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A vector perpendicular to the plane containing the vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $3 \hat{i}-2 \hat{j}-\hat{k}$ is inclined to the vector $\hat{i}+\hat{j}+\hat{k}$ at an angle
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Verified Answer
The correct answer is:
$\tan ^{-1} \sqrt{14}$
A vector perpendicular to the plane is
$\begin{aligned}
& (\hat{i}-2 \hat{j}-\hat{k}) \times(3 \hat{i}-2 \hat{j}-\hat{k}) \\
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & -1 \\
3 & -2 & -1
\end{array}\right|=-2 \hat{j}+4 \hat{k} \\
& \Rightarrow \text { unit vector } \hat{a}=\frac{-2 \hat{j}+4 \hat{k}}{\sqrt{4+16}}=\frac{-2 \hat{j}+4 \hat{k}}{2 \sqrt{5}}
\end{aligned}$
Angle between the unit vector and $\vec{r}=\hat{i}+\hat{j}+\hat{k}$
$=\cos ^{-1} \frac{\vec{r} \cdot \hat{a}}{|\vec{r}| \cdot|\hat{a}|}=\cos ^{-1} \frac{1}{\sqrt{15}}=\tan ^{-1} \sqrt{14}$
$\begin{aligned}
& (\hat{i}-2 \hat{j}-\hat{k}) \times(3 \hat{i}-2 \hat{j}-\hat{k}) \\
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & -1 \\
3 & -2 & -1
\end{array}\right|=-2 \hat{j}+4 \hat{k} \\
& \Rightarrow \text { unit vector } \hat{a}=\frac{-2 \hat{j}+4 \hat{k}}{\sqrt{4+16}}=\frac{-2 \hat{j}+4 \hat{k}}{2 \sqrt{5}}
\end{aligned}$
Angle between the unit vector and $\vec{r}=\hat{i}+\hat{j}+\hat{k}$
$=\cos ^{-1} \frac{\vec{r} \cdot \hat{a}}{|\vec{r}| \cdot|\hat{a}|}=\cos ^{-1} \frac{1}{\sqrt{15}}=\tan ^{-1} \sqrt{14}$
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