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Q.
If $a, b, c$ are three non-parallel unit vectors such that $a \times(b \times c)=\frac{1}{2} b$, then the angles which a makes with $b$ and $c$ are
Vector Algebra
Solution:
We have, $a \times(b \times c)=\frac{1}{2} b$
$\therefore (a \cdot c) b-(a \cdot b) c=\frac{1}{2} b $
$\Rightarrow \left(a \cdot c-\frac{1}{2}\right) b-(a \cdot b) c=0\,\,\,(1)$
Since $b$ and $c$ are non-parallel , (Given)
$\therefore $ (i) exists if coefficients of $b$ and $c$ vanish separately
i.e., $a \cdot c-\frac{1}{2}=0$ and $a \cdot b=0$.
$\Rightarrow a \cdot c=\frac{1}{2}$ and $a \cdot b=0$
Let $\theta_{1}$ and $\theta_{2}$ be the angles which $a$ makes with $b$ and $c$, respectively.
Also $a, b, c$ are unit vectors,
$\therefore \cos \theta_{2}=\frac{1}{2}$ and $\cos \theta_{1}=0$
i.e., $\theta_{2}=60^{\circ}$ and $\theta_{1}=90^{\circ}$
Hence, $\theta_{1}=90^{\circ}$ and $\theta_{2}=60^{\circ}$.