Question

If $\overrightarrow{A}=\overrightarrow{B}+\overrightarrow{C}$ and the magnitude $A,B$ and $C$ are $5,4$ and $3$ units, the angle between $\overrightarrow{A}$ and $\overrightarrow{C}$ is

• A. $co{s}^{-1}\left(\frac{3}{5}\right)$
• B. $co{s}^{-1}\left(\frac{4}{5}\right)$
• C. $\frac{\pi }{2}$
• D. $si{n}^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is A $co{s}^{-1}\left(\frac{3}{5}\right)$

A is the resultant of B and C.

Further B is perpendicular to C since $B^2+C^2=A^2$.

If $\theta$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{C}$ then:

$cos\theta =\frac{\overrightarrow{A}.\overrightarrow{C}}{|\overrightarrow{A}|.|\overrightarrow{C}|}=\frac{(\overrightarrow{B}+\overrightarrow{C}).\overrightarrow{C}}{5.3}=\frac{\overrightarrow{B}.\overrightarrow{C}+{\overrightarrow{C}}^2}{5.3}=\frac{0+3^2}{5.3}$ $\Rightarrow \theta =co{s}^{-1}\left(\frac{3}{5}\right)$