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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 start by expanding the left-hand side (LHS) of the equation. ### Step 1: Expand the LHS 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 can expand this as follows: ...
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