Question

Let $a , b , c$, be three vectors such that $c \neq 0$ and $a \cdot b=2 a \cdot c ,| a |=| c |=1,| b |=4$ and $| b \times c |=\sqrt{15}$, if $b -2 c=\lambda \alpha$. 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)
$\Rightarrow | b \| c | \sin \alpha=\sqrt{15}$
$\Rightarrow \sin \alpha=\frac{\sqrt{15}}{4}$
$\therefore \cos \alpha=\frac{1}{4}$
Also, $b -2 c =\lambda a$
$\Rightarrow (b-2 c)^{2}=\lambda^{2}(a)^{2}$
$\Rightarrow | b |^{2}+4| c |^{2}-4 b \cdot c =\lambda^{2}| a |^{2}$
$\Rightarrow 16+4-4\{| b \| c | \cos \alpha\}=\lambda^{2}$
$\Rightarrow \lambda^{2}=16$
$\Rightarrow \lambda=\pm 4$