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

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

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 $A$