If three vectors $\vec a, \vec b, \vec c$ are such that $\vec a≠ \vec 0$ and $\vec a × \vec b=2(\vec a × \vec c), |\vec a|=|\vec c|=1,|\vec b|=4$ and the angle between $\vec b$ and $\vec c$ is $\cos^{-1}(\frac{1}{4})$, then $\vec b-2\vec c=λ\vec a$ where λ is equal to
Answer & explanation
Correct answer: option 1
We have,
$\vec a × \vec b = 2(\vec a × \vec c) = \vec a × (\vec b-2\vec c)=\vec 0⇒ \vec a$ is parallel to $\vec b-2\vec c$
Now,
$(\vec b-2\vec c)=λ\vec a$
$⇒|\vec b-2\vec c|^2=λ^2|\vec a|^2$
$⇒|\vec b|^2 + 4|\vec c|^2 - 4 (\vec b.\vec c)=λ^2|\vec a|^2$
$⇒16+4-4×|\vec b||\vec c|×\frac{1}{4}=λ^2$
$⇒20-4=λ^2⇒λ=±4$