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Home  >>  TN XII Math  >>  Vector Algebra

Prove that $( \overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}= \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})$ if $ \overrightarrow{a}$ and $ \overrightarrow{c}$ are collinear. (Where vector triple product is non-zero).

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  • Vector triple product $(\overrightarrow a \times \overrightarrow b) \times \overrightarrow c=(\overrightarrow a.\overrightarrow c)\overrightarrow b-(\overrightarrow b.\overrightarrow c)\overrightarrow a$ $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c)= (\overrightarrow a.\overrightarrow c)\overrightarrow b - (\overrightarrow a.\overrightarrow b)\overrightarrow c$
Let $ (\overrightarrow a \times \overrightarrow b) \times \overrightarrow c=\overrightarrow a \times (\overrightarrow b \times \overrightarrow c)$
$ \Rightarrow (\overrightarrow a.\overrightarrow c)\overrightarrow b - (\overrightarrow b.\overrightarrow c)\overrightarrow a=(\overrightarrow a. \overrightarrow c)\overrightarrow b-(\overrightarrow a.\overrightarrow b)\overrightarrow c$
$ (\overrightarrow b.\overrightarrow c)\overrightarrow a = (\overrightarrow a.\overrightarrow b)\overrightarrow c$
$ \therefore \overrightarrow a = \large\frac{(\overrightarrow a.\overrightarrow b)} {(\overrightarrow b.\overrightarrow c)}\overrightarrow c \Rightarrow \overrightarrow a, \overrightarrow c$ are collinear.
Now $(\overrightarrow b.\overrightarrow c)\overrightarrow a-(\overrightarrow a.\overrightarrow b)\overrightarrow c=\overrightarrow 0$
$ \Rightarrow (\overrightarrow c \times \overrightarrow a ) \times \overrightarrow b = \overrightarrow 0$ But
Conversely let $ \overrightarrow a,\overrightarrow c$ be collinear $ \Rightarrow \overrightarrow c=k\overrightarrow a$ for same scalar k
Then $(\overrightarrow a \times \overrightarrow b) \times \overrightarrow c = (\overrightarrow a \times \overrightarrow b) \times k\overrightarrow a = k[(\overrightarrow a.\overrightarrow a)\overrightarrow b-(\overrightarrow a.\overrightarrow b)\overrightarrow a]$ and $ \overrightarrow a \times (\overrightarrow b \times \overrightarrow c) = \overrightarrow a \times (\overrightarrow b \times k\overrightarrow a)=k[(\overrightarrow a-\overrightarrow a)\overrightarrow b-(\overrightarrow a.\overrightarrow b)\overrightarrow a]$
It can be seen that $ (\overrightarrow a \times \overrightarrow b ) \times \overrightarrow c=\overrightarrow a \times (\overrightarrow b \times \overrightarrow c)$


answered Jun 9, 2013 by thanvigandhi_1
edited Jun 23, 2013 by thanvigandhi_1

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