A) \[\frac{\mathbf{a}\times (\mathbf{b}\times \mathbf{c})}{|\mathbf{a}\times (\mathbf{b}\times \mathbf{c})|}\]
B) \[\frac{\mathbf{b}\times (\mathbf{c}\times \mathbf{a})}{|\mathbf{b}\times (\mathbf{c}\times \mathbf{a})|}\]
C) \[\frac{\mathbf{c}\times (\mathbf{a}\times \mathbf{b})}{|\mathbf{c}\times (\mathbf{a}\times \mathbf{b})|}\]
D) None of these
Correct Answer: C
Solution :
Any vector \[(\mathbf{r})\] in plane of \[\mathbf{a},\,\mathbf{b}\] must be in form of linear combination of \[\mathbf{a}\] and \[\mathbf{b}\] \[\overrightarrow{r}=x\mathbf{a}+y\mathbf{b}\] Such combination is possible in alternate \[(c).\] As \[\mathbf{c}\times (\mathbf{a}\times \mathbf{b})=(\mathbf{c}\,.\,\mathbf{b})\mathbf{a}-(\mathbf{c}\,.\,\mathbf{a})\mathbf{b}\] ?..(i) Also (i) is perpendicular to \[\mathbf{c}\] As \[\mathbf{c}.\,\{(\mathbf{c}\,.\,\mathbf{b})\,\mathbf{a}-(\mathbf{c}\,.\,\mathbf{a})\,\mathbf{b}\}\]\[=(\mathbf{c}\,.\,\mathbf{a})(\mathbf{c}\,.\,\mathbf{b})-(\mathbf{c}\,.\,\mathbf{b})(\mathbf{c}\,.\,\mathbf{a})=0\] Thus unit vector perpendicular to \[\mathbf{c}\] and coplanar with \[\mathbf{a},\,\mathbf{b}\] is, \[\mathbf{a}|\,|\mathbf{c}\,\,,\,\,\,\,\therefore \,\,\,\mathbf{a}\,.\,\mathbf{c}=1\]. Other similar concepts: (1) Unit vector perpendicular to \[\mathbf{a}\] and coplanar with \[\mathbf{b}\] and \[\mathbf{c}\] is \[\mathbf{r}=\frac{\mathbf{a}\times (\mathbf{b}\times \mathbf{c})}{|\mathbf{a}\times (\mathbf{b}\times \mathbf{c})|}\]. (2) Unit vector perpendicular to \[\mathbf{b}\] and coplanar with \[\mathbf{c}\] and \[\mathbf{a}\] is\[\mathbf{r}=\frac{\mathbf{b}\times (\mathbf{c}\times \mathbf{a})}{|\mathbf{b}\times (\mathbf{c}\times \mathbf{a})|}\] .
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