Vectors 2 Question 18
18. The unit vector perpendicular to the plane determined by $P(1,-1,2), Q(2,0,-1)$ and $R(0,2,1)$ is . .
(1983, 2M)
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Answer:
Correct Answer: 18. (b, d)
Solution:
- A unit vector perpendicular to the plane determined by
$$ \begin{aligned} P, Q, R \quad & = \pm \frac{(\overrightarrow{\mathbf{P Q}}) \times(\overrightarrow{\mathbf{P R}})}{|\overrightarrow{\mathbf{P Q}} \times \overrightarrow{\mathbf{P R}}|} \\ \therefore \quad \text { Unit vector } & = \pm \frac{(\overrightarrow{\mathbf{P Q}}) \times(\overrightarrow{\mathbf{P R}})}{|\overrightarrow{\mathbf{P Q}} \times \overrightarrow{\mathbf{P R}}|} \end{aligned} $$
where, $\quad \overrightarrow{\mathbf{P Q}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
and $\overrightarrow{\mathbf{P R}}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\therefore \quad \overrightarrow{\mathbf{P Q}} \times \overrightarrow{\mathbf{P R}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 1 & 1 & -3 \ -1 & 3 & -1\end{array}\right|$
$$ =\hat{\mathbf{i}}(-1+9)-\hat{\mathbf{j}}(-1-3)+\hat{\mathbf{k}}(3+1) $$
$$ =8 i+4 j+4 \hat{k} $$
$\Rightarrow \quad|\overrightarrow{\mathbf{P Q}} \times \overrightarrow{\mathbf{P R}}|=4 \sqrt{4+1+1}=4 \sqrt{6}$
$\therefore \quad \frac{\overrightarrow{\mathbf{P Q}} \times \overrightarrow{\mathbf{P R}}}{|\overrightarrow{\mathbf{P Q}} \times \overrightarrow{\mathbf{P R}}|}= \pm \frac{4(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{4 \sqrt{6}}$
$$ = \pm \frac{(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{6}} $$
