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Let $\mathbf{x}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{y}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{k}}$. Then, the vector $\mathbf{r}$ of magnitude $\sqrt{21}$ satisfying $\mathbf{r} \times \mathbf{x}=\mathbf{y} \times \mathbf{x}$ and $\mathbf{r} \times \mathbf{y}=\mathbf{x} \times \mathbf{y}$, is
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Verified Answer
The correct answer is:
$4 \hat{i}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
Given,
$\begin{aligned}
& \mathbf{x}=\hat{\mathbf{i}}+\hat{\mathfrak{j}} y=3 \hat{\hat{i}}-2 \hat{\mathbf{k}} \\
& \mathbf{r} \times \mathbf{x}=\mathbf{y} \times \mathbf{x} \text { and } \mathbf{r} \times \mathbf{y}=\mathbf{x} \times \mathbf{y} \\
& (\mathrm{r}-\mathbf{y}) \times \mathbf{x}=0
\end{aligned}$
$\mathrm{r}-\mathrm{y}$ is parallel to $\mathrm{x}$.
$\begin{array}{llll}
& \therefore & \mathbf{r} & =\mathbf{y}+\lambda \mathrm{x}=3 i-2 k+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \\
& =(3+\lambda) \hat{\mathbf{i}}+\lambda \hat{\mathrm{j}}-2 \hat{\mathbf{k}} \\
& & |\mathrm{r}| & =\sqrt{2 \mathrm{l}} \\
& \therefore & & \sqrt{(3+\lambda)^2+\lambda^2+4}=\sqrt{2 \mathrm{l}}
\end{array}$
$\begin{array}{ll}
\Rightarrow & 9+6 \lambda+2 \lambda^2+4=21 \\
\Rightarrow & \lambda^2+3 \lambda-4=0 \\
\Rightarrow & (\lambda+4)(\lambda-1)=0 \Rightarrow \lambda=1 \\
\therefore & \mathrm{r}=4 \hat{\hat{\mathrm{i}}}+\hat{\mathrm{j}}-2 \hat{\mathbf{k}}
\end{array}$
$\begin{aligned}
& \mathbf{x}=\hat{\mathbf{i}}+\hat{\mathfrak{j}} y=3 \hat{\hat{i}}-2 \hat{\mathbf{k}} \\
& \mathbf{r} \times \mathbf{x}=\mathbf{y} \times \mathbf{x} \text { and } \mathbf{r} \times \mathbf{y}=\mathbf{x} \times \mathbf{y} \\
& (\mathrm{r}-\mathbf{y}) \times \mathbf{x}=0
\end{aligned}$
$\mathrm{r}-\mathrm{y}$ is parallel to $\mathrm{x}$.
$\begin{array}{llll}
& \therefore & \mathbf{r} & =\mathbf{y}+\lambda \mathrm{x}=3 i-2 k+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \\
& =(3+\lambda) \hat{\mathbf{i}}+\lambda \hat{\mathrm{j}}-2 \hat{\mathbf{k}} \\
& & |\mathrm{r}| & =\sqrt{2 \mathrm{l}} \\
& \therefore & & \sqrt{(3+\lambda)^2+\lambda^2+4}=\sqrt{2 \mathrm{l}}
\end{array}$
$\begin{array}{ll}
\Rightarrow & 9+6 \lambda+2 \lambda^2+4=21 \\
\Rightarrow & \lambda^2+3 \lambda-4=0 \\
\Rightarrow & (\lambda+4)(\lambda-1)=0 \Rightarrow \lambda=1 \\
\therefore & \mathrm{r}=4 \hat{\hat{\mathrm{i}}}+\hat{\mathrm{j}}-2 \hat{\mathbf{k}}
\end{array}$
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