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

Answer

$\text{Let}\ \ \vec{a}=\hat{i}-\hat{j}=\hat{i}-\hat{j}+0\hat {k}\ \text{and}\ \vec{b}=\hat{i}+\hat{j}$ $=\hat{i}+\hat{j}+0\hat {k}$
Projection of vector  $\vec{a}\ \text{and}\ \vec{b}=\frac{\vec{a}.\vec{b}}{\big|\vec{b}\big|}=\frac{(1)(1)+(-1)(1)+0(0)}{\sqrt{(1)^2+(1)^2+(0)^2}}$ $=\frac{1-1+0}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0$
If projection of vector $\vec{a}\ \text{and}\ \vec{b}$ is zero, then vector $\vec{a}$ is perpendicular to $\vec{b}.$

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