Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

Choose the correct answer in the following : let \( \overrightarrow a\) and \( \overrightarrow b\) be two unit vectors and θ is the angle between them. Then \( \overrightarrow a\ + \overrightarrow b\) is a unit vector if

\[ \begin{array} (A)\; \theta = \frac{\pi}{4} \quad & (B)\; \theta = \frac{\pi}{3} \quad & (C) \;\theta = \frac{\pi}{2} \quad &(D) \;\theta = \frac{2\pi}{3} \end{array} \]
Can you answer this question?

1 Answer

0 votes
  • $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
  • $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
  • If $\cos\theta<0$ then $\theta$ lies in the II quadrant (i.e)$(\pi-\theta)$
Step 1:
Given $|\overrightarrow a|=1$ and $|\overrightarrow b|=1$
$|\overrightarrow a+\overrightarrow b|=1$ or $|\overrightarrow a+\overrightarrow b|^2=1$
Let us consider $(\overrightarrow a+\overrightarrow b)^2=1$
(i.e) $(\overrightarrow a+\overrightarrow b)(\overrightarrow a+\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2+\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow a.$
But $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
Therefore $(\overrightarrow a+\overrightarrow b)(\overrightarrow a+\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b=1.$
Step 2:
But $|\overrightarrow a|=|\overrightarrow b|=1$
$1^2+1^2+2\times 1\times 1\cos\theta=1$
Therefore $\cos\theta=-\large\frac{1}{2}$
Step 3:
If $\cos\theta$ is negative,then it lies in the II quadrant.
Hence D is the correct option.
answered May 24, 2013 by sreemathi.v

Related questions

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