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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If $ \overrightarrow a, \overrightarrow b, \overrightarrow c, $are unit vectors such that $ \overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow 0$, find the value of $ \overrightarrow a ⋅ \overrightarrow b + \overrightarrow b ⋅ \overrightarrow c + \overrightarrow c ⋅ \overrightarrow a$.

$\begin{array}{1 1}(A) \frac{1}{\sqrt 2} (a^2+b^2+c^2)\\(B) \frac{-1}{\sqrt 2}(a^2+b^2+c^2) \\ (C) \frac{1}{ 2}(a^2+b^2+c^2) \\ (D) \frac{-1}{ 2}(a^2+b^2+c^2) \end{array} $

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1 Answer

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Toolbox:
  • $\overrightarrow a.\overrightarrow a=|\overrightarrow a|^2$
  • $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
  • $\Rightarrow$ Vector multiplication is commutative.
Step 1:
Given : $a+b+c=0$------(1)
Pre multiplying eq(1) by $\overrightarrow a$ we get
$\overrightarrow a.\overrightarrow b+\overrightarrow a.\overrightarrow c=-\overrightarrow a.\overrightarrow a$
But $\overrightarrow a.\overrightarrow a=|\overrightarrow a|^2$
Therefore $\overrightarrow a.\overrightarrow b+\overrightarrow a.\overrightarrow c=-\overrightarrow a^2$-------(2)
Step 2:
Pre multiplying eq(1) by $\overrightarrow b$ we get
$\overrightarrow b.\overrightarrow a+\overrightarrow b.\overrightarrow b+\overrightarrow b.\overrightarrow c=0$
$\Rightarrow \overrightarrow b.\overrightarrow a+\overrightarrow b.\overrightarrow c=-b^2$------(3)
Step 3:
Pre multiplying eq(1) by $\overrightarrow c$ we get
$\overrightarrow c.\overrightarrow a+\overrightarrow c.\overrightarrow b+\overrightarrow c.\overrightarrow c=0$
$\Rightarrow \overrightarrow c.\overrightarrow a+\overrightarrow c.\overrightarrow b=-c^2$------(4)
Step 4:
Add equ(2),(3) and (4)
$-(a^2+b^2+c^2)=\overrightarrow a.\overrightarrow b+\overrightarrow a.\overrightarrow c+\overrightarrow b.\overrightarrow a+\overrightarrow b.\overrightarrow c+\overrightarrow c.\overrightarrow a+\overrightarrow c.\overrightarrow b$
But we know $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$,$\overrightarrow b.\overrightarrow c=\overrightarrow c.\overrightarrow b$ and $\overrightarrow a.\overrightarrow c=\overrightarrow c.\overrightarrow a$
Hence $-(a^2+b^2+c^2)=2(\overrightarrow a.\overrightarrow b)+2(\overrightarrow b.\overrightarrow c)+2(\overrightarrow c.\overrightarrow a)$
Step 5:
Therefore $-(a^2+b^2+c^2)=2[\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow c+\overrightarrow c.\overrightarrow a]$
$\Rightarrow \overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow c+\overrightarrow c.\overrightarrow a=\large\frac{-1}{2}$$(a^2+b^2+c^2)$
answered May 21, 2013 by sreemathi.v
 

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