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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the magnitude of two vectors \( \overrightarrow a\) and \( \overrightarrow b\) , having the same magnitude and such that the angle between them is 60\(^{\omicron}\) and their scalar product is \(\large \frac{1}{2}.\)

$\begin{array}{1 1} |\overrightarrow a|=|\overrightarrow b| = 0 \\ |\overrightarrow a|=|\overrightarrow b| = 1 \\ |\overrightarrow a|=|\overrightarrow b| =\large \frac{1}{2} \\ |\overrightarrow a|=|\overrightarrow b| = \large\frac{\sqrt3}{2}\end{array} $

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1 Answer

–1 vote
Toolbox:
  • Scalar product of two vectors $\overrightarrow a$ and $\overrightarrow b$ is $ \overrightarrow a.\overrightarrow b=|\overrightarrow a|.|\overrightarrow b| cos \theta$
  • $\cos 60^{\large\circ} =\large\frac{1}{2}$
Step 1:
Angle between the vectors=$60^{\large\circ}$
Given: $\overrightarrow a.\overrightarrow b=\large\frac{1}{2}$ and $|\overrightarrow a|=|\overrightarrow b|$
We know $ \overrightarrow a.\overrightarrow b=|\overrightarrow a|.|\overrightarrow b| cos \theta$
Also given that $\theta=60^{\large\circ}$
$\Rightarrow \cos 60^{\large\circ}=\large\frac{1}{2}$
Step 2:
$ \overrightarrow a.\overrightarrow b=|\overrightarrow a|.|\overrightarrow b| cos \theta$
But $\overrightarrow a.\overrightarrow b=\large\frac{1}{2}$
$\therefore\:\large\frac{1}{2}$$=|\overrightarrow a||\overrightarrow b|.\large\frac{1}{2}$
$\Rightarrow |\overrightarrow a|.|\overrightarrow b|=1$
Step 3:
But it is given that $\overrightarrow a$ and $\overrightarrow b$ have same magnitude.
Hence $|\overrightarrow a|$ and $|\overrightarrow b|=1$
answered May 20, 2013 by sreemathi.v
edited Feb 12, 2014 by rvidyagovindarajan_1
 

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