Q. If $\overrightarrow{\mathbf{a}} = 4\widehat{\mathbf{i}} + 6\widehat{\mathbf{j}}\ and\ \overrightarrow{\mathbf{b}} = 3\widehat{\mathbf{j}} + 4\widehat{\mathbf{k}}$, the vector form of the component of $\overrightarrow{\mathbf{a}}\ along\ \overrightarrow{\mathbf{b}}$ is
A) $\frac{18}{5}(3\widehat{\mathbf{i}} + 4\widehat{\mathbf{k}}\ )$
B) $\frac{18}{25}(3\widehat{\mathbf{j}} + 4\widehat{\mathbf{k}}\ )$
C) $\frac{36}{25}(3\widehat{\mathbf{j}} + 4\widehat{\mathbf{k}}\ )$
D) $\frac{19}{18}(2\widehat{\mathbf{i}} + 3\widehat{\mathbf{j}}\ )$
Explanation
(b)
$\because\ \overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}} = 18\ and\ |\overrightarrow{\mathbf{b}}| = 5$
$\therefore$ Vector component of $\overrightarrow{\mathbf{a}}\ along\ \overrightarrow{\mathbf{b}}$
$= \left( \frac{\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}}}{\mathbf{\ |}\overrightarrow{\mathbf{b}}\left. \ \mathbf{} \right|^{2}} \right)\overrightarrow{\mathbf{b}} = \frac{18}{25}\left( 3\widehat{\mathbf{j}} + 4\widehat{\mathbf{k}}\ \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.