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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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True-or-False: If $\mid \overrightarrow{a}+\overrightarrow{b}\mid=\mid \overrightarrow{a}-\overrightarrow{b}\mid$,then the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are orthogonal.

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Toolbox:
  • $|\overrightarrow a+\overrightarrow b|^2=(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)=|\;a\;|^2+|\;b\;|^2+2 \overrightarrow a.\overrightarrow b$
  • $|\overrightarrow a-\overrightarrow b|^2=(\overrightarrow a-\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\;a\;|^2+|\;b\;|^2-2 \overrightarrow a.\overrightarrow b$
We know that $|\overrightarrow a+\overrightarrow b|^2=(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)$
$= |\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b$ and
$|\overrightarrow a-\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2-2\overrightarrow a.\overrightarrow b$
Given that $ |\overrightarrow a+\overrightarrow b|=|\overrightarrow a-\overrightarrow b|$
$\Rightarrow\: |\overrightarrow a+\overrightarrow b|^2=|\overrightarrow a-\overrightarrow b|^2$
Substituting the values of $|\overrightarrow a+\overrightarrow b|^2$ and $|\overrightarrow a-\overrightarrow b|^2$ we get
$ \Rightarrow |\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b=|\overrightarrow a|^2+|\overrightarrow b|^2-2\overrightarrow a.\overrightarrow b$
$ \Rightarrow 4\overrightarrow a.\overrightarrow b=0 $
$ \Rightarrow \overrightarrow a.\overrightarrow b=0 \Rightarrow \overrightarrow a\: and\: \overrightarrow b \: are \perp$.
Hence it is a $True$ statement
answered May 30, 2013 by meena.p
 

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