Question

Let $\vec{a} , \vec{b} \, and \, \vec{c}$be three unit vectors, out of which vectors $\vec{b}$ and $\vec{c}$ are non-parallel. If $\alpha$ and $\beta$ are the angles which vector a $\vec{a}$ makes with vectors $\vec{b} \, and \, \vec{c}$respectively and $\vec{a} \times \, (\vec{b} \times \vec{c}) \, = \, \frac{1}{2} \vec{b}$,then $|\alpha - \beta|$is equal to :

• A. $60^{\circ}$
• B. $30^{\circ}$
• C. $90^{\circ}$
• D. $45^{\circ}$

Answer 1

Answer: (B)

$(\vec{a}.\vec{c})\vec{b} - (\vec{a}.\vec{b}).\vec{c} \, = \, \frac{1}{2} \, \vec{b}$
$\because \, \vec{b}$ & $\vec{c}$ are linearly independent
$\therefore \, \, \vec{a}. \vec{c} \, = \, \frac{1}{2} $& $\vec{a} . \vec{b} \, = \, 0$
(All given vectors are unit vectors)
$\therefore \, \, \vec{a} \hat{} \vec{c} \, = 60^{\circ} $ & $\vec{a} \hat{} \vec{b} \, = 90^{\circ}$
$\therefore \, \, |\alpha - \beta| = 30^{\circ}$