Question
If $\vec{\text{A}},\vec{\text{B}},\vec{\text{C}}$ are mutually perpendicular, show that $\vec{\text{C}}\times(\vec{\text{A}}\times\vec{\text{B}})=0.$ Is the converse true?
$\therefore$ Angle between $\vec{\text{C}}$ and $\vec{\text{A}}\times\vec{\text{B}}$ is $0^{\circ}$ or $180^{\circ}$
So, $\vec{\text{C}}\times(\vec{\text{A}}\times\vec{\text{B}})=0$ The converse is not true. For example, if two of the vector are parallel, then also$\vec{\text{C}}\times(\vec{\text{A}}\times\vec{\text{B}})=0$
So, they need not be mutually perpendicular.
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