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?
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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.
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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