Question

Let $ \vec{a} = \hat{i} + 4\hat{j} + 3\hat{k} $ and $ \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} $, and $ \vec{c} $ is a vector perpendicular to $ \vec{a} $ and lies in the plane of $ \vec{a} $, $ \vec{b} $. Then $ \vec{c} $ is equal to:

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

Hint:

For problems involving vectors perpendicular to multiple vectors, use the cross product to find a vector perpendicular to both.

Answer 1

The Correct Option is B

We are given \( \vec{a} = \hat{i} + 4\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), and need to find \( \vec{c} \), which is perpendicular to \( \vec{a} \) and lies in the plane of \( \vec{a} \), \( \vec{b} \).
Step 1: Perpendicular Condition Since \( \vec{c} \) is perpendicular to \( \vec{a} \), we have: \[ \vec{a} \cdot \vec{c} = 0 \]
Step 2: Plane Condition Since \( \vec{c} \) lies in the plane of \( \vec{a} \) and \( \vec{b} \), we can express \( \vec{c} \) as: \[ \vec{c} = \lambda \vec{a} + \mu \vec{b} \] where \( \lambda \) and \( \mu \) are constants.
Step 3: Dot Product For \( \vec{c} \) to be perpendicular to \( \vec{a} \), we set: \[ \vec{a} \cdot \vec{c} = \vec{a} \cdot (\lambda \vec{a} + \mu \vec{b}) = 0 \] Expanding: \[ \lambda (\vec{a} \cdot \vec{a}) + \mu (\vec{a} \cdot \vec{b}) = 0 \] Using the values for the dot products: \[ \vec{a} \cdot \vec{a} = 26, \quad \vec{a} \cdot \vec{b} = 26 \] This gives the equation: \[ \lambda \times 26 + \mu \times 26 = 0 \quad \Rightarrow \quad \lambda + \mu = 0 \]
Step 4: Final Vector Thus, \( \mu = -\lambda \), and we have: \[ \vec{c} = \lambda (\vec{a} - \vec{b}) \] Simplifying \( \vec{a} - \vec{b} \): \[ \vec{a} - \vec{b} = -\hat{i} + \hat{j} - \hat{k} \] Thus, \( \vec{c} = -\hat{i} + \hat{j} - \hat{k} \). Therefore, the correct answer is \( \vec{c} = -\hat{i} + \hat{j} - \hat{k} \).