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. 8
• C. -6
• 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

Given vectors:

\( \mathbf{a} = 2\hat{i} + 2\hat{j} + 3\hat{k} \)

\( \mathbf{b} = -\hat{i} + 2\hat{j} + \hat{k} \)

\( \mathbf{c} = 3\hat{i} + \hat{j} \)

Let \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \)

\( \mathbf{r} = (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k} \)

Since \( \mathbf{r} \) is perpendicular to \( \mathbf{c} \), their dot product is zero:

\( \mathbf{r} \cdot \mathbf{c} = 0 \)

Calculate the dot product:

\( [(2 - \lambda) \cdot 3] + [(2 + 2\lambda) \cdot 1] + [(3 + \lambda) \cdot 0] = 0 \)

Simplify the equation:

\( 3(2 - \lambda) + (2 + 2\lambda) = 0 \)

\( 6 - 3\lambda + 2 + 2\lambda = 0 \)

\( 8 - \lambda = 0 \)

Solving for \( \lambda \):

\( \lambda = 8 \)

Final Answer:

\( \boxed{\lambda = 8} \)