Question

\(\vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}),\; \vec{b}=\frac{1}{7}(3\hat{i}-\lambda \hat{j}+2\hat{k})\). If \(\vec{a}\perp \vec{b}\), find \(\lambda\)

• A. 2
• B. -1
• C. 6
• D. -6

Hint:

Common scalar factors do not affect perpendicular condition.

Answer 1

The Correct Option is C

Concept: If two vectors are perpendicular: \(\vec{a}\cdot \vec{b} = 0\).

Step 1: Ignore common factor \(\frac{1}{49}\). \[ (2\hat{i}+3\hat{j}+6\hat{k}) \cdot (3\hat{i}-\lambda \hat{j}+2\hat{k}) = 0 \]

Step 2: Compute dot product. \[ 2 \cdot 3 + 3(-\lambda) + 6 \cdot 2 = 0 \] \[ 6 - 3\lambda + 12 = 0 \] \[ 18 - 3\lambda = 0 \Rightarrow \lambda = 6 \]