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The magnitude of the projection of the vector $2 \hat{i}+\hat{j}+\hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{6}}$
The vector perpendicular to both vectors containing $(\hat{i}+\hat{j}+\hat{k})$ and $(\hat{i}+2 \hat{j}+3 \hat{k})$ is
$\begin{aligned}
& =(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}+2 \hat{j}+3 \hat{k}) \\
& =\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
1 & 2 & 3
\end{array}\right| \\
& =\hat{i}-2 \hat{j}+\hat{k}
\end{aligned}$
Therefore, the magnitude of the projection of vector $(2 \hat{i}+\hat{j}+\hat{k})$ on $(\hat{i}-2 \hat{j}+\hat{k})$ is
$\begin{aligned}
& =\left|\frac{(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{1^2+(-2)^2+(1)^2}}\right| \\
& =\left|\frac{2-2+1}{\sqrt{6}}\right| \\
& =\frac{1}{\sqrt{6}}
\end{aligned}$
$\begin{aligned}
& =(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}+2 \hat{j}+3 \hat{k}) \\
& =\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
1 & 2 & 3
\end{array}\right| \\
& =\hat{i}-2 \hat{j}+\hat{k}
\end{aligned}$
Therefore, the magnitude of the projection of vector $(2 \hat{i}+\hat{j}+\hat{k})$ on $(\hat{i}-2 \hat{j}+\hat{k})$ is
$\begin{aligned}
& =\left|\frac{(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{1^2+(-2)^2+(1)^2}}\right| \\
& =\left|\frac{2-2+1}{\sqrt{6}}\right| \\
& =\frac{1}{\sqrt{6}}
\end{aligned}$
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