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?

Answer

Given that $\vec{\text{A}},\vec{\text{B}}$ and $\vec{\text{C}}$ are mutually perpendicular$\vec{\text{A}}\times\vec{\text{B}}$ is a vector which direction is perpendicular to the plane containing $\vec{\text{A}}$ and $\vec{\text{B}}.$

Also $\vec{\text{C}}$ is perpendicular to $\vec{\text{A}}$ and $\vec{\text{B}}$

$\therefore$ 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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