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If $|\vec{a}| = 3$, $|\vec{b}| = 2$, then $(2\vec{a} + 3\vec{b})\cdot(2\vec{a} - 3\vec{b})$ is equal to
If $|\vec{a}| = 3$, $|\vec{b}| = 2$, then $(2\vec{a} + 3\vec{b})\cdot(2\vec{a} - 3\vec{b})$ is equal to
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$(A+B)\cdot(A-B) = |A|^2 - |B|^2$ holds for vectors just as $(a+b)(a-b)=a^2-b^2$ does for scalars. The cross-terms cancel because the dot product is commutative.
Updated On: Apr 25, 2026
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