Prove that,#(vecA+2vecB).(2vecA-3vecB)=2A^2+ABcostheta -6B^2#thanks?

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Answer 1 · 2017-06-11T06:10:30

See the proof below

We need

#cos0=1#

#vecA*vecA=A*A*cos0=A^2#

#vecB*vecB=B*B*cos0=B^2#

#vecA*vecB=ABcostheta#

Where #theta# is the angle between #vecA# and #vecB#
Therefore

#(vecA+2vecB)(2vecA-3vecB)#

#=vecA*2vecA-vecA*3vecB+2vecA*2vecB-2vecB*3vecB#

#=2A^2+vecA*vecB-6B^2#

#=2A^2+ABcostheta-6B^2#

#QED#