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

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Aicharya · Accepted Answer

$\begin{aligned} & (\bar{a}+\lambda \bar{b}) \cdot \bar{c}=0 \ & 	herefore[(1-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}] \cdot[3 \hat{i}+\hat{j}]=0 \ & 	herefore(1-\lambda)(3)+(2+2 \lambda)(1)=0 \Rightarrow \lambda=5\end{aligned}$

If $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{\imath}+2 \hat{\jmath}+\hat{k}, \vec{c}=3…

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