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Questions  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Q)

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0,$show that $\overrightarrow{a} \times \overrightarrow{b}=\overrightarrow{b} \times \overrightarrow{c} =\overrightarrow{c} \times \overrightarrow{a}$.Interpret the result geometrically?

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A)
Toolbox:
  • $\overrightarrow{a} \times \overrightarrow{a}=\overrightarrow{b} \times \overrightarrow{b} =\overrightarrow{c} \times \overrightarrow{c}=0$.
  • $\overrightarrow{a} \times \overrightarrow{b}=-(\overrightarrow{b} \times \overrightarrow{a})$
Step 1:
Given $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
we have to prove that
$\overrightarrow{a} \times \overrightarrow{b}=\overrightarrow{b} \times \overrightarrow{c} =\overrightarrow{c} \times \overrightarrow{a}$.
Consider $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
Take cross product with $\overrightarrow{a}$
$\overrightarrow{a} \times \overrightarrow{a}+\overrightarrow{a} \times \overrightarrow{b} +\overrightarrow{a} \times \overrightarrow{c}=0$.
But we know $\overrightarrow{a} \times \overrightarrow{a}=0$
Therefore $\overrightarrow{a} \times \overrightarrow{b}+\overrightarrow{a} \times \overrightarrow{c} =0$
But $\overrightarrow{a} \times \overrightarrow{c}=-\overrightarrow{c} \times \overrightarrow{a} $
Hence $\overrightarrow{a} \times \overrightarrow{b}-\overrightarrow{c} \times \overrightarrow{a} =0$
Therefore $\overrightarrow{a} \times \overrightarrow{b}=\overrightarrow{c} \times \overrightarrow{a} $-----(1)
Step 2:
Again consider $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
Take cross product with $\overrightarrow{b}$
$\overrightarrow{b} \times \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{b} +\overrightarrow{b} \times \overrightarrow{c}=0$.
But we know $\overrightarrow{b} \times \overrightarrow{b}=0$
Therefore $\overrightarrow{b} \times \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{c} =0$
But $\overrightarrow{b} \times \overrightarrow{a}=-\overrightarrow{a} \times \overrightarrow{b} $
Therefore $-(\overrightarrow{a} \times \overrightarrow{b})+\overrightarrow{b} \times \overrightarrow{c} =0$
$=>\overrightarrow{b} \times \overrightarrow{c}=\overrightarrow{a} \times \overrightarrow{b} $-----(2)
Hence from equ(1) and equ(2) we get,
$\overrightarrow{a} \times \overrightarrow{b}=\overrightarrow{b} \times \overrightarrow{c} =\overrightarrow{c} \times \overrightarrow{a}$.
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