Question

Let $a,b,c$ be three vectors such that $a\ne 0$ and $a\times b=2a\times c,\left|a\right|=\left|c\right|=1,$ $\left|b\right|=4$ and $|b\times c|=\sqrt{15}$, if $b-2c=\lambda a$, then $\lambda$ equals

• A. $1$
• B. $-1$
• C. $2$
• D. $-4$

Answer 1

The correct option is D $-4$

If the angle between $b$ and $c$ is $\alpha$ then $|b\times c|=\sqrt{15}$

$\Rightarrow \left|b\right|\left|c\right|\sin \alpha =\sqrt{15}$

$\Rightarrow \sin \alpha =\frac{\sqrt{15}}{4}$

$\therefore \cos \alpha =\frac{1}{4}$

Now, $b-2c=\lambda a$

$\Rightarrow {(b-2c)}^2={\lambda }^2{\left(a\right)}^2$

$\Rightarrow {\left(b\right)}^2+4{\left(c\right)}^2-4b.c=\lambda {\left(a\right)}^2$

$\Rightarrow 16+4-4\left\{\left|b\right|\left|c\right|\cos \alpha \right\}={\lambda }^2$

$\therefore {\lambda }^2=16\Rightarrow \lambda =\pm 4$