Question

If $a\times (b\times c)+(a\cdot b)b=(4-2\beta -\sin \alpha )b+\left({\beta }^2-1\right)c$ and $(c\cdot c)a=c$, while $b$ and $c$ are non-collinear, then

• A. $\alpha =\frac{\pi }{2},\beta =-1$
• B. $\alpha =\frac{\pi }{2},\beta =1$
• C. $\alpha =\frac{\pi }{3},\beta =-1$
• D. $\alpha =\frac{\pi }{3},\beta =-1$

Answer 1

Answer: (B)

We have, $a\times (b\times c)+(a\cdot b)b$
$=(4-2\beta -\sin \alpha )b+\left(b^2-1\right)c(1)$
and $(c\cdot c)a=c(2)$
where $b$ and $c$ are non-collinear vectors and $\alpha ,\beta$ are scalars
From (2), $(c\cdot c)a\cdot c=c\cdot c$
$\therefore a\cdot c=1(3)$
From (1), we get
$(a\cdot c)b-(a\cdot b)c+(a\cdot b)b$
$=(4-2\beta -\sin \alpha )b+\left({\beta }^2-1\right)c$
or $[1+(a\cdot b)]b-(a\cdot b)c$
$=(4-2\beta -\sin \alpha )b+\left({\beta }^2-1\right)c$
$\Rightarrow 1+(a\cdot b)=4-2\beta -\sin \alpha (4)$
and $a\cdot b=-\left(b^2-1\right)(5)$
$\therefore \sin \alpha =1+(1-\beta {)}^2$
$\Rightarrow \beta =1,\sin \alpha =1$
i.e., $\alpha =\frac{\pi }{2}+2n\pi ,n\in I$