Discussion
If \(|\vec{a}| = 3, |\vec{b}| = 2\), then the value of \((2\vec{a} + 3\vec{b}).(2\vec{a} - 3\vec{b})\) is equal to
Answer: E
The value is 0.
The product is a difference of squares, and \(\vec{a}\cdot\vec{a} = |\vec{a}|^{2}\):
\[ \begin{aligned} (2\vec{a} + 3\vec{b})\cdot(2\vec{a} - 3\vec{b}) &= 4|\vec{a}|^{2} - 9|\vec{b}|^{2} \\ &= 4(9) - 9(4) = 0 \end{aligned} \]
The cross terms \(\pm 6\,\vec{a}\cdot\vec{b}\) cancel, so the angle between the vectors is not needed. Options \(\pm 6\) and \(\pm 12\) come from using \(|\vec{a}|\) instead of \(|\vec{a}|^{2}\) or mixing the coefficients.
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