Question

If \( \mathbf{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}, \mathbf{b} = -\hat{i} + 2\hat{j} + \hat{k} \) and \( \mathbf{c} = 3\hat{i} + \hat{j} \) are the vectors such that \( \mathbf{a} + \lambda \mathbf{b} \) is perpendicular to \( \mathbf{c} \), then the value of \( \lambda \) is:

• A. 6
• B. -6
• C. 8
• D. -8

Hint:

In vector problems involving dot products and perpendicular vectors, remember that the dot product of two perpendicular vectors is zero. Use this property to solve the problem step by step.

Answer 1

The Correct Option is B

We are given that \( \mathbf{a} + \lambda \mathbf{b} \) is perpendicular to \( \mathbf{c} \). This means that the dot product of \( \mathbf{a} + \lambda \mathbf{b} \) with \( \mathbf{c} \) must be zero: \[ (\mathbf{a} + \lambda \mathbf{b}) \cdot \mathbf{c} = 0 \] Substitute the given vectors into this equation: \[ (2\hat{i} + 2\hat{j} + 3\hat{k} + \lambda(-\hat{i} + 2\hat{j} + \hat{k})) \cdot (3\hat{i} + \hat{j}) = 0 \] Now, simplify the left-hand side: \[ \left( (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k} \right) \cdot (3\hat{i} + \hat{j}) \] Now perform the dot product: \[ (2 - \lambda)(3) + (2 + 2\lambda)(1) + (3 + \lambda)(0) = 0 \] This simplifies to: \[ 3(2 - \lambda) + (2 + 2\lambda) = 0 \] \[ 6 - 3\lambda + 2 + 2\lambda = 0 \] \[ 8 - \lambda = 0 \] \[ \lambda = -6 \] Thus, the value of \( \lambda \) is \(-6\).