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Prove that (vecA+2vecB).(2 vecA-3vecB) =...

Prove that `(vecA+2vecB).(2 vecA-3vecB) = 2A^2 + ABcos theta - 6B^2` .

Text Solution

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We are given:
L.H.S.= `(vecA + 2vecB).(2vecA - 3vecB)`
`{:(= vecA. 2vecA - vecA . 3vecB + 2vecB. 2vecA - 2vecB - 3vecB),(= 2A^2 - 3vecA - vecB + 4vecA - vecB - 6B^2):} " "({:(because vecA. vecA = A^2),(vecA. vecB = vecB. vecA):})`
`implies 2A^2 + vecA. vecB - 6 B^2`
`= 2A^2 + AB cos theta - 6B^2 = R.H.S.` Hence `(vecA+ 2vecB) . (2vecA - 3vecB) = 2A^2 + AB cos theta - 6B^2`
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