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Question
If \[\vec{a,} \vec{b,} \vec{c}\] are three non-coplanar mutually perpendicular unit vectors, then \[\left[ \vec{a} \vec{b} \vec{c} \right],\] is
Options
± 1
0
-2
2
Solution
\[\pm 1\]
We have
\[\left[ \vec{a} \vec{b} \vec{c} \right]\]
\[ = \left( \vec{a} \times \vec{b} \right) . \vec{c} \]
\[ = \left| \vec{a} \times \vec{b} \right|\left| \vec{c} \right|\cos 0^\circ \text { or }\left| \vec{a} \times \vec{b} \right|\left| \vec{c} \right|cos { 180 }^\circ \left( \because \vec{a} , \vec{b,} \vec{c} \text { are perpendicular to each other } \right)\]
\[ = \left| \vec{a} \times \vec{b} \right| or - \left| \vec{a} \times \vec{b} \right| \left( \because \left| \vec{c} \right| = 1, cos 0^\circ = 1 \text { and } \cos {180}^\circ= - 1 \right) \]
\[ = \left| \vec{a} \right|\left| \vec{b} \right|\sin {90}^\circ\text {or} - \left| \vec{a} \right|\left| \vec{b} \right|\sin {90}^\circ \left( \because \vec{a}^{} \text { is perpendicular to } \vec{b} \right)\]
\[ = 1 \text { or } - 1 \left( \because \left| \vec{a} \right| = 1 \text { and } \left| \vec{b} \right| = 1 \right)\]
\[ = \pm 1\]
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