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If $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}$ 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:
$\frac{-11}{5}$
Let $\overline{\mathrm{d}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$
$\begin{aligned}
\overline{\mathrm{d}} & =(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\
& =2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+2 \lambda \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}} \\
& =(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}
\end{aligned}$
Now, $\overline{\mathrm{d}}$ is perpendicular to $\overline{\mathrm{c}}$.
$\begin{aligned}
& \therefore \quad \overline{\mathrm{c}} \cdot \overline{\mathrm{d}}=0 \\
& \Rightarrow(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot[(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}]=0 \\
& \Rightarrow 1(2 \lambda+2)+3(3+\lambda)=0 \\
& \Rightarrow 2 \lambda+2+9+3 \lambda=0 \\
& \Rightarrow 5 \lambda+11=0 \\
& \Rightarrow \lambda=\frac{-11}{5}
\end{aligned}$
$\begin{aligned}
\overline{\mathrm{d}} & =(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\
& =2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+2 \lambda \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}} \\
& =(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}
\end{aligned}$
Now, $\overline{\mathrm{d}}$ is perpendicular to $\overline{\mathrm{c}}$.
$\begin{aligned}
& \therefore \quad \overline{\mathrm{c}} \cdot \overline{\mathrm{d}}=0 \\
& \Rightarrow(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot[(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}]=0 \\
& \Rightarrow 1(2 \lambda+2)+3(3+\lambda)=0 \\
& \Rightarrow 2 \lambda+2+9+3 \lambda=0 \\
& \Rightarrow 5 \lambda+11=0 \\
& \Rightarrow \lambda=\frac{-11}{5}
\end{aligned}$
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