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: Understanding the Question:
We are given two conditions to find vector \( \bar{p} \). One is a dot product constraint, and the other is a cross product relationship.
Step 2: Key Formula or Approach:
1. \( \bar{p} \times \bar{b} = \bar{c} \times \bar{b} \implies (\bar{p} - \bar{c}) \times \bar{b} = 0 \). This means \( \bar{p} - \bar{c} \) is parallel to \( \bar{b} \).
2. So, \( \bar{p} = \bar{c} + \lambda \bar{b} \) for some scalar \( \lambda \).
3. Use \( \bar{p} \cdot \bar{a} = 0 \) to find \( \lambda \).
Step 3: Detailed Explanation:
Substitute \( \bar{p} = \bar{c} + \lambda \bar{b} \) into \( \bar{p} \cdot \bar{a} = 0 \):
\[ (\bar{c} + \lambda \bar{b}) \cdot \bar{a} = 0 \implies \bar{c} \cdot \bar{a} + \lambda (\bar{b} \cdot \bar{a}) = 0 \] Calculate the dot products:
\[ \bar{c} \cdot \bar{a} = (3, -1, 1) \cdot (1, 1, 0) = 3 - 1 = 2 \] \[ \bar{b} \cdot \bar{a} = (2, 0, -1) \cdot (1, 1, 0) = 2 + 0 = 2 \] Now solve for \( \lambda \):
\[ 2 + \lambda(2) = 0 \implies 2\lambda = -2 \implies \lambda = -1 \] Find \( \bar{p} \):
\[ \bar{p} = \bar{c} - 1\bar{b} = (3\hat{i} - \hat{j} + \hat{k}) - (2\hat{i} - \hat{k}) = \hat{i} - \hat{j} + 2\hat{k} \] Step 4: Final Answer: The vector is \( \hat{i} - \hat{j} + 2\hat{k} \).