Question

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

• A. -1
• B. 1
• C. -4
• D. 2

Answer 1

Answer: (C)

Let angle between $b$ and $c$ is $\alpha$, then
$|b\times c|=\sqrt{15}$ (given)
$\Rightarrow |b\Vert c|\sin \alpha =\sqrt{15}$
$\Rightarrow \sin \alpha =\frac{\sqrt{15}}{4}$
$\therefore \cos \alpha =\frac{1}{4}$
Also, $b-2c=\lambda a$
$\Rightarrow (b-2c{)}^2={\lambda }^2(a{)}^2$
$\Rightarrow |b{|}^2+4|c{|}^2-4b\cdot c={\lambda }^2|a{|}^2$
$\Rightarrow 16+4-4\{|b\Vert c|\cos \alpha \}={\lambda }^2$
$\Rightarrow {\lambda }^2=16$
$\Rightarrow \lambda =\pm 4$