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Find the angle between the vectors \(( \overrightarrow a + \overrightarrow b)\: and \: ( \overrightarrow a - \overrightarrow b)\: if \: \overrightarrow a = (2i-j+3k)\: and\: \overrightarrow b = (3\hat i + \hat j -2\hat k).\)

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  • $\cos\theta=\large\frac{(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)}{\mid \overrightarrow a+\overrightarrow b\mid\mid\overrightarrow a-\overrightarrow b\mid}$
Step 1:
Given :
$\overrightarrow a=2\hat i-\hat j+3\hat k$
$\overrightarrow b=3\hat i+\hat j-2\hat k$
$\overrightarrow a+\overrightarrow b=5\hat i+\hat k$
$\overrightarrow a-\overrightarrow b=-\hat i-2\hat j+5\hat k$
$\mid\overrightarrow a+\overrightarrow b\mid=\sqrt{5^2+1^2}$
$\Rightarrow \sqrt {26}$
$\mid\overrightarrow a-\overrightarrow b\mid=\sqrt{1^2+2^2+5^2}$
$\Rightarrow \sqrt {30}$
Step 2:
$\cos\theta=\large\frac{(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)}{\mid \overrightarrow a+\overrightarrow b\mid\mid\overrightarrow a-\overrightarrow b\mid}$
$\qquad=\large\frac{(5\hat i+\hat k).(-\hat i-2\hat j+5\hat k)}{\mid5\hat i+\hat k\mid.\mid -\hat i-2\hat j+5\hat k)}$
$\qquad=\large\frac{-5+5}{\sqrt{5^2+1^2}\sqrt{1^2+2^2+5^2}}$
$\qquad=0$
$\Rightarrow \cos\theta=0$
$\Rightarrow \theta=\cos^{-1}0$
$\cos^{-1}0=\large\frac{\pi}{2}$
$\theta=\large\frac{\pi}{2}$
answered Sep 30, 2013 by sreemathi.v
 

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