If vectors $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=-3…
If vectors $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=-3 \hat{i}+\hat{j}+2 \hat{k}$ are such that, $\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ is perpendicular to $\overline{\mathrm{c}}$, then $\lambda=$
-14
14
2
-2
Solution
From given data, we write
$\bar{a}+\lambda \bar{b}=(2-\lambda)
\hat{i}+(2+2 \lambda)
\hat{j}+(3+\lambda)
\hat{k}$
Since (1) is $\perp$ er to $\bar{c}$, we write
$\begin{aligned}
& (2-\lambda)(3)+(2+2 \lambda)(1)+(3+\lambda)(2)=0 \\
& \therefore 6-3 \lambda+2+2 \lambda+6+2 \lambda=0 \Rightarrow \lambda=-14
\end{aligned}$