JEE Main · 2010 · 11 Apr 2010 · Offline · Vector Algebra

Mutually orthogonal vectors have every pairwise dot product zero. From $\vec{a}\cdot \vec{c}=0$ with $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$ we get $\lambda -1+2\mu =0$, and from $\vec{b}\cdot \vec{c}=0$ with $\vec{b}=2\hat{i}+4\hat{j}+\hat{k}$ we get $2\lambda +4+\mu =0$. The first gives $\lambda =1-2\mu$; substituting, $2(1-2\mu )+4+\mu =0$ leads to $6-3\mu =0$, so $\mu =2$ and $\lambda =1-4=-3$. One checks $\vec{a}\cdot \vec{b}=2-4+2=0$ as well, so the three are indeed mutually orthogonal, and $(\lambda ,\mu )=(-3,2)$.