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Write the angle between two vectors \( \overrightarrow a \: and \: \overrightarrow b\) with magnitudes \( \sqrt 3 \) and 2 respectively having \( \overrightarrow a.\overrightarrow b=\sqrt 6 \).

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  • The scalar product of two vectors is $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
Step 1:
Given : The magnitude of $\overrightarrow a=\sqrt 3$ and $\overrightarrow b=2$
$\Rightarrow \mid \overrightarrow a\mid=\sqrt 3$ and $\mid \overrightarrow b\mid=2$
Also $\overrightarrow a.\overrightarrow b=\sqrt 6$
We know $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
$\cos \theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
Step 2:
Now substituting the values we get,
$\cos\theta=\large\frac{\sqrt 6}{\sqrt 3\times 2}$
This can be written as
$\cos\theta=\large\frac{\sqrt 2\times \sqrt 3}{\sqrt 3\times \sqrt 2\times \sqrt 2}$
$\qquad=\large\frac{1}{\sqrt 2}$
Step 3:
Therefore $\theta=\cos^{-1}\big(\large\frac{1}{\sqrt 2}\big)$
Hence the angle between $\overrightarrow a$ and $\overrightarrow b$ is $\large\frac{\pi}{4}$
answered Nov 14, 2013 by sreemathi.v

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