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Let $\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$, $\mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. The vector $\mathbf{x}$ such that $\mathbf{x} \cdot \mathbf{c}=60$ and perpendicular to both $\mathbf{a}, \mathbf{b}$ is
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Verified Answer
The correct answer is:
$\hat{i}+34 \hat{j}+25 \hat{k}$
We have,
$\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Since, $\mathbf{a} \perp \mathbf{x}$ and $\mathbf{b} \perp \mathbf{x}$
$\therefore \quad \mathbf{x}=\lambda(\mathbf{a} \times \mathbf{b})=\lambda\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & -3 & 4 \\
7 & 2 & -3
\end{array}\right|=\lambda(\hat{\mathbf{i}}+34 \hat{\mathbf{j}}+25 \hat{\mathbf{k}})$
Now, we have $\mathbf{x} \cdot \mathbf{c}=60$
$\begin{array}{lc}
\Rightarrow & (\lambda \hat{\mathbf{i}}+34 \lambda \hat{\mathbf{j}}+25 \lambda \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=60 \\
\Rightarrow & \lambda+34 \lambda+25 \lambda=60 \Rightarrow 60 \lambda=60 \Rightarrow \lambda=1 \\
\therefore & \mathbf{x}=\hat{\mathbf{i}}+34 \hat{\mathbf{j}}+25 \hat{\mathbf{k}}
\end{array}$
$\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Since, $\mathbf{a} \perp \mathbf{x}$ and $\mathbf{b} \perp \mathbf{x}$
$\therefore \quad \mathbf{x}=\lambda(\mathbf{a} \times \mathbf{b})=\lambda\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & -3 & 4 \\
7 & 2 & -3
\end{array}\right|=\lambda(\hat{\mathbf{i}}+34 \hat{\mathbf{j}}+25 \hat{\mathbf{k}})$
Now, we have $\mathbf{x} \cdot \mathbf{c}=60$
$\begin{array}{lc}
\Rightarrow & (\lambda \hat{\mathbf{i}}+34 \lambda \hat{\mathbf{j}}+25 \lambda \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=60 \\
\Rightarrow & \lambda+34 \lambda+25 \lambda=60 \Rightarrow 60 \lambda=60 \Rightarrow \lambda=1 \\
\therefore & \mathbf{x}=\hat{\mathbf{i}}+34 \hat{\mathbf{j}}+25 \hat{\mathbf{k}}
\end{array}$
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