Find the projection of the vector $\hat i - \hat j$ on the vector $\hat i + \hat j$.

Let $\vec a = \hat i - \hat j$ and $\vec b = \hat i + \hat j$ , $|\vec b| = \sqrt {{1^2} + {1^2}} = \sqrt {1 + 1} = \sqrt 2$

Also,. $\vec a\vec b = (\hat i - \hat j) \cdot (\hat i + \hat j) = (1)(1) + ( - 1)(1) = 1 - 1 = 0$
$\therefore$ Projection of $\vec a$ on $\vec b$ $= \cfrac{{\vec a \cdot \vec b}}{{|\vec b|}} = \cfrac{0}{{\sqrt 2 }} = 0$.