Question

Let \( \bar{a} = \hat{i} + \hat{j}, \bar{b} = 2\hat{i} - \hat{k}, \bar{c} = 3\hat{i} - \hat{j} + \hat{k} \), then vector \( \bar{p} \) satisfying \( \bar{p} \cdot \bar{a} = 0 \) and \( \bar{p} \times \bar{b} = \bar{c} \times \bar{b} \) is

• A. \( \hat{i} - \hat{j} + \hat{k} \)
• B. \( \hat{i} - 2\hat{j} + \hat{k} \)
• C. \( -\hat{i} + \hat{j} + \hat{k} \)
• D. \( \hat{i} - \hat{j} + 2\hat{k} \)

Hint:

$\bar{u} \times \bar{v} = \bar{w} \times \bar{v}$ implies $\bar{u} - \bar{w} = \lambda \bar{v}$.

Answer 1

The Correct Option is C

Step 1: Simplify Cross Product
$(\bar{p} - \bar{c}) \times \bar{b} = 0 \implies (\bar{p} - \bar{c})$ is parallel to $\bar{b}$.
$\bar{p} - \bar{c} = \lambda \bar{b} \implies \bar{p} = \bar{c} + \lambda \bar{b}$.
Step 2: Use Dot Product
$\bar{p} \cdot \bar{a} = (\bar{c} + \lambda \bar{b}) \cdot \bar{a} = 0$.
$\bar{c} \cdot \bar{a} + \lambda (\bar{b} \cdot \bar{a}) = 0$.
$(3-1) + \lambda (2+0) = 0 \implies 2 + 2\lambda = 0 \implies \lambda = -1$.
Step 3: Find \(\bar{p}\)
$\bar{p} = \bar{c} - \bar{b} = (3\hat{i} - \hat{j} + \hat{k}) - (2\hat{i} - \hat{k})$.
Step 4: Conclusion
$\bar{p} = \hat{i} - \hat{j} + 2\hat{k}$. (Check options: D matches this calculation).
Final Answer:(D)