Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar unit vectors such that $\overrightarrow{a}\times (\overrightarrow{b} \times\overrightarrow{c}) =\frac{(\overrightarrow{b} +\overrightarrow{c})}{\sqrt 2}$ then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

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

Answer 1

Answer: (A)

$\frac{3p}{4}$

Answer 2

Since, $\, \, \, \, \, \, \, \overrightarrow{a}\times (\overrightarrow{b} \times\overrightarrow{c}) =\frac{(\overrightarrow{b} +\overrightarrow{c})}{\sqrt 2}$
$\Rightarrow \, \, \, \, \, \, (\overrightarrow{a}.\overrightarrow{c}) \overrightarrow{b}- (\overrightarrow{a} .\overrightarrow{b}) \overrightarrow{c} =\frac{1}{\sqrt 2}\overrightarrow{b} +\frac{1}{\sqrt 2}\overrightarrow{c}$
On equating the coefficient of $\overrightarrow{c}$ we get
\hspace20mm \overrightarrow{a}.\overrightarrow{b}= - \frac{1}{\sqrt 2} \Rightarrow \, \, \, |\overrightarrow{a}|| \overrightarrow{b}| \, cos \theta =\frac{1}{\sqrt 2}
$\therefore$ \hspace15mm cos \theta = - \frac{1}{\sqrt 2} \Rightarrow \, \theta =\frac{3p}{4}