Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
0 votes

If the magnitudes of vectors $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$ are 3,4 and $\theta$ units respectively and if $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C},$ then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

$\begin{array}{1 1}(A)\;\large\frac{\pi}{4} \\(B)\;\large\frac{\pi}{2} \\(C)\;\tan^{-1}(0.75)\\(D)\;\cos^{-1}(0.6) \end{array} $

Can you answer this question?

1 Answer

0 votes
Let the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ be $\theta$ . Then
$c^2=A62+B^2 +2AB \cos \theta$
$5^2 =3^2 +4^2 +(2)(3)(4) \cos \theta$
Solving $\theta=90^{\circ}$ .
So, the triangle representing vectors $\overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$ is a right-angled triangle with $\overrightarrow{C}$ as the hypotenuse. Hence
$\cos \beta = \large\frac{ |\overrightarrow{A}|}{|\overrightarrow{C}|}$
$\qquad= \large\frac{3}{5}$
$\qquad= 0.6$
Hence D is the correct answer.
answered Jul 16, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App