JEE Main 2022 (Online) 29th July Evening ShiftQuestion 1
Let $\vec a,\vec b,\vec c$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is $14$ and $(\vec a\times\vec b)\cdot(\vec b\times\vec c)+(\vec b\times\vec c)\cdot(\vec c\times\vec a)+(\vec c\times\vec a)\cdot(\vec a\times\vec b)=168$, then $|\vec a|+|\vec b|+|\vec c|$ is equal to :
Vector Algebra
Solution
Coplanar, concurrent, equal angles $\Rightarrow$ $120^\circ$ pairwise. Let magnitudes $p,q,r$, $pqr=14$. $(\vec a\times\vec b)\cdot(\vec b\times\vec c)=(\vec a\cdot\vec b)(\vec b\cdot\vec c)-(\vec a\cdot\vec c)(\vec b\cdot\vec b)=\frac{pq^2r}4+\frac{pq^2r}2=\frac34pq^2r$. Cyclically the sum is $\frac34pqr(p+q+r)=\frac34\cdot14\,(p+q+r)=168\Rightarrow p+q+r=16$. Option (C).