Question

Based on dot product of vectors:
Prove that(A+2B). (2A-3B) =2A^2+AB cos¢ -6B^2.

Answer 1

$(\vec{A}+2\vec{B}) . (2 \vec{A}-3\vec{B})\\\\=\vec{A}.2\vec{A}-\vec{A}.3\vec{B}+2\vec{B}.2\vec{A}-2\vec{B}.3\vec{B}\\\\=2|A|^2-3\vec{A}.\vec{B}+4\vec{B}{A}-6|B|^2\\\\=2|A|^2+|A|*|B|\ Cos\ \phi-6|B|^2$

we use the following in the above proof:

$\vec{A}.\vec{B}=|A|*|B|* Cos\ \phi,\ where\ \phi=angle\ between\ vectors\ \vec{A}\ and\ \vec{B}\\\vec{A}.\vec{B}=\vec{B}.\vec{A}\\\vec{nA}=n*\vec{A}$

Answer 2

Explanation:

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