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2023 Paper 2 Q8
D: 1500.0 B: 1500.0
vectors dot product tetrahedron geometric proof isosceles tetrahedron centroid angle between vectors

A tetrahedron is called isosceles if each pair of edges which do not share a vertex have equal length.

  1. Prove that a tetrahedron is isosceles if and only if all four faces have the same perimeter.
Let \(OABC\) be an isosceles tetrahedron and let \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{c}\).
  1. By considering the lengths of \(OA\) and \(BC\), show that \[2\mathbf{b}.\mathbf{c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2.\] Show that \[\mathbf{a}.(\mathbf{b}+\mathbf{c}) = |\mathbf{a}|^2.\]
  2. Let \(G\) be the centroid of the tetrahedron, defined by \(\overrightarrow{OG} = \frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c})\). Show that \(G\) is equidistant from all four vertices of the tetrahedron.
  3. By considering the length of the vector \(\mathbf{a}-\mathbf{b}-\mathbf{c}\), or otherwise, show that, in an isosceles tetrahedron, none of the angles between pairs of edges which share a vertex can be obtuse. Can any of them be right angles?

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2005 Paper 2 Q7
D: 1600.0 B: 1501.0
vectors position vectors circular motion dot product trigonometric angle between vectors inequality parametric 3D geometry

The position vectors, relative to an origin \(O\), at time \(t\) of the particles \(P\) and \(Q\) are $$\cos t \; {\bf i} + \sin t\;{\bf j} + 0 \; {\bf k} \text{ and } \cos (t+\tfrac14{\pi})\, \big[{\tfrac32}{\bf i} + { \tfrac {3\sqrt{3}}2} {\bf k}\big] + 3\sin(t+\tfrac14{\pi}) \; {\bf j}\;,$$ respectively, where \(0\le t \le 2\pi\,\).

  1. Give a geometrical description of the motion of \(P\) and \(Q\).
  2. Let \(\theta\) be the angle \(POQ\) at time \(t\) that satisfies \(0\le\theta\le\pi\,\). Show that \[ \cos\theta = \tfrac{3\surd2}{8} -\tfrac14 \cos( 2t +\tfrac14 \pi)\;. \]
  3. Show that the total time for which \(\theta \ge \frac14 \pi\) is \(\tfrac32 \pi\,\).

Solution:

  1. \(P\) is travelling in a unit circle about the origin in the \(\mathbf{i}-\mathbf{j}\) plane. \(Q\) is travelling in a circle (also about the origin, but in a different plane with radius \(3\)).
  2. \(\,\) \begin{align*} && \mathbf{p}\cdot \mathbf{q} &= |\mathbf{p}||\mathbf{q}| \cos \theta \\ \Rightarrow && \cos \theta &= \frac{\tfrac32\cos t \cos(t + \tfrac{\pi}4)+3\sin t \sin (t + \tfrac{\pi}{4})}{3} \\ &&&= \tfrac12\cos t \cos(t + \tfrac{\pi}4)+\sin t \sin (t + \tfrac{\pi}{4}) \\ &&&= \tfrac14 (\cos (2t + \tfrac{\pi}{4}) + \cos(\tfrac{\pi}{4} ))+\tfrac12(\cos(\tfrac{\pi}{4})-\cos(2t + \tfrac{\pi}{4})) \\ &&&= \tfrac{3\sqrt{2}}8 - \tfrac14 \cos ( 2t +\tfrac{\pi}{4}) \end{align*}
  3. If \(\theta \geq \frac14\pi\), then \(\cos \theta \leq \frac{\sqrt{2}}2\) \begin{align*} && \frac{\sqrt{2}}2 & \geq \frac{3\sqrt{2}}8 - \frac14 \cos ( 2t +\tfrac{\pi}{4}) \\ \Rightarrow && \frac{\sqrt{2}}2 &\geq -\cos(2t + \tfrac{\pi}{4}) \\ \Rightarrow && \cos(2t + \tfrac{\pi}{4}) &\geq -\frac{1}{\sqrt{2}} \\ \Rightarrow && 2t + \tfrac{\pi}{4} &\not\in (\tfrac{3\pi}{4},\tfrac{5\pi}{4}) \cup (\tfrac{11\pi}{4},\tfrac{13\pi}{4}) \\ \Rightarrow && t &\not\in (\tfrac{\pi}{4}, \tfrac{\pi}{2})\cup (\tfrac{5\pi}{4}, \tfrac{3\pi}{2}) \end{align*} which is is a time of \(\frac{\pi}{2}\), therefore the left over time is \(\frac32\pi\)

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