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Question
If \[\vec{A} , \vec{B} , \vec{C}\] are mutually perpendicular, show that \[\vec{C} \times \left( \vec{A} \times \vec{B} \right) = 0\] Is the converse true?
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Solution
Given: \[\vec{A} , \vec{B} \text{ and }\vec{C}\] are mutually perpendicular. \[\vec{A} \times \vec{B}\] is a vector with its direction perpendicular to the plane containing \[\vec{A} \text{ and } \vec{B}\]
∴ The angle between \[\vec{C} \text{ and } \vec{A} \times \vec{B}\] is either 0° or 180°.
i.e., \[\vec{C} \times \left( \vec{A} \times \vec{B} \right) = 0\] However, the converse is not true. For example, if two of the vectors are parallel, then also, \[\vec{C} \times \left( \vec{A} \times \vec{B} \right) = 0\] 
So, they need not be mutually perpendicular.
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