JEE Main 2025 (Online) 29th January Morning ShiftQuestion 1
Let $\vec a=2\hat i-\hat j+3\hat k$, $\vec b=3\hat i-5\hat j+\hat k$ and $\vec c$ be a vector such that $\vec a\times\vec c=\vec c\times\vec b$ and $(\vec a+\vec c)\cdot(\vec b+\vec c)=168$. Then the maximum value of $|\vec c|^2$ is :
Vector Algebra
Solution
$\vec a\times\vec c=\vec c\times\vec b$ gives $(\vec a+\vec b)\times\vec c=0$, so $\vec c=k(5,-6,4)$. $(\vec a+\vec c)\cdot(\vec b+\vec c)=14+77k+77k^2=168$ gives $k^2+k-2=0$, $k=1,-2$. Max $|\vec c|^2=77\cdot4=308$.