If the vector $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3…

If the vector $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$ are such that $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$, then the value of $\lambda$ is
  1. -8
  2. 10
  3. 8
  4. $\frac{8}{3}$

Solution

$\begin{aligned} & \because \vec{a}+\lambda \vec{b} \perp \vec{c} \\ & \Rightarrow(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0 \\ & \Rightarrow\{(2 \widehat{i}+2 \widehat{j}+3 \widehat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})\} \cdot(3 \hat{i}+\hat{j})=0 \\ & \Rightarrow\{(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \widehat{k}\} \cdot(3 \hat{i}+\hat{j})=0 \\ & \Rightarrow(2-\lambda) \times 3+(2+2 \lambda) \times 1+(3+\lambda) \times 0=0 \\ & \Rightarrow \lambda=8\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

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