Three vectors $\vec a, \vec b, \vec c$ are such that $\vec a×\vec b=3(\vec a × \vec c)$. Also, $|\vec a|=|\vec b|=1, |\vec c|=\frac{1}{3}$. If the angle between $\vec b$ and $\vec c$ is 60°, then
Answer & explanation
Correct answer: option 2
The correct answer is Option 2: $\vec b = 3\vec c -\vec a$
We have,
$\vec a×\vec b=3(\vec a × \vec c)$
$⇒\vec a×\vec b-\vec a×3\vec c=\vec 0$
$⇒\vec a×(\vec b-3\vec c)=\vec 0$
$⇒\vec a||\vec b-3\vec c$
$⇒\vec b-3\vec c=λ\vec a$
$⇒|\vec b-3\vec c|^2=λ^2|\vec a|^2$
$⇒|\vec b|^2+9|\vec c|^2-6(\vec b.\vec c)=λ^2|\vec a|^2$
$⇒2-6×\frac{1}{3}\cos 60° = λ^2$ $[∵|\vec a|=|\vec b|=1,|\vec c|=1/3\,and\,\vec b.\vec c=|\vec b||\vec c|\cos 60°]$
$⇒λ=±1$
Hence, $\vec b-3\vec c=±\vec a ⇒\vec b=3\vec c + \vec a$ and $\vec b = 3\vec c-\vec a$.