If $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$ such that $\bar{b}+\lambda \bar{a}$ is perpendicular to $\bar{c}$,…

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If $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$ such that $\bar{b}+\lambda \bar{a}$ is perpendicular to $\bar{c}$, then $\lambda$ is

Aicharya · Accepted Answer

According to the given condition, we get $\begin{aligned} & (\overline{\mathrm{b}}+\dot{\lambda} \overline{\mathrm{a}}) \cdot \overrightarrow{\mathrm{c}}=0 \ & \Rightarrow[(-1+2 \lambda) \hat{\mathrm{i}}+(2+2 \lambda) \hat{\mathrm{j}}+(1+3 \lambda) \hat{\mathrm{k}}] \cdot(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})=0 \ & \Rightarrow 3(-1+2 \lambda)+(2+2 \lambda)=0 \ & \Rightarrow-3+6 \lambda+2+2 \lambda=0 \ & \Rightarrow \lambda=\frac{1}{8} \end{aligned}$

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

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