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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Prove that \( (\overrightarrow a + \overrightarrow b) ⋅ (\overrightarrow a + \overrightarrow b) =| \overrightarrow a |^2 + |\overrightarrow b |^2\), if and only if \(\overrightarrow a,\: \overrightarrow b\) are perpendicular, given \(\overrightarrow a ≠ 0, \overrightarrow b ≠ 0\).

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  • $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
  • If $\overrightarrow a\perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$
Step 1:
Let us first determine $(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)$
$(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)=\overrightarrow a(\overrightarrow a+\overrightarrow b)\overrightarrow b(\overrightarrow a+\overrightarrow b)$
$\qquad\qquad\qquad\quad=\overrightarrow a.\overrightarrow a+\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow a+\overrightarrow b.\overrightarrow b$
$\qquad\qquad\qquad\quad\;\;=|\overrightarrow a|^2+2\overrightarrow a.\overrightarrow b+|\overrightarrow b|^2$
Step 2:
Let us consider if $\overrightarrow a\perp \overrightarrow b$
If $(\overrightarrow a \perp \overrightarrow b)$ then $\overrightarrow a.\overrightarrow b=0$
Therefore $(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2$
Step 3:
If $\overrightarrow a$ is not perpendicular to $\overrightarrow b$ then,$\overrightarrow a.\overrightarrow b\neq 0$,Similarly $\overrightarrow b.\overrightarrow a\neq 0$
Then $(\overrightarrow a+\overrightarrow b).(\overrightarrow a.\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b\neq |\overrightarrow a|^2+|\overrightarrow b|^2$
Hence only if $\overrightarrow a\perp \overrightarrow b,(\overrightarrow a+\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2$
answered May 24, 2013 by sreemathi.v
edited May 24, 2013 by sreemathi.v

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