Prove that


Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ 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 $\vec{a}$ makes with vectors

$\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ respectively and $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{1}{2} \overrightarrow{\mathrm{b}}$, then $|\alpha-\beta|$ is equal to :

  1. $60^{\circ}$

  2. $30^{\circ}$

  3. $90^{\circ}$

  4. $45^{\circ}$

Correct Option: , 2


$(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \cdot \vec{c}=\frac{1}{2} \vec{b}$

$\because \quad \vec{b} \& \vec{c}$ are linearly independent

$\therefore \quad \vec{a} \cdot \vec{c}=\frac{1}{2} \& \vec{a} \cdot \vec{b}=0$

(All given vectors are unit vectors)

$\therefore \quad \overrightarrow{\mathrm{a}} \wedge \overrightarrow{\mathrm{c}}=60^{\circ} \quad \& \quad \overrightarrow{\mathrm{a}} \wedge \overrightarrow{\mathrm{b}}=90^{\circ}$


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