Question

Let $\overrightarrow{ A }$ be vector parallel to line of intersection of planes $P_{1}$ and $P_{2}$ through origin. $P_{1}$ is parallel to the vectors $2 \hat{ j }+3 \hat{ k }$ and $4 \hat{ j }-3 \hat{ k }$ and $P_{2}$ is parallel to $\hat{ j }-\hat{ k }$ and $3 \hat{ i }+3 \hat{ j }$, then the angle between vector $\overrightarrow{ A }$ and $2 \hat{ i }+\hat{ j }-2 \hat{ k }$ is

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (B), (D)

Let vector $\overrightarrow{ A O }$ be parallel to line of intersection of planes $P_{1}$ and $P_{2}$ through origin.
Normal to plane $p_{1}$ is
$\overrightarrow{ n _{1}}=[(2 \hat{ j }+3 \hat{ k }) \times(4 \hat{ j }-3 \hat{ k })]=-18 \hat{ i }$
Normal to plane $p_{2}$ is
$\overrightarrow{ n }_{2}=(\hat{ j }-\hat{ k }) \times(3 \hat{ i }+3 \hat{ j })=3 \hat{ i }-3 \hat{ j }-3 \hat{ k }$
So, $\overrightarrow{ OA }$ is parallel to $\pm\left(\overrightarrow{ n _{1}} \times \overrightarrow{ n _{2}}\right)=54 \hat{ j }-54 \hat{ k }$.
$\therefore$ Angle between $54(\hat{ j }-\hat{ k })$ and $(2 \hat{ i }+\hat{ j }-2 \hat{ k })$ is
$\cos \theta=\pm\left(\frac{54+108}{3 \cdot 54 \cdot \sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}}$
$\therefore \theta=\frac{\pi}{4}, \frac{3 \pi}{4}$
Hence, (b) and (d) are correct answers.