Projection vector of $\overrightarrow{a}$ on $\overrightarrow{b}$ is

Question

\[(A)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{b}\mid^2}.\overrightarrow{b}\quad(B)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{b}\mid}\quad(C)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{a}\mid}\quad(D)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{a}\mid^2}.\overrightarrow{b}\]

Answer 1

Toolbox:

• Projection of $\overrightarrow a$ on $\overrightarrow b$ is $\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|} $$\hat b$

Projection of $\overrightarrow a$ on $\overrightarrow b$ is $\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|^2} $$\overrightarrow b $

If geometically interpreted, the scalar product of two vectors in the product of modulus of either vector and the projection of the other in its direction.

Thus, the projection of $\overrightarrow a$ on $\overrightarrow b$ is the dot product of $\overrightarrow a$ with the unit vector along $ \overrightarrow b$. is $\large\frac{\overrightarrow b}{|\overrightarrow b|}$

Therefore projection of $\overrightarrow a$ on $\overrightarrow b$ is $\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|^2}$$\:\hat b$

$=\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|^2}$$\:\overrightarrow b$

Hene the correct option is $A$