Specialist Mathematics · Unit 2 · Circle and geometric proofs · Geometric proofs using vectors
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A circle has centre \(O\) at the origin and radius \(r\). Points \(P\) and \(Q\) lie on the circle such that \(PQ\) is a diameter. Point \(R\) is any other point on the circle. The position vectors of \(P\), \(Q\), and \(R\) are \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) respectively. Given that \(\vec{q} = -\vec{p}\) and \(|\vec{p}| = |\vec{r}| = r\), use vector methods to prove that the angle \(\angle PRQ = 90^\circ\).
A circle has centre \(O\) at the origin and radius \(r\). Points \(A\) and \(B\) lie on the circle such that \(AB\) is a diameter, with \(A = (r, 0)\) and \(B = (-r, 0)\). Point \(P\) lies on the circle at position \((r\cos(\theta), r\sin(\theta))\) where \(0 < \theta < \pi\). Use vectors to prove that \(\angle APB = 90°\).
A circle has centre \(O\) and diameter \(AB\) where \(A = (2, 1)\), \(B = (8, 9)\), and \(P\) is any point on the circumference such that \(P \neq A\) and \(P \neq B\). Use vector methods to prove that \(\angle APB = 90^\circ\).