Q. If $\overrightarrow{\mathbf{a}}\mathbf{=}2\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}2\widehat{\mathbf{k}}\mathbf{\ }and\ \overrightarrow{\mathbf{b}}\mathbf{=}5\widehat{\mathbf{i}}\mathbf{-}3\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}$, then the projection of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}$ is

A) 3
B) 4
C) 5
D) 6
Explanation

(a)
Given that, $\overrightarrow{\mathbf{a}} = 2\widehat{\mathbf{i}} + \widehat{\mathbf{j}} + 2\widehat{\mathbf{k}}\ and\ \overrightarrow{\mathbf{b}} = 5\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} + \widehat{\mathbf{k}}$
The projection of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}} = \frac{\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}}}{\mathbf{|}\overrightarrow{\mathbf{a}}\mathbf{|\ }}$
$= \frac{\left( 2\widehat{\mathbf{i}} + \widehat{\mathbf{j}} + 2\widehat{\mathbf{k}} \right) \bullet (5\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} + \widehat{\mathbf{k}})\ \ }{\sqrt{(2)^{2} + (1)^{2} + (2)^{2}}}$
$= \frac{10 - 3 + 2}{\sqrt{9}} = \frac{9}{3} = 3$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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