If $\vec{a} imes(\vec{b} imes \vec{c})+(\vec{a} \cdot \vec{b}) \vec{b}=(4-2 \beta-\sin \alpha) \

Question

If $\vec{a} 	imes(\vec{b} 	imes \vec{c})+(\vec{a} \cdot \vec{b}) \vec{b}=(4-2 \beta-\sin \alpha) \vec{b}+\left(\beta^2-1ight) \vec{c}$ and $(\vec{c} \cdot \vec{c}) \vec{a}=\vec{c}$ where $\vec{b}$ and $\vec{c}$ are noncollinear and $\alpha, \beta$ are scalars then $\beta=$ $\_\_\_\_$

Professor PC Thomas Classes · Answer

$\vec{a} 	imes(\vec{b} 	imes \vec{c})+(\vec{a} \cdot \vec{b}) \vec{b}=(4-2 \beta-\sin \alpha) \vec{b}+\left(\beta^2-1ight) \vec{c}$ ...(1)
$\Rightarrow(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=(4-2 \beta-\sin \alpha) \vec{b}+\left(\beta^2-1ight) \vec{c}$ 
$\Rightarrow \vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=4-2 \beta-\sin \alpha$    ...(2)
$\vec{a} \cdot \vec{b}=1-\beta^2$    ...(3)
 As $(\vec{c} \cdot \vec{c}) \vec{a}=\vec{c} \Rightarrow \vec{a} \cdot \vec{c}=1 \ldots$ (4)
$\Rightarrow 2-\beta^2=4-2 \beta-\sin \alpha \Rightarrow \sin \alpha=\beta^2-2 \beta+2$ \ $\Rightarrow \sin \alpha=(\beta-1)^2+1 \Rightarrow \beta=1, \alpha=(4 n+1) \frac{\pi}{2}$