Question

A=2I+3j-k and B=4I+6j-2k Angle

Answer 1

cos. theta=|A.B|/|A|.|B|

( 2i+3j-k).(4i+6j-2k)/√4+9+1.√16+36+4

8+18+2/√14.√56

28/√784

cos theta=28/28=1

theta=0°

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