Discussion
Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be the sides of a triangle ABC such that \(\vec{BC} = \vec{a}, \vec{CA} = \vec{b}\) and \(\vec{AB} = \vec{c}\).If BC=AC=3 and \(\vec{b}.\vec{c} = -9\) then \(\vec{a}.\vec{b}\) is equal to
Answer: D
\(\vec{a} \cdot \vec{b}\) equals 0.
Going round the triangle, \(\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}\), so \(\vec{a} = -(\vec{b} + \vec{c})\). With \(|\vec{b}| = CA = 3\):
\[ \begin{aligned} \vec{a} \cdot \vec{b} &= -(\vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{b}) \\ &= -(9 - 9) = 0 \end{aligned} \]
So BC is perpendicular to CA: the triangle is right-angled at C, which fits BC = AC = 3 and AB = \(3\sqrt{2}\). 9 drops the \(\vec{c} \cdot \vec{b}\) term (keeping only \(|\vec{b}|^{2}\)), and 27 and the \(3\sqrt{3}\) values do not follow from \(\vec{a} = -(\vec{b} + \vec{c})\) at all.
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