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

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

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

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

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

$QED$

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