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
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
Related Videos
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change