JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Vector Algebra

Since $\vec{a}\times \vec{b}$ is perpendicular to $\vec{b}$, their dot product vanishes: $(13\hat{i}-\hat{j}-4\hat{k})\cdot (\hat{i}+\hat{j}+\lambda \hat{k})=13-1-4\lambda =0$, giving $\lambda =3$ and $\vec{b}=\hat{i}+\hat{j}+3\hat{k}$ with $|\vec{b}{|}^2=11$. To recover $\vec{a}$, use $\vec{b}\times (\vec{a}\times \vec{b})=|\vec{b}{|}^2\vec{a}-(\vec{a}\cdot \vec{b})\vec{b}$. Computing $\vec{b}\times (13\hat{i}-\hat{j}-4\hat{k})=-\hat{i}+43\hat{j}-14\hat{k}$ and using $\vec{a}\cdot \vec{b}=-21$ gives $11\vec{a}=(-\hat{i}+43\hat{j}-14\hat{k})-21\vec{b}=-22\hat{i}+22\hat{j}-77\hat{k}$, so $\vec{a}=-2\hat{i}+2\hat{j}-7\hat{k}$. Then $\vec{b}-\vec{a}=3\hat{i}-\hat{j}+10\hat{k}$ and $\vec{b}+\vec{a}=-\hat{i}+3\hat{j}-4\hat{k}$. The first, dotted with $\hat{k}-\hat{j}$, gives $(-1)(-1)+(10)(1)=11$, and the second, dotted with $\hat{i}-\hat{k}$, gives $(-1)(1)+(-4)(-1)=3$, so the required expression equals $11+3=14$.