If 'a' is any vector then evaluate |(axj)|^2 + |(axi)|^2 + |(axk)|^2

Asked by DIVYA GARG | 6th Jun, 2013, 01:33: PM

Expert Answer:

Let a = mi + nj+ok
 
Then, |(axi)|^2 + |(axj)|^2 + |(axk)|^2 = 
|((mi + nj+ok) x i)|^2 + |((mi + nj+ok) x j)|^2 + |((mi + nj+ok) x k)|^2
= |(-nk + oj)|^2 + |(mk -oi)|^2 + |(-mj+ni)|^2
= n^2 + o^2 + m^2+o^2 + m^2 +n^2
= 2(m^2 + n^2+o^2)
= 2|a|^2

Answered by  | 7th Jun, 2013, 02:07: AM

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