Q114 / 170 · Manipal (MET) 2021 · Mathematics
$\overrightarrow{\mathbf{a}}=\frac{1}{7}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}), \overrightarrow{\mathbf{b}}=\frac{1}{7}(3 \hat{\mathbf{i}}-\lambda \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$. If $\overrightarrow{\mathbf{a}}$ and $\vec{b}$ are mutually perpendicular, then value of $\lambda$ is
- A2
- B-1
- Correct answerC6
- D-6
Solution
Given, $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \vec{b}=\frac{1}{7}(3 \hat{i}-\lambda \hat{j}+2 \hat{k})$
Since, the vectors $\vec{a}$ and $\vec{b}$ are perpendicular.
$\therefore \quad \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0$
$\begin{array}{ll}\Rightarrow \frac{1}{7}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \cdot \frac{1}{7}(3 \hat{\mathbf{i}}-\lambda \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0 \\ \Rightarrow & \frac{1}{49}(6-3 \lambda+12)=0 \\ \Rightarrow & 3 \lambda=18 \Rightarrow \lambda=6\end{array}$
Hence, the correct answer is option: 'C'.