priscillianaw1

2022-10-07

Prove that: ${‖\stackrel{\to }{v}×\stackrel{\to }{w}‖}^{2}=det\left(\begin{array}{cc}<\stackrel{\to }{v},\stackrel{\to }{v}>& <\stackrel{\to }{v},\stackrel{\to }{w}>\\ <\stackrel{\to }{w},\stackrel{\to }{v}>& <\stackrel{\to }{w},\stackrel{\to }{w}>\end{array}\right)$
What is the relation between cross and inner product so that I can conclude the above equality?

Gabriella Hensley

I think I figured it out
${‖\stackrel{\to }{v}×\stackrel{\to }{w}‖}^{2}={‖\stackrel{\to }{v}‖}^{2}{‖\stackrel{\to }{w}‖}^{2}-<\stackrel{\to }{v},\stackrel{\to }{w}{>}^{2}={‖\stackrel{\to }{v}‖}^{2}{‖\stackrel{\to }{w}‖}^{2}-<\stackrel{\to }{v},\stackrel{\to }{w}><\stackrel{\to }{w},\stackrel{\to }{v}>=<\stackrel{\to }{v},\stackrel{\to }{v}><\stackrel{\to }{w},\stackrel{\to }{w}>-<\stackrel{\to }{v},\stackrel{\to }{w}><\stackrel{\to }{w},\stackrel{\to }{v}>=|\begin{array}{cc}<\stackrel{\to }{v},\stackrel{\to }{v}>& <\stackrel{\to }{v},\stackrel{\to }{w}>\\ <\stackrel{\to }{w},\stackrel{\to }{v}>& <\stackrel{\to }{w},\stackrel{\to }{w}>\end{array}|$

Do you have a similar question?

Recalculate according to your conditions!