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

Question

Find the unit vector perpendicular to both \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \).

Answer 1

Step 1: Understanding the Concept:
Two non-zero vectors are perpendicular if and only if their scalar (dot) product is equal to zero.
Step 2: Detailed Explanation:
Given vectors are mutually perpendicular, so \( \vec{a} \cdot \vec{b} = 0 \). \[ \left( \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) \right) \cdot \left( \frac{1}{7}(3\hat{i} - \lambda\hat{j} + 2\hat{k}) \right) = 0 \] The scalars \( 1/7 \times 1/7 = 1/49 \) can be factored out and removed since the RHS is zero: \[ (2)(3) + (3)(-\lambda) + (6)(2) = 0 \] \[ 6 - 3\lambda + 12 = 0 \] \[ 18 - 3\lambda = 0 \] \[ 3\lambda = 18 \implies \lambda = 6 \].
Step 3: Final Answer:
The value of \( \lambda \) is 6.

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

Show Hint

The Correct Option is C

Solution and Explanation

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