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Show that the vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar if and only if $\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}$ are coplanar.

1 Answer

  • Scalar triple product
  • $ ( \overrightarrow a \times \overrightarrow b).\overrightarrow c = (\overrightarrow b \times \overrightarrow c)\overrightarrow a = (\overrightarrow c \times \overrightarrow a).\overrightarrow b = [ \overrightarrow a, \overrightarrow b, \overrightarrow c ]$ Also $ [ \overrightarrow a\: \overrightarrow b\: \overrightarrow c] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$ = volume of the parallelopiped formed by three non-coplanar vectors $ \overrightarrow a, \overrightarrow b, \overrightarrow c$.
Step 1
$ \overrightarrow a , \overrightarrow b, \overrightarrow c$ be coplanar
Then $ [ \overrightarrow a\: \overrightarrow b\: \overrightarrow c ]=0$
Step 2
To show that $ \overrightarrow a+ \overrightarrow b, \: \overrightarrow b+ \overrightarrow c, \: \overrightarrow c+ \overrightarrow a$ are also coplanar.
Consider $ [ \overrightarrow a+ \overrightarrow b\: \overrightarrow b+ \overrightarrow c\: \overrightarrow c+ \overrightarrow a]$
$ = ( \overrightarrow a \times \overrightarrow b) \times ( \overrightarrow b + \overrightarrow c). ( \overrightarrow c+ \overrightarrow a)$
$ = ( \overrightarrow a \times \overrightarrow b+ \overrightarrow a \times \overrightarrow c + \overrightarrow b \times b+ \overrightarrow b \times \overrightarrow c).( \overrightarrow c + \overrightarrow a)$
$ = ( \overrightarrow a \times \overrightarrow b+ \overrightarrow a \times \overrightarrow c+ \overrightarrow 0+ \overrightarrow b \times \overrightarrow c).( \overrightarrow c+ \overrightarrow a)$
$[ \overrightarrow a \: \overrightarrow b\: \overrightarrow c]+[ \overrightarrow a\: \overrightarrow c\: \overrightarrow c]+[ \overrightarrow b\: \overrightarrow c\: \overrightarrow c]+[ \overrightarrow a\: \overrightarrow b\: \overrightarrow c]+[ \overrightarrow a\: \overrightarrow c\: \overrightarrow a]+[ \overrightarrow b\: \overrightarrow c\: \overrightarrow a]$
$ = [ \overrightarrow a\: \overrightarrow b\: \overrightarrow c\: ] +0+0+0+0+ [ \overrightarrow a\: \overrightarrow b\: \overrightarrow c]$
$ = 2[ \overrightarrow a\: \overrightarrow b\: \overrightarrow c]=0$
Step 3
Since $ [ \overrightarrow a+ \overrightarrow b\: \overrightarrow b+ \overrightarrow c\: \overrightarrow c+ \overrightarrow a] = 0,$ it follows that they are coplanar.
Step 4
Conversely, if $ \overrightarrow a+ \overrightarrow b, \: \overrightarrow b+ \overrightarrow c,\: \overrightarrow c+ \overrightarrow a$ are coplanar then $ [ \overrightarrow a+ \overrightarrow b\: \overrightarrow b+ \overrightarrow c\: \overrightarrow c+ \overrightarrow a]=0$
Step 5
From the above, it can be seen that $ [ \overrightarrow a+ \overrightarrow b\: \overrightarrow b+ \overrightarrow c\: \overrightarrow c+ \overrightarrow a]=2[ \overrightarrow a\: \overrightarrow b\: \overrightarrow c]$ It follows that $ [ \overrightarrow a\: \overrightarrow b\: \overrightarrow c]=0 \Rightarrow \overrightarrow a, \overrightarrow b, \overrightarrow c$ are coplanar.
answered Jun 8, 2013 by thanvigandhi_1

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