Question

let $\overline{a},\overline{b},\overline{c}$ be unit vectors. Let $\overline{b}$ and $\overline{c}$ are inclined to $\overline{a}$ each at $\theta$. If the angle between $\overline{b}$ and $\overline{c}$ is $\frac{\pi }{3}$ then which is/are correct, given that $x=(\overline{a}\times \overline{b}).(\overline{a}\times \overline{c})).y=|\overline{a}\times \overline{b}-\overline{a}\times \overline{c}|$

• A. $\theta =\frac{\pi }{6}$ gives $x=-1/4$
• B. $\theta =\frac{\pi }{6}$ gives $y=2/3$
• C. $\theta =\frac{\pi }{2}$ gives $x=1/3$
• D. $\theta =\frac{\pi }{2}$ gives $y=1$

Answer 1

The correct option is D $\theta =\frac{\pi }{2}$ gives $y=1$

$|a|=|b|=|c|=1$ $\overrightarrow{b}.\overrightarrow{c}=\frac{1}{2}$

$x=(\overrightarrow{a}\times \overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{c})$ $y=|\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}|$

$|\overrightarrow{a}\times \overrightarrow{b}|=|a||b|\sin \theta$ $=|\overrightarrow{a}\times (\overrightarrow{b}-\overrightarrow{c})|$

$|\overrightarrow{a}\times \overrightarrow{c}|=|a||c|\sin \theta$ $y^2=|a\times b{|}^2+|a\times c{|}^2-2)\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c})$

$\theta =\frac{\pi }{2}$ $y^2=1$

$y=1$