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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If \(\overrightarrow a = \overrightarrow b + \overrightarrow c\), then is it true that \(| \overrightarrow a | = | \overrightarrow b | + | \overrightarrow c |\)? Justify your answer.

$\begin{array}{1 1} True \\ False \end{array} $

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  • $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
  • $\overrightarrow a.\overrightarrow a=|\overrightarrow a|^2$
Step 1:
Given : $\overrightarrow a=\overrightarrow b+\overrightarrow c$
Hence the magnitude is $|\overrightarrow a|=|\overrightarrow b+\overrightarrow c|$
Now squaring on both sides we get
$|\overrightarrow a|^2=|\overrightarrow b+\overrightarrow c|^2$
$\quad\quad=(\overrightarrow b+\overrightarrow c).(\overrightarrow b+\overrightarrow c)$
Step 2:
On expanding we get
$\quad\quad=\overrightarrow b.\overrightarrow b+\overrightarrow b.\overrightarrow c+\overrightarrow c\overrightarrow b+\overrightarrow c.\overrightarrow c$
$\quad\quad=|\overrightarrow b|^2+2\overrightarrow b.\overrightarrow c+|\overrightarrow c|^2$
We know $\overrightarrow b.\overrightarrow c=|\overrightarrow b||\overrightarrow c|\cos\theta$
When $\theta=0;\cos\theta=1$
Therefore $|\overrightarrow a|^2=|\overrightarrow b|^2+|\overrightarrow c|^2+2|\overrightarrow b||\overrightarrow c|$
Step 3:
This is of the form $(a+b)^2$
Therefore $|\overrightarrow a|^2=(|\overrightarrow b|+|\overrightarrow c|)^2$
Taking square root on both sides,
$|\overrightarrow a|=|\overrightarrow b|+|\overrightarrow c|$
If $\theta\neq 0$ then $|\overrightarrow a|\neq |\overrightarrow b|+|\overrightarrow c|$
answered May 23, 2013 by sreemathi.v

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