Question

Let $\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$, $\vec{b} = 4\hat{i} + \hat{j} + 7\hat{k}$, and $\vec{c} = \hat{i} - 3\hat{j} + 4\hat{k}$ be three vectors.
If a vector $\vec{p}$ satisfies $\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{p} \cdot \vec{a} = 0$, then $\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$ is equal to

• A. 24
• B. 36
• C. 28
• D. 32

Answer 1

Answer: (D)

32

Answer 2

Given:

$\vec{p} \times \vec{b} - \vec{c} \times \vec{b} = 0 \quad \implies \quad (\vec{p} - \vec{c}) \times \vec{b} = 0$

This implies:

$\vec{p} - \vec{c} = \lambda \vec{b} \quad \implies \quad \vec{p} = \vec{c} + \lambda \vec{b}$

Given that $\vec{p} \cdot \vec{a} = 0$, we have:

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

Substituting values:

$\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$ $(3 - 3 - 8) + \lambda (12 + 1 - 14) = 0 \quad \implies \quad \lambda = -8$

Thus:

$\vec{p} = \vec{c} - 8\vec{b} = -31\hat{i} - 11\hat{j} - 52\hat{k}$

Now, compute:

$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$ $= (-31)(1) + (-11)(-1) + (-52)(-1)$ $= -31 + 11 + 52 = 32$