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If $\overrightarrow{a}\times \overrightarrow{b}= \overrightarrow{C}\times \overrightarrow{d}$ and $\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}\times\overrightarrow{d},$ show that $\overrightarrow{a}- \overrightarrow{d}$ and $\overrightarrow{b}-\overrightarrow{c}$ are parallel.

1 Answer

  • $ \overrightarrow a \times \overrightarrow b = -(\overrightarrow b \times \overrightarrow a)\: or \: \overrightarrow b \times \overrightarrow a = -(\overrightarrow a \times \overrightarrow b)$ and $ \overrightarrow a \times ( \overrightarrow b + \overrightarrow c) = \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c$
  • If $ \overrightarrow a \: and \: \overrightarrow c $ are parallel vectors then $\overrightarrow a \times \overrightarrow c = \overrightarrow 0$
Consider $ (\overrightarrow a-\overrightarrow d) \times (\overrightarrow b-\overrightarrow c) $
$ = \overrightarrow a \times ( \overrightarrow b - \overrightarrow c)-\overrightarrow d \times (\overrightarrow b-\overrightarrow c)$
$ = \overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c - \overrightarrow d \times \overrightarrow b + \overrightarrow d \times \overrightarrow c$
$ = \overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow d - \overrightarrow c \times \overrightarrow d$
$ = \overrightarrow a \times \overrightarrow b-\overrightarrow c \times \overrightarrow d + \overrightarrow b \times \overrightarrow d - \overrightarrow a \times \overrightarrow c$
$ \overrightarrow 0$ [ Since $ \overrightarrow a \times \overrightarrow b = \overrightarrow c \times \overrightarrow d \: and \: \overrightarrow b \times \overrightarrow d = \overrightarrow a \times \overrightarrow c]$
Now $(\overrightarrow a-\overrightarrow d) \times ( \overrightarrow b-\overrightarrow c)=\overrightarrow 0 \Rightarrow (\overrightarrow a-\overrightarrow d) \parallel ( \overrightarrow b-\overrightarrow c)$
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