JEE Main 2025 Jan 29 Shift 1, Mathematics Q5: Scalar and Vector Triple Product

From $\vec{a}\times \vec{c}=\vec{c}\times \vec{b}$ we get $(\vec{a}+\vec{b})\times \vec{c}=\vec{0}$, so $\vec{c}$ is parallel to $\vec{a}+\vec{b}$.

Compute $\vec{a}+\vec{b}=5\hat{i}-6\hat{j}+4\hat{k}$, with $|\vec{a}+\vec{b}{|}^2=25+36+16=77.$ Write $\vec{c}=\lambda (\vec{a}+\vec{b})$, so $|\vec{c}{|}^2=77{\lambda }^2.$

Compute $\vec{a}\cdot \vec{b}=6+5+3=14.$ Expanding the dot product condition:

$(\vec{a}+\vec{c})\cdot (\vec{b}+\vec{c})=\vec{a}\cdot \vec{b}+\vec{c}\cdot (\vec{a}+\vec{b})+|\vec{c}{|}^2=14+77\lambda +77{\lambda }^2=168.$

So $77{\lambda }^2+77\lambda -154=0\Rightarrow {\lambda }^2+\lambda -2=0\Rightarrow \lambda =1$ or $\lambda =-2.$

$|\vec{c}{|}^2$ is maximised at $|\lambda |=2$: $|\vec{c}{|}_{\max }^2=77\cdot 4=308.$