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Prove that (vec(A)+2vec(B)).(2vec(A)-3ve...

Prove that `(vec(A)+2vec(B)).(2vec(A)-3vec(B))=2A^(2)+AB cos theta-6B^(2)`.

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To prove that \((\vec{A} + 2\vec{B}) \cdot (2\vec{A} - 3\vec{B}) = 2A^2 + AB \cos \theta - 6B^2\), we will expand the left-hand side and simplify it step by step. ### Step 1: Expand the Left-Hand Side We start with the expression: \[ (\vec{A} + 2\vec{B}) \cdot (2\vec{A} - 3\vec{B}) \] Using the distributive property of the dot product, we expand this: ...
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