Dot Product (Component Form)

Given,

$$\vec a=2\hat i-\hat j+2\hat k$$ and $$\vec b=\hat i+2\hat j-2\hat k$$

First compute

$$\vec a\times\vec b$$

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

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

$$=-2\hat i+6\hat j+5\hat k$$

Now,

$$\vec a+2\vec b$$

$$=(2,-1,2)+2(1,2,-2)$$

$$=(4,3,-2)$$

Let

$$\vec p=\vec a\times\vec b=(-2,6,5)$$

and

$$\vec q=\vec a+2\vec b=(4,3,-2)$$

Using

$$|\vec p\times\vec q|^2=|\vec p|^2|\vec q|^2-(\vec p\cdot\vec q)^2,$$

first compute:

$$|\vec p|^2=(-2)^2+6^2+5^2$$

$$=4+36+25$$

$$=65$$

Also,

$$|\vec q|^2=4^2+3^2+(-2)^2$$

$$=16+9+4$$

$$=29$$

Now,

$$\vec p\cdot\vec q=(-2)(4)+6(3)+5(-2)$$

$$=-8+18-10$$

$$=0$$

Therefore,

$$|\vec p\times\vec q|^2=65\times29$$

$$=1885$$

Since $$\vec{d}$$ is coplanar with $$\vec{a}$$ and $$\vec{b}$$, it can be written as a linear combination of them:

$$\vec{d} = \lambda \vec{a} + \mu \vec{b}$$

Substitute the given vectors $$\vec{a}$$ and $$\vec{b}$$:

$$\vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$$ $$\vec{d} = (\lambda + \mu)\hat{i} + \lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}$$

We are given that $$\vec{d}$$ is perpendicular to $$\vec{a}$$ ($$\vec{d} \cdot \vec{a} = 0$$):

$$(\lambda \vec{a} + \mu \vec{b}) \cdot \vec{a} = 0$$ $$\lambda |\vec{a}|^2 + \mu (\vec{b} \cdot \vec{a}) = 0$$

Let's calculate the required dot products:

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

Substitute these values back into the condition:

$$3\lambda + \mu(1) = 0 \implies \mu = -3\lambda$$

Now, update our expression for $$\vec{d}$$ in terms of $$\lambda$$ only:

$$\vec{d} = \lambda \vec{a} - 3\lambda \vec{b} = \lambda (\vec{a} - 3\vec{b})$$

Let's compute $$\vec{a} - 3\vec{b}$$:

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

So, the vector $$\vec{d}$$ simplifies to:

$$\vec{d} = \lambda(-2\hat{i} + 4\hat{j} - 2\hat{k})$$

We are given $$\vec{d} \cdot \vec{c} = 6$$, where $$\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$$:

$$\lambda(-2\hat{i} + 4\hat{j} - 2\hat{k}) \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 6$$ $$\lambda \left[ (-2)(1) + (4)(2) + (-2)(-1) \right] = 6$$ $$\lambda \left[ -2 + 8 + 2 \right] = 6$$ $$8\lambda = 6 \implies \lambda = \frac{3}{4}$$

Now we can write the exact vector $$\vec{d}$$:

$$\vec{d} = \frac{3}{4}(-2\hat{i} + 4\hat{j} - 2\hat{k}) = -\frac{3}{2}\hat{i} + 3\hat{j} - \frac{3}{2}\hat{k}$$

Finally, compute the square of its magnitude:

$$|\vec{d}|^2 = \left(-\frac{3}{2}\right)^2 + (3)^2 + \left(-\frac{3}{2}\right)^2$$ $$|\vec{d}|^2 = \frac{9}{4} + 9 + \frac{9}{4}$$ $$|\vec{d}|^2 = \frac{18}{4} + 9 = \frac{9}{2} + 9 = \frac{27}{2}$$

Correct Answer: C. $$\frac{27}{2}$$