If $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}, \bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}, \bar{c}=\hat{i}+3…
If $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}, \bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}, \bar{c}=\hat{i}+3 \hat{j}+\hat{k}$ and $\bar{a}+\lambda \bar{b}$ is perpendicular to $\overline{\mathrm{c}}$, then $\lambda=$
-2
4
-4
2
Solution
$\begin{aligned}
& \bar{a}+\lambda \bar{b}=(\hat{i}+2 \hat{j}-3 \hat{k})+\lambda(3 \hat{i}-\hat{j}+2 \hat{k}) \\
& =(1+3 \lambda) \hat{i}+(2-\lambda) \hat{j}+(-3+2 \lambda) \hat{k}
\end{aligned}$
Since $\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ is $\perp$ er to $\overline{\mathrm{c}}$, we write
$\begin{aligned}
& (1)(1+3 \lambda)+(3)(2-\lambda)+(1)(-3+2 \lambda) \\
& \therefore 1+3 \lambda+6-3 \lambda-3+2 \lambda=0 \Rightarrow \lambda=-2
\end{aligned}$