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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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True-or-False: The formula $(\overrightarrow{a}+\overrightarrow{b})^2=\mid{\overrightarrow{a}\mid}^2+\mid{\overrightarrow{b}\mid}^2\;+2\mid\overrightarrow{a}\mid.\mid\overrightarrow{b}\mid$ is valid for non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}.$

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  • $(\overrightarrow a+\overrightarrow b)^2 =(\overrightarrow a+\overrightarrow b).(\overrightarrow a.\overrightarrow b)$
  • $=|\;\overrightarrow a|^2+|\overrightarrow b|^2+2.\overrightarrow a.\overrightarrow b$
  • $\;\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b| \cos \theta$
We know that $(a+b)^2=|\;a\;|^2+|\;b\;|^2+2 \;\overrightarrow a.\;\overrightarrow b$
$\qquad\qquad\qquad=|\;a\;|^2+|\;b\;|^2+ 2 |\;a\;||\;b\;| \cos \theta$
But it is given that $(a+b)^2=|\;a\;|^2+|\;b\;|^2+2 |\overrightarrow a||\;\overrightarrow b\;|$
This implies $\cos \theta=1 (ie) \theta=0$
This implies that the angle between the vectors is zero. That is they are parallel vector.
Hence $|\;a\;|^2+|\;b\;|^2+2 |\overrightarrow a||\;\overrightarrow b\;|$ is valid for non zero vectors. $\overrightarrow a$ and $\overrightarrow b$ only if they are parallel
Hence the statment is $False$
answered May 30, 2013 by meena.p

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