Vectors 2 Question 12
12. Let $\overrightarrow{\mathbf{A}}$ be vector parallel to line of intersection of planes $P _1$ and $P _2$ through origin. $P _1$ is parallel to the vectors $2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $P _2$ is parallel to $\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$, then the angle between vector $\overrightarrow{\mathbf{A}}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ is
(2006, 5M)
(a) $\frac{\pi}{2}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{6}$
(d) $\frac{3 \pi}{4}$
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Answer:
Correct Answer: 12. (a, c)
Solution:
- Let vector $\overrightarrow{\mathbf{A O}}$ be parallel to line of intersection of planes $P _1$ and $P _2$ through origin.
Normal to plane $p _1$ is
$$ \overrightarrow{\mathbf{n} _1}=[(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})]=-18 \hat{\mathbf{i}} $$
Normal to plane $p _2$ is
$$ \overrightarrow{\mathbf{n}} _2=(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})=3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $$
So, $ \overrightarrow{\mathbf{O A}}$ is parallel to $\pm\left(\overrightarrow{\mathbf{n}} _1 \times \overrightarrow{\mathbf{n}} _2\right)=54 \hat{\mathbf{j}}-54 \hat{\mathbf{k}}$.
$\therefore$ Angle between $54(\hat{\mathbf{j}}-\hat{\mathbf{k}})$ and $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ is
$$ \begin{aligned} \cos \theta & = \pm\left(\frac{54+108}{3 \cdot 54 \cdot \sqrt{2}}\right)= \pm \frac{1}{\sqrt{2}} \\ \therefore \quad \theta & =\frac{\pi}{4}, \frac{3 \pi}{4} \end{aligned} $$
Hence, (b) and (d) are correct answers.
