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If $\bar{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}, \bar{b}=2 \hat{i}+\hat{j}-\hat{k}$ and $\bar{c}=3 \hat{i}-\hat{j} \quad$ are such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$, then the value of $\lambda$ is
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Verified Answer
The correct answer is:
$\frac{-3}{5}$
According to the given condition, we get
$\begin{array}{ll}
& (\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}) \cdot \overline{\mathrm{c}}=0 \\
\therefore \quad & {[(2+2 \lambda) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}] \cdot(3 \hat{\mathrm{i}}-\hat{\mathrm{j}})=0} \\
\therefore \quad & 3(2+2 \lambda)-(3+\lambda)=0 \\
\therefore \quad & 6+6 \lambda-3-\lambda=0 \\
\therefore \quad & 3+5 \lambda=0 \\
\therefore \quad & \lambda=\frac{-3}{5}
\end{array}$
$\begin{array}{ll}
& (\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}) \cdot \overline{\mathrm{c}}=0 \\
\therefore \quad & {[(2+2 \lambda) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}] \cdot(3 \hat{\mathrm{i}}-\hat{\mathrm{j}})=0} \\
\therefore \quad & 3(2+2 \lambda)-(3+\lambda)=0 \\
\therefore \quad & 6+6 \lambda-3-\lambda=0 \\
\therefore \quad & 3+5 \lambda=0 \\
\therefore \quad & \lambda=\frac{-3}{5}
\end{array}$
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