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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra

The angle between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ with magnitudes $\sqrt 3$ and 4,respectively,and $\overrightarrow{a}.\overrightarrow{b}=2\sqrt 3$ is

\[(A)\;\frac{\pi}{6}\quad(B)\;\frac{\pi}{3}\quad(C)\;\frac{\pi}{2}\quad(D)\;\frac{\pi}{4}\]

1 Answer

Toolbox:
  • $\cos \theta =\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Let $|\overrightarrow a|=\sqrt 3$ and $|\overrightarrow b|=4$
$\overrightarrow a.\overrightarrow b=2 \sqrt 3$
we know $\cos \theta =\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Now substituting the values we get,
$\cos \theta= \large\frac{2 \sqrt 3}{\sqrt 3.4}=\frac{1}{2}$
Therefore $\theta=\cos ^{-1} \bigg(\large\frac{1}{2}\bigg)$
But $\cos ^{-1} \bigg(\large\frac{1}{2}\bigg)$ is $\large\frac{\pi}{3}$
$=\large\frac{\pi}{3}$
Hence the correct option is $B$
answered May 28, 2013 by meena.p
 

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