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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Show that \(| \overrightarrow a | \overrightarrow b + | \overrightarrow b | \overrightarrow a\) is perpendicular to \(| \overrightarrow a | \overrightarrow b - | \overrightarrow b | \overrightarrow a\), for any two nonzero vectors \(\overrightarrow a\) and \(\overrightarrow b\).

$\begin{array}{1 1}(A) \overrightarrow a.\overrightarrow b=0 \\(B) \overrightarrow a=\overrightarrow b \\ (C) \overrightarrow a.\overrightarrow b=1 \\(D) None \;of\; the\; above \end{array} $

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1 Answer

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Toolbox:
  • If $\overrightarrow a\perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$
  • $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
  • $|\overrightarrow a|^2=\overrightarrow a.\overrightarrow a$
Step 1:
Let $\overrightarrow p=|\overrightarrow a|\overrightarrow b+|\overrightarrow b|\overrightarrow a$
$\overrightarrow q=|\overrightarrow a|\overrightarrow b-|\overrightarrow b|\overrightarrow a$
Then $\overrightarrow p.\overrightarrow q=[|\overrightarrow a|\overrightarrow b+|\overrightarrow b|\overrightarrow a].[|\overrightarrow a|\overrightarrow b-|\overrightarrow b|\overrightarrow a]$
$\qquad\qquad=|\overrightarrow a|\overrightarrow b.[|\overrightarrow a|\overrightarrow b-|\overrightarrow b|\overrightarrow a]+|\overrightarrow b|\overrightarrow a.[|\overrightarrow a|\overrightarrow b-|\overrightarrow b|\overrightarrow a]$
$\qquad\qquad\;=|\overrightarrow a|^2\overrightarrow b.\overrightarrow b-|\overrightarrow a||\overrightarrow b|\overrightarrow b.\overrightarrow a+|\overrightarrow b||\overrightarrow a||\overrightarrow a.\overrightarrow b|-|\overrightarrow a.\overrightarrow b|-|\overrightarrow b|^2|\overrightarrow a.\overrightarrow a|$
Step 2:
But we know $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$ and $\overrightarrow a.\overrightarrow a=|\overrightarrow a|^2$ and $\overrightarrow b.\overrightarrow b=|\overrightarrow b|^2$
Therefore $\overrightarrow p.\overrightarrow q=|\overrightarrow a|^2|\overrightarrow b|^2-|\overrightarrow a||\overrightarrow b|\overrightarrow a.\overrightarrow b+|\overrightarrow a||\overrightarrow b|\overrightarrow a.\overrightarrow b-|\overrightarrow b|^2|\overrightarrow a|^2$
$\qquad\qquad\qquad=|\overrightarrow a|^2|\overrightarrow b|^2-|\overrightarrow a|^2|\overrightarrow b|^2=0$
Step 3:
Since the scalar product is 0.
$\overrightarrow p\perp \overrightarrow q$ (i.e) $\overrightarrow p.\overrightarrow q=0$
Hence $|\overrightarrow a|\overrightarrow b+|\overrightarrow b|\overrightarrow a$ and $|\overrightarrow a|\overrightarrow b-|\overrightarrow b|\overrightarrow a$ are perpendicular to each other.
If $\overrightarrow a\perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$
answered May 21, 2013 by sreemathi.v
 

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