Question

For non-coplanar vectors $A,B$ and $C,\left|\right(A\times B)\cdot C|=\left|A\right|\left|B\right|\left|C\right|$ holds if and only if

• A. $A\cdot B=B\cdot C=C.\cdot A=0$
• B. $A\cdot B=0=B\cdot C$
• C. $A\cdot B=0=C\cdot A$
• D. $B\cdot C=0=C\cdot A$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $A\cdot B=B\cdot C=C.\cdot A=0$
B $A\cdot B=0=B\cdot C$
C $B\cdot C=0=C\cdot A$
D $A\cdot B=0=C\cdot A$

Let $\theta$ be the angle between $A$ and $B$

and $\varphi$ the angle between $C$ and the vector $A\times B$,

that is the angle between $C$ and the perpendicular to the plane containing $A$ and $B$.

Then $|(A\times B).C|=|A\times B|\left|C\right|\cos \varphi =\left|A\right|\left|B\right|\left|C\right|\sin \theta \cos \varphi$
So if the given relation holds, we must have $\sin \theta \cos \varphi =1$,

which means both $\sin \theta$ and $\cos \varphi$ must be $1.$
That is, we must have $\theta =\frac{\pi }{2}$ and $\varphi =0$
The former implies that $A$ and $B$ are perpendicular,

So that $A.B=0$.
On the other hand, if $\varphi$ is zero, $C$ must be perpendicular to both $A$ and $B$,

So that $B.C=0=A.C$