If $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2…

If $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}$ are such that $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$, then the value of $\lambda$ is
  1. $\frac{5}{11}$
  2. $\frac{11}{5}$
  3. $\frac{-11}{5}$
  4. $\frac{-5}{11}$

Solution

Let $\overline{\mathrm{d}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ $\begin{aligned} \overline{\mathrm{d}} & =(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+2 \lambda \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}} \\ & =(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}} \end{aligned}$ Now, $\overline{\mathrm{d}}$ is perpendicular to $\overline{\mathrm{c}}$. $\begin{aligned} & \therefore \quad \overline{\mathrm{c}} \cdot \overline{\mathrm{d}}=0 \\ & \Rightarrow(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot[(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}]=0 \\ & \Rightarrow 1(2 \lambda+2)+3(3+\lambda)=0 \\ & \Rightarrow 2 \lambda+2+9+3 \lambda=0 \\ & \Rightarrow 5 \lambda+11=0 \\ & \Rightarrow \lambda=\frac{-11}{5} \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 1)

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