Let $\vec a,\vec b,\vec c$ be three vectors satisfying $\vec a×\vec b=2(\vec a×\vec c), |\vec a|=|\vec c|=1,|\vec b|=4$ and $|\vec b×\vec c| =\sqrt{15}$. If $\vec b-2\vec c=λ\vec a$. Then, λ equals
Answer & explanation
Correct answer: option 4
Let θ be the angle between $\vec b$ and $\vec c$. Then,
$|\vec b×\vec c| =\sqrt{15}$
$⇒|\vec b||\vec c|\sin θ=\sqrt{15}⇒\sin θ=\frac{\sqrt{15}}{4}⇒\cos θ=\frac{1}{4}$
Now,
$\vec b-2\vec c=λ\vec a$
$⇒|\vec b-2\vec c|^2=λ^2|\vec a|^2$
$⇒|\vec b|^2+4|\vec c|^2-4(\vec b.\vec c)=λ^2|\vec a|^2$
$⇒16+4-4|\vec b||\vec c|\cos θ=λ^2$
$⇒20-16 \cos θ=λ^2⇒20-4=λ^2⇒λ=±4$.