Question

Prove that `|axxb|^2 =a^2b^2 - (a.b)^2`

Answer 1

`Let |a|=a,|b|=b`
and `theta` be the angle between them.
`|axxb|^(2)=(ab"sin"theta)^(2)=a^(2)b^(2)sin^(2)theta`
`=a^(2)b^(2)(1-cos^(2)theta)=a^(2)b^(2)-(a.bcostheta)^(2)`
`=a^(2)b^(2)-(a.b)^(2)=RHS`