logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

If the projection of $\overrightarrow{a} $ on $\overrightarrow{b} $ and the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ are equal then the angle between $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ is

\[\begin{array}{1 1}(1)\frac{\pi}{2} &(2)\frac{\pi}{3}\\(3)\frac{\pi}{4}&(4)\frac{2\pi}{3}\end{array}\]

Can you answer this question?
 
 

1 Answer

0 votes
Projection of $\overrightarrow{a}$ on $\overrightarrow{b}= \large\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|}$
Projection of $\overrightarrow{b}$ on $\overrightarrow{a}= \large\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|}$
Given $\large\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|} =\large\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|}$
$=>|\overrightarrow{a}|= |\overrightarrow{b}|$
$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=\overrightarrow{a}^2- \overrightarrow{b}^2$
$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=|\overrightarrow{a}|^2- |\overrightarrow{b}|^2$
$\qquad=0$
$\qquad= \large\frac{\pi}{2} $
Hence 1 is the correct answer.
answered May 12, 2014 by meena.p
 

Related questions

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