $$O(0,0,0)$$.

Step 1  Direction vector
Any vector parallel to the $$x$$-axis has non-zero $$x$$-component only. Choose $$\vec d=(1,0,0)$$ (the unit vector $$\hat i$$).

Step 2  Point on the line
The required line passes through the origin, whose position vector is $$\vec 0=(0,0,0)$$.

Step 3  Equation of the line
The general vector form of a line through $$\vec a$$ and parallel to $$\vec d$$ is $$\vec r = \vec a + \lambda\vec d,\; \lambda\in\mathbb R$$. Substituting $$\vec a = \vec 0$$ and $$\vec d=(1,0,0)$$:

$$\vec r = \lambda(1,0,0),\; \lambda\in\mathbb R.$$

Parametric form
If $$\vec r=(x,y,z)$$, then

$$x = \lambda,\qquad y = 0,\qquad z = 0,\; \lambda\in\mathbb R.$$

Thus the required line is the $$x$$-axis itself.

The two lines are written in symmetric form.

For a line expressed as $$\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$$ the triplet $$(a,b,c)$$ gives a set of direction ratios.

Therefore

• Line 1 has direction ratios $$(-3,\,2k,\,2)$$ because $$\dfrac{x-1}{-3}=\dfrac{y-2}{2k}=\dfrac{z-3}{2}.$$
• Line 2 has direction ratios $$(3k,\,1,\,-5)$$ because $$\dfrac{x-1}{3k}=\dfrac{y-1}{1}=\dfrac{z-6}{-5}.$$

The lines are perpendicular ⇔ the dot product of their direction vectors is zero:

$$(-3,2k,2)\cdot(3k,1,-5)=0$$

Compute the dot product:

$$(-3)(3k)+(2k)(1)+2(-5)=0$$

$$-9k+2k-10=0$$

$$-7k-10=0$$

$$-7k=10$$

$$k=-\dfrac{10}{7}$$

Hence the required value of $$k$$ is $$-\dfrac{10}{7}.$$

Given lines

Line 1 : $$\vec r = 6\hat i + 2\hat j + 2\hat k + \lambda\,(\hat i - 2\hat j + 2\hat k)$$
Line 2 : $$\vec r = -4\hat i - \hat k + \mu\,(3\hat i - 2\hat j - 2\hat k)$$

Write them in standard vector form
$$\vec r = \vec a_1 + \lambda\,\vec b_1,\qquad \vec a_1 = 6\hat i + 2\hat j + 2\hat k,\; \vec b_1 = \hat i - 2\hat j + 2\hat k$$ $$\vec r = \vec a_2 + \mu\,\vec b_2,\qquad \vec a_2 = -4\hat i + 0\hat j - \hat k,\; \vec b_2 = 3\hat i - 2\hat j - 2\hat k$$

Step 1 · Check that the lines are skew

Their direction vectors are not parallel because
$$\vec b_1 = (1,-2,2),\; \vec b_2 = (3,-2,-2)$$ are not scalar multiples (easy to verify).
To see they do not intersect, solve $$\vec a_1+\lambda\vec b_1=\vec a_2+\mu\vec b_2$$ :

From the second equation $$\mu = \lambda-1$$.
Substitute in the first: $$6+\lambda = -4+3(\lambda-1) \Rightarrow 13-2\lambda =0 \Rightarrow \lambda=\tfrac{13}{2},\; \mu=\tfrac{11}{2}$$.
Check in the third: $$2+2\lambda = 15\neq -1-2\mu = -12$$, so no common point exists. Hence the lines are skew.

Step 2 · Formula for shortest distance

For skew lines $$\vec r=\vec a_1+\lambda\vec b_1$$ and $$\vec r=\vec a_2+\mu\vec b_2$$,
shortest distance d is
$$d = \frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$$

Step 3 · Evaluate the cross-product

$$\vec b_1 = (1,-2,2),\; \vec b_2 = (3,-2,-2)$$

$$\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\ 1&-2&2\\ 3&-2&-2\end{vmatrix}$$

$$\qquad = \hat i[(-2)(-2)-2(-2)] - \hat j[1(-2)-2(3)] + \hat k[1(-2)-(-2)(3)]$$ $$\qquad = 8\hat i + 8\hat j + 4\hat k$$

Magnitude:
$$|\vec b_1\times\vec b_2| = \sqrt{8^2+8^2+4^2}=\sqrt{64+64+16}=\sqrt{144}=12$$

Step 4 · Evaluate the scalar triple product

$$\vec a_2-\vec a_1 = (-4-6,\;0-2,\;-1-2)=(-10,-2,-3)$$

Dot product:
$$ (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) = (-10)(8)+(-2)(8)+(-3)(4) = -80-16-12 = -108$$

Step 5 · Shortest distance

$$ d = \frac{|{-108}|}{12} = \frac{108}{12}=9 $$

Therefore, the shortest distance between the given lines is $$9\text{ units}.$$

Let the required line be L. We need

• a point on L: given point $$A(1,2,-4)$$, so $$\vec a = \hat i+2\hat j-4\hat k$$.
• a direction vector $$\vec d$$ that is perpendicular to both of the given lines.

For each given line, read its direction vector from the equal–ratio form.

1. First line
   $$\dfrac{x-8}{3}=\dfrac{y+19}{-16}=\dfrac{z-10}{7}\;\;\Longrightarrow\;\;\vec d_1 = 3\hat i-16\hat j+7\hat k.$$
2. Second line
   $$\dfrac{x-15}{3}=\dfrac{y-29}{8}=\dfrac{z-5}{-5}\;\;\Longrightarrow\;\;\vec d_2 = 3\hat i+8\hat j-5\hat k.$$

The direction vector of L must be perpendicular to both $$\vec d_1$$ and $$\vec d_2$$, hence parallel to their cross product:

$$\vec d = \vec d_1\times\vec d_2.$$

Compute the cross product through the determinant

$$ \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -16 & 7\\ 3 & 8 & -5 \end{vmatrix} = \hat i\,( (-16)(-5) - 7\cdot 8 ) -\hat j\,( 3(-5) - 7\cdot 3 ) +\hat k\,( 3\cdot 8 - (-16)\cdot 3 ). $$

Simplify each component:

• $$\hat i: \;80-56 = 24$$
• $$\hat j: -\bigl(-15-21\bigr) = 36$$
• $$\hat k: 24+48 = 72$$

Thus $$\vec d = 24\hat i + 36\hat j + 72\hat k.$$ Divide by $$12$$ (common factor) for a shorter vector:

$$\vec d = 2\hat i + 3\hat j + 6\hat k.$$

Now write the vector equation of the required line through $$A(1,2,-4)$$ with direction $$\vec d$$:

$$\vec r = (\hat i + 2\hat j - 4\hat k) + \lambda\,(2\hat i + 3\hat j + 6\hat k), \;\lambda\in\mathbb R.$$

Equivalent parametric form:

$$x = 1 + 2\lambda,\; y = 2 + 3\lambda,\; z = -4 + 6\lambda.$$

Equivalent symmetric form:

$$\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z+4}{6}.$$