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2021 Paper 3 Q4
D: 1500.0 B: 1500.0
vectors dot product projection angle bisector unit vectors trigonometric identities

Let \(\mathbf{n}\) be a vector of unit length and \(\Pi\) be the plane through the origin perpendicular to \(\mathbf{n}\). For any vector \(\mathbf{x}\), the projection of \(\mathbf{x}\) onto the plane \(\Pi\) is defined to be the vector \(\mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\,\mathbf{n}\). The vectors \(\mathbf{a}\) and \(\mathbf{b}\) each have unit length and the angle between them is \(\theta\), which satisfies \(0 < \theta < \pi\). The vector \(\mathbf{m}\) is given by \(\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})\).

  1. Show that \(\mathbf{m}\) bisects the angle between \(\mathbf{a}\) and \(\mathbf{b}\).
  2. The vector \(\mathbf{c}\) also has unit length. The angle between \(\mathbf{a}\) and \(\mathbf{c}\) is \(\alpha\), and the angle between \(\mathbf{b}\) and \(\mathbf{c}\) is \(\beta\). Both angles are acute and non-zero. Let \(\mathbf{a}_1\) and \(\mathbf{b}_1\) be the projections of \(\mathbf{a}\) and \(\mathbf{b}\), respectively, onto the plane through the origin perpendicular to \(\mathbf{c}\). Show that \(\mathbf{a}_1 \cdot \mathbf{c} = 0\) and, by considering \(|\mathbf{a}_1|^2 = \mathbf{a}_1 \cdot \mathbf{a}_1\), show that \(|\mathbf{a}_1| = \sin\alpha\). Show also that the angle \(\varphi\) between \(\mathbf{a}_1\) and \(\mathbf{b}_1\) satisfies \[ \cos\varphi = \frac{\cos\theta - \cos\alpha\cos\beta}{\sin\alpha\sin\beta}. \]
  3. Let \(\mathbf{m}_1\) be the projection of \(\mathbf{m}\) onto the plane through the origin perpendicular to \(\mathbf{c}\). Show that \(\mathbf{m}_1\) bisects the angle between \(\mathbf{a}_1\) and \(\mathbf{b}_1\) if and only if \[ \alpha = \beta \qquad \text{or} \qquad \cos\theta = \cos(\alpha - \beta). \]

Solution:

  1. \(\,\) \begin{align*} && \cos \angle MOB &= \frac{\mathbf{m} \cdot \mathbf{b}}{|\mathbf{m}||\mathbf{b}|} \\ &&&= \frac{\cos \theta + 1}{2\sqrt{\frac14(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})}} \\ &&&= \frac{\cos \theta + 1}{\sqrt{1+1+2\cos \theta}} \\ &&&= \frac{1 + \cos \theta}{\sqrt{2(1+\cos \theta})} \\ &&&= \frac1{\sqrt{2}} \sqrt{1+\cos \theta} \\ &&&= \cos \tfrac{\theta}{2} \end{align*} Since \(0 < \theta < \pi\) we must have \(\angle MOB = \tfrac{\theta}{2}\) ie it is the angle bisector.
  2. The plane through the origin perpendicular to \(\mathbf{c}\) has \(\mathbf{x} \cdot \mathbf{c} = 0\), so \begin{align*} && \mathbf{a}_1 \cdot \mathbf{c} &= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot \mathbf{c} \\ &&&= \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{c} \\ &&&= 0 \\ \\ && |\mathbf{a}_1|^2 &= \mathbf{a}_1 \cdot \mathbf{a}_1 \\ &&&= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \\ &&&= 1 - 2(\mathbf{a} \cdot \mathbf{c})^2 + \mathbf{a} \cdot \mathbf{c} \\ &&&= (1-\cos^2 \alpha) \\ &&&= \sin^2 \alpha \\ \Rightarrow && |\mathbf{a}_1| &= \sin \alpha \\ \Rightarrow && |\mathbf{b}_1| &= \sin \beta \tag{changing a and b} \\ \\ && \cos \phi &= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1}{|\mathbf{a}_1||\mathbf{b}_1|} \\ &&&= \frac{(\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{b} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{c})}{\sin \alpha \sin \beta} \\ &&&= \frac{\mathbf{a} \cdot \mathbf{b} - 2(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})+(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})}{\sin \alpha \sin \beta} \\ &&&= \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta} \end{align*}
  3. Note that \(\mathbf{m}_1 = \tfrac12(\mathbf{a}_1 + \mathbf{b}_1)\) either by expanding or by noting that projection is linear \begin{align*} && \cos \angle M_1OB_1 &= \frac{\mathbf{m}_1 \cdot \mathbf{b}_1}{|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{(\mathbf{a}_1 + \mathbf{b}_1) \cdot \mathbf{b}_1}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1 + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{|\mathbf{a}_1 || \mathbf{b}_1| \cos \phi + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{|\mathbf{a}_1 |\cos \phi + |\mathbf{b}_1|}{2|\mathbf{m}_1|} \\ &&&= \frac{\sin \alpha \cos \phi + \sin \beta}{2\sin \frac{\theta}{2}} \\ &&&= \frac{\sin \alpha \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta} + \sin \beta}{2\sin \frac{\theta}{2}} \\ &&&= \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} \\ \Rightarrow && \cos \angle M_1OA_1 &= \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} \end{align*} \(M_1\) is a bisector iff these two cosines are equal, ie \begin{align*} && \cos \angle M_1OB_1 &= \cos \angle M_1OA_1 \\ \Leftrightarrow && \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} &= \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} \\ \Leftrightarrow && \cos \theta (\sin \alpha - \sin \beta) &= \cos \alpha \cos \beta(\sin \alpha - \sin \beta) + \sin \alpha \sin \beta (\sin \alpha - \sin \beta) \\ \Leftrightarrow &&0 &= (\sin \alpha - \sin \beta)( \cos \theta - (\cos \alpha \cos \beta + \sin \alpha \sin \beta)) \\ &&&= (\sin \alpha - \sin \beta) (\cos \theta - \cos (\alpha - \beta)) \end{align*} From which the result immediately follows

