JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Vector Algebra

From $\hat{w}\cdot (\hat{u}\times \vec{v})=1$ and $|\hat{u}\times \vec{v}|=1$, we have $|\hat{w}|\cdot 1\cdot \cos \alpha =1$, so $\cos \alpha =1$ (since $|\hat{w}|=1$), meaning $\hat{w}$ is parallel to $\hat{u}\times \vec{v}$. Thus $\hat{u}\times \vec{v}=\hat{w}$. This means $\hat{w}\perp \hat{u}$ and $\hat{w}\perp \vec{v}$. For such $\vec{v}$ to exist, $\hat{u}$ must be perpendicular to $\hat{w}$. Given this, there are infinitely many $\vec{v}$ satisfying the condition (any vector in the plane perpendicular to $\hat{w}$ and making the right angle with $\hat{u}$), confirming (B). If $\hat{u}=u_1\hat{i}+u_2\hat{j}$ (in $xy$-plane), then $\hat{u}\cdot \hat{w}=0$ gives $u_1+u_2=0$, so $|u_1|=|u_2|$, confirming (C). If $\hat{u}=u_1\hat{i}+u_3\hat{k}$ (in $xz$-plane), then $u_1+2u_3=0$, so $|u_1|=2|u_3|$, not $2|u_1|=|u_3|$, so (D) is false.