Question

Let the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\pi }{6}$, between vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi }{4}$ and between vectors $\overrightarrow{c}$ and $\overrightarrow{a}$ is $\frac{\pi }{3}$. The angle, the vector $\overrightarrow{a}$ makes with the plane containing vectors $\overrightarrow{b}$ and $\overrightarrow{c}$, is

• A. ${\cos }^{-1}\sqrt{1-\sqrt{\frac{2}{3}}}$
• B. ${\cos }^{-1}\sqrt{2-\sqrt{\frac{3}{2}}}$
• C. ${\sin }^{-1}\sqrt{\sqrt{\frac{2}{3}}-1}$
• D. ${\sin }^{-1}\sqrt{\sqrt{\frac{2}{3}}+1}$

Answer 1

Answer: (B), (C)

The correct options are
B

${\cos }^{-1}\sqrt{2-\sqrt{\frac{3}{2}}}$

C

${\sin }^{-1}\sqrt{\sqrt{\frac{2}{3}}-1}$

$|\hat{a}\times (\hat{b}\times \hat{c}){|}^2=|(\hat{a}.\hat{c})\hat{b}-(\hat{a}.\hat{b})\hat{c}{|}^2$

Let the angle, vector $\overrightarrow{a}$ makes with the plane containing vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ be $\theta$.

$\Rightarrow {\sin }^2(90-\theta ).{\sin }^2\frac{\pi }{4}={|\left(\cos \frac{\pi }{3}\right)\hat{b}-\left(\cos \frac{\pi }{6}\right)\hat{c}|}^2$

$\Rightarrow \frac{{\cos }^2\theta }{2}=\frac{1}{4}+\frac{3}{4}-\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}=1-\frac{\sqrt{3}}{2\sqrt{2}}$

$\Rightarrow {\cos }^2\theta =2-\sqrt{\frac{3}{2}}$ ... (1)

$\Rightarrow \cos \theta =\sqrt{2-\sqrt{\frac{3}{2}}}$

$\Rightarrow \theta ={\cos }^{-1}\sqrt{2-\sqrt{\frac{3}{2}}}$

Also, from (1), ${\sin }^2\theta =1-{\cos }^2\theta =\sqrt{\frac{3}{2}}-1$

$\Rightarrow \sin \theta =\sqrt{\sqrt{\frac{3}{2}}-1}$

$\Rightarrow \theta ={\sin }^{-1}\sqrt{\sqrt{\frac{3}{2}}-1}$