CUET Preparation Today
The projection of $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$ is: |
$\sqrt{2}$ $\frac{1}{\sqrt{2}}$ 2 ×0 |
×0 |
The correct answer is Option (4) → 0 Let $\vec a$ and $\vec b$ be two vectors. Projection of $\vec a$ on $\vec b$ = $\frac{a.b}{|b|^2}\hat b$ $a=\hat i-\hat j$ $b=\hat i+\hat j$ then, $a.b=1×1-1×1=0$ ∴ Projection of $\vec a$ on $\vec b$ = 0 |
