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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 $\large \frac{1}{2}.$

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

–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 $|\overrightarrow b|=1$
edited Feb 12, 2014