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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the angle between the vectors $2\hat i\;-\hat j\;-\hat k$ and $3\hat i\;+4\hat j\;+\hat k$.

$\begin{array}{1 1} cos^{-1} \bigg(\large \frac{1}{\sqrt{156}} \bigg) \\ cos^{-1} \bigg(\large \frac{9}{\sqrt{156}} \bigg) \\ cos^{-1} \bigg(\large \frac{11}{\sqrt{156}} \bigg) \\ cos^{-1} \bigg(\large \frac{3}{\sqrt{156}} \bigg) \end{array} $

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  • $ \overrightarrow a . \overrightarrow b=| \;a\; |\; |\; b\; | \cos \theta$
  • Therefore $ cos \theta= \large\frac{ \overrightarrow a.\overrightarrow b}{| \;a\;||\; b\;|}$
Let $\overrightarrow a=2\hat i-\hat j-\hat k\:and\:\overrightarrow b=3\hat i+4\hat j+\hat k$
$ \overrightarrow a . \overrightarrow b=| \;a\; |\; |\; b\; | \cos \theta$
$ cos \theta= \large\frac{ \overrightarrow a.\overrightarrow b}{| \;a\;||\; b\;|}$
$\qquad=\large\frac{ (2\hat i-\hat j-\hat k).(3\hat i+4\hat j+\hat k)}{\sqrt {2^2+1^2+1^2}\sqrt {3^2+4^2+1^2}}$
$\qquad =\large\frac{6-4-1}{\sqrt {4+1+1}\sqrt {9+16+1}}$
$ \cos \theta =\large\frac{1}{\sqrt6\:\sqrt{26}}$
$\cos \theta =\large \frac{1}{\sqrt{156}} $
Therefore $ \theta =\cos ^{-1}\bigg(\large \frac{1}{\sqrt{156}} \bigg) $
answered May 27, 2013 by meena.p

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