Q33 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
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
- Correct answerA$\tan ^{-1} \sqrt{14}$
- B$\sec ^{-1} \sqrt{14}$
- C$\tan ^{-1} \sqrt{15}$
- D
None of these
Solution
A vector perpendicular to the plane is
$\begin{aligned}<br />& (\hat{i}-2 \hat{j}-\hat{k}) \times(3 \hat{i}-2 \hat{j}-\hat{k}) \\<br />& =\left|\begin{array}{ccc}<br />\hat{i} & \hat{j} & \hat{k} \\<br />1 & -2 & -1 \\<br />3 & -2 & -1<br />\end{array}\right|=-2 \hat{j}+4 \hat{k} \\<br />& \Rightarrow \text { unit vector } \hat{a}=\frac{-2 \hat{j}+4 \hat{k}}{\sqrt{4+16}}=\frac{-2 \hat{j}+4 \hat{k}}{2 \sqrt{5}}<br />\end{aligned}$
Angle between the unit vector and $\vec{r}=\hat{i}+\hat{j}+\hat{k}$
$=\cos ^{-1} \frac{\vec{r} \cdot \hat{a}}{|\vec{r}| \cdot|\hat{a}|}=\cos ^{-1} \frac{1}{\sqrt{15}}=\tan ^{-1} \sqrt{14}$
$\begin{aligned}<br />& (\hat{i}-2 \hat{j}-\hat{k}) \times(3 \hat{i}-2 \hat{j}-\hat{k}) \\<br />& =\left|\begin{array}{ccc}<br />\hat{i} & \hat{j} & \hat{k} \\<br />1 & -2 & -1 \\<br />3 & -2 & -1<br />\end{array}\right|=-2 \hat{j}+4 \hat{k} \\<br />& \Rightarrow \text { unit vector } \hat{a}=\frac{-2 \hat{j}+4 \hat{k}}{\sqrt{4+16}}=\frac{-2 \hat{j}+4 \hat{k}}{2 \sqrt{5}}<br />\end{aligned}$
Angle between the unit vector and $\vec{r}=\hat{i}+\hat{j}+\hat{k}$
$=\cos ^{-1} \frac{\vec{r} \cdot \hat{a}}{|\vec{r}| \cdot|\hat{a}|}=\cos ^{-1} \frac{1}{\sqrt{15}}=\tan ^{-1} \sqrt{14}$
Hence, the correct answer is option: 'A'.