A) 1
B) ? 1
C) \[|\mathbf{a}||\mathbf{b}||\mathbf{c}|\]
D) 0
Correct Answer: C
Solution :
We have \[\mathbf{a}.\,\mathbf{b}=\mathbf{b}.\mathbf{c}=\mathbf{c}.\,\mathbf{a}=0\]and the scalar triple product of three vectors is that, \[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=(\mathbf{a}\times \mathbf{b}).\mathbf{c}\] \[\because \mathbf{a}.\,\mathbf{b}=0,\,\,\,\,\,\,\therefore \mathbf{a}\,\bot \,\mathbf{b}\] So, angle between \[\mathbf{a}\]and \[\mathbf{b}\]is \[\theta =90{}^\text{o}\]. Similarly, \[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=|\mathbf{a}||\mathbf{b}|\mathbf{\hat{n}}.\,\mathbf{c}\], where \[\mathbf{\hat{n}}\] is a normal vector \ \[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=|\mathbf{a}||\mathbf{b}|\mathbf{\hat{n}}\,\,\,\mathbf{c}\] \[\because \mathbf{\hat{n}}\] and \[\mathbf{c}\]are parallel to each other \ \[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=|\mathbf{a}||\mathbf{b}||\mathbf{\hat{n}}|\,\,\,\,|\mathbf{c}|\cos \theta \]\[=|\mathbf{a}||\mathbf{b}||\mathbf{c}|\].
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