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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If $ \bar{a}\;and\;\bar{b}$ are vectors,prove that $ |\bar{a} \times \bar{b}|^2+(\bar{a}.\bar{b})^2=|\bar{a}|^2.|\bar{b}|^2$

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1 Answer

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Toolbox:
  • $\large|\overrightarrow a\times\overrightarrow b|=|\overrightarrow a||\overrightarrow b|sin\theta$
  • $\large\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$
  • $\large\:sin^2\theta+cos^2\theta=1$
We know that $\large|\overrightarrow a\times\overrightarrow b|=|\overrightarrow a||\overrightarrow b|sin\theta$ and
$\large\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$
$\Rightarrow\large|\overrightarrow a\times\overrightarrow b|^2=|\overrightarrow a|^2|\overrightarrow b|^2sin^2\theta$ and
$\large(\overrightarrow a.\overrightarrow b)^2=|\overrightarrow a|^2|\overrightarrow b|^2cos^2\theta$
$\Rightarrow\large|\overrightarrow a\times\overrightarrow b|^2+(\overrightarrow a.\overrightarrow b)^2=$
$\large|\overrightarrow a|^2|\overrightarrow b|^2sin^2\theta+|\overrightarrow a|^2|\overrightarrow b|^2cos^2\theta$
$=\large\big(|\overrightarrow a|^2+|\overrightarrow b|^2\big)\big(sin^2\theta+cos^2\theta\big)$
$=\large|\overrightarrow a|^2+|\overrightarrow b|^2$
Hence proved.
answered Apr 19, 2013 by rvidyagovindarajan_1
 

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