JEE Advanced 2011 Paper 1Question 1
Let $\vec a=\hat i+\hat j+\hat k$, $\vec b=\hat i-\hat j+\hat k$ and $\vec c=\hat i-\hat j-\hat k$ be three vectors. A vector $\vec v$ in the plane of $\vec a$ and $\vec b$, whose projection on $\vec c$ is $\frac{1}{\sqrt3}$, is given by
Vector Algebra
Solution
In the plane of $\vec a,\vec b$: $\vec v=(x+y,\,x-y,\,x+y)$. Projection on $\vec c$: $\frac{(x+y)-(x-y)-(x+y)}{\sqrt3}=\frac{y-x}{\sqrt3}=\frac1{\sqrt3}$, so $y-x=1$. $(3,-1,3)$ has $x+y=3$, $x-y=-1$, so $y=2,x=1$ and fits.