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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Prove the triangle inequality $\bigg|\overrightarrow{a}+\overrightarrow{b}\bigg|\leq\bigg|\overrightarrow{a}\bigg|+\bigg|\overrightarrow{b}\bigg|$

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Toolbox:
  • $\large|\overrightarrow a+\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2+|2\overrightarrow a.\overrightarrow b|$
  • $\large\:|cos\theta|\in[0,1]$
We know that $\large|\overrightarrow a+\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2+|2\overrightarrow a.\overrightarrow b|$
But $\large\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$
$\Rightarrow \:\large|\overrightarrow a+\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2+2|\overrightarrow a||\overrightarrow b||cos\theta|$
We know that $\large|cos\theta|\in[0,1]$
$\Rightarrow\:$ maximum value of $cos\theta$ is 1
$\Rightarrow\:\large2|\overrightarrow a||\overrightarrow b||cos\theta|\leq\:2|\overrightarrow a||\overrightarrow b|$
$\Rightarrow \:\large|\overrightarrow a+\overrightarrow b|^2\leq|\overrightarrow a|^2+|\overrightarrow b|^2+2|\overrightarrow a||\overrightarrow b|$
But $\large |\overrightarrow a|^2+|\overrightarrow b|^2+2|\overrightarrow a||\overrightarrow b|=(|\overrightarrow a|+|\overrightarrow b|)^2$
$\Rightarrow \:\large|\overrightarrow a+\overrightarrow b|^2\leq(|\overrightarrow a|+|\overrightarrow b|)^2$
Taking squareroot on both the sides we get the result
$\large|\overrightarrow a+\overrightarrow b|\leq|\overrightarrow a|+|\overrightarrow b|$
answered Apr 17, 2013 by rvidyagovindarajan_1
 

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