Question

If \[ \vec{a} = 3\hat{i} + \hat{j} - 2\hat{k} \quad \text{and} \quad \vec{b} = \hat{i} + \lambda \hat{j} - 3\hat{k} \] are perpendicular to each other, then the value of \( \lambda \) is:

• A. -3
• B. -6
• C. -9
• D. -1

Hint:

For two vectors to be perpendicular, their dot product must equal zero. Always expand the dot product carefully and solve for the unknown variable.

Answer 1

The Correct Option is B

Two vectors are perpendicular if their dot product is zero. So, we calculate the dot product \( \vec{a} \cdot \vec{b} \) and set it equal to zero. The dot product \( \vec{a} \cdot \vec{b} \) is given by: \[ \vec{a} \cdot \vec{b} = (3\hat{i} + \hat{j} - 2\hat{k}) \cdot (\hat{i} + \lambda \hat{j} - 3\hat{k}) \] Using the distributive property of the dot product: \[ \vec{a} \cdot \vec{b} = 3\hat{i} \cdot \hat{i} + 3\hat{i} \cdot \lambda \hat{j} + 3\hat{i} \cdot (-3\hat{k}) + \hat{j} \cdot \hat{i} + \hat{j} \cdot \lambda \hat{j} + \hat{j} \cdot (-3\hat{k}) + (-2\hat{k}) \cdot \hat{i} + (-2\hat{k}) \cdot \lambda \hat{j} + (-2\hat{k}) \cdot (-3\hat{k}) \] Simplifying the dot products: \[ = 3(1) + 3(\lambda)(0) + 3(-3)(0) + 1(0) + 1(\lambda)(1) + 1(-3)(0) + (-2)(0) + (-2)(\lambda)(0) + (-2)(-3)(1) \] \[ = 3 + 0 + 0 + 0 + \lambda + 0 + 0 + 0 + 6 \] \[ = 9 + \lambda \] For the vectors to be perpendicular, the dot product must be zero: \[ 9 + \lambda = 0 \] Solving for \( \lambda \): \[ \lambda = -9 \] Thus, the correct answer is: \[ \boxed{-9} \]