Question

. If a, b, c are three unit vectors such that $a\times (b\times c)=\frac{1}{2}$b then the angles between a, b and a, c are

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

Answer 1

The correct option is B ${90}^{\circ },{60}^{\circ }$

Suppose if possible b, c are collinear. Then $b\times c=0\Rightarrow a\times (b\times c)=0\Rightarrow \frac{1}{2}b=0\Rightarrow b=0$
It is a contradiction to the fact that b is a unit vector. $\therefore b,c$ are non collinear.
$a\times (b\times c)=\frac{1}{2}b\Rightarrow (a.c)b-(a.b)c=\frac{1}{2}b\Rightarrow (a.c-\frac{1}{2})b-(a.b)c=0$
$\Rightarrow a.c-\frac{1}{2}=0,a.b=0$ [Since b, c are not collinear]
$a.b=0\Rightarrow (a,b)={90}^{\circ }$
$a.c-\frac{1}{2}=0\Rightarrow a.c=\frac{1}{2}\Rightarrow |a||c|cos(a,c)=\frac{1}{2}\Rightarrow cos(a,c)=\frac{1}{2}\Rightarrow (a,c)={60}^{\circ }$