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2019 Paper 2 Q7
D: 1500.0 B: 1500.0
vectors scalar product unit vectors geometric proof equilateral triangle regular tetrahedron quadrilateral dot product identity symmetry

  1. The points \(A\), \(B\) and \(C\) have position vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), respectively. Each of these vectors is a unit vector (so \(\mathbf{a} \cdot \mathbf{a} = 1\), for example) and $$\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}.$$ Show that \(\mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}\). What can be said about the triangle ABC? You should justify your answer.
  2. The four distinct points \(A_i\) (\(i = 1, 2, 3, 4\)) have unit position vectors \(\mathbf{a}_i\) and $$\sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}.$$ Show that \(\mathbf{a}_1 \cdot \mathbf{a}_2 = \mathbf{a}_3 \cdot \mathbf{a}_4\).
    1. Given that the four points lie in a plane, determine the shape of the quadrilateral with vertices \(A_1\), \(A_2\), \(A_3\) and \(A_4\).
    2. Given instead that the four points are the vertices of a regular tetrahedron, find the length of the sides of this tetrahedron.

Solution:

  1. Given \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\), we can form the following results: \begin{align*} && \begin{cases} \mathbf{a} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{b} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{c} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= 0 \\ \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= 0 \\ \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} +\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -\frac12 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{b} = -\frac12 \\ \mathbf{a} \cdot \mathbf{c} = -\frac12 \\ \mathbf{b} \cdot \mathbf{c} = -\frac12 \\ \end{cases} \end{align*} The triangle must be equilateral since the angles between each vertex are the same.
  2. We have \(\displaystyle \sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}\) so \(\displaystyle \mathbf{a}_i \cdot \sum_{i=1}^{4} \mathbf{a}_i = 0\) or for each \(i\), \(\displaystyle \sum_{j \neq i} \mathbf{a}_i \cdot \mathbf{a}_j = -1\). \begin{align*} && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_1 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ && \text{adding the first two, subtracting the last two} \\ \Rightarrow && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 +\cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ \Rightarrow && 2 (\mathbf{a}_1 \cdot \mathbf{a}_2) - 2(\mathbf{a}_3 \cdot \mathbf{a}_4) = 0 \end{align*} Rather than adding the first two and last two, we could have done any pair, resulting in the relations: \begin{align*} \mathbf{a}_1 \cdot \mathbf{a}_2 &= \mathbf{a}_3 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 &= \mathbf{a}_2 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_4 &= \mathbf{a}_2 \cdot \mathbf{a}_3 \end{align*}
    1. The shape must be a parallelogram (from the angle requirement, but also cyclic quadrilateral (since all vectors are unit length), therefore it must be a rectangle
    2. Given it's a regular tetrahedron, \(\mathbf{a}_i \cdot \mathbf{a}_j\) must be the same for all \(i \neq j\), ie \(-\frac13\). We are interested in \(|\mathbf{a}_i - \mathbf{a}_j|\) so consider, \begin{align*} |\mathbf{a}_i - \mathbf{a}_j|^2 &= (\mathbf{a}_i - \mathbf{a}_j) \cdot (\mathbf{a}_i - \mathbf{a}_j) \\ &= \mathbf{a}_i \cdot \mathbf{a}_i - 2 \mathbf{a}_i \cdot \mathbf{a}_j + \mathbf{a}_j \cdot \mathbf{a}_j \\ &= 1 - \frac23 + 1 \\ &= \frac43 \end{align*} Therefore the unit side lengths are \(\frac{2}{\sqrt{3}}\)

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2011 Paper 2 Q11
D: 1600.0 B: 1487.5
vectors 3D geometry statics equilibrium tension unit vectors trigonometry suspended particle resolution of forces

Three non-collinear points \(A\), \(B\) and \(C\) lie in a horizontal ceiling. A particle \(P\) of weight \(W\) is suspended from this ceiling by means of three light inextensible strings \(AP\), \(BP\) and \(CP\), as shown in the diagram. The point \(O\) lies vertically above \(P\) in the ceiling.

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The angles \(AOB\) and \(AOC\) are \(90^\circ+\theta\) and \(90^\circ+\phi\), respectively, where \(\theta\) and \(\phi\) are acute angles such that \(\tan\theta = \sqrt2\) and \(\tan\phi =\frac14\sqrt2\). The strings \(AP\), \(BP\) and \(CP\) make angles \(30^\circ\), \(90^\circ-\theta\) and \(60^\circ\), respectively, with the vertical, and the tensions in these strings have magnitudes \(T\), \(U\) and \(V\) respectively.
  1. Show that the unit vector in the direction \(PB\) can be written in the form \[ -\frac13\, {\bf i} - \frac{\sqrt2\,}3\, {\bf j} + \frac{\sqrt2\, }{\sqrt3 \,} \,{\bf k} \,,\] where \(\bf i\,\), \(\, \bf j\) and \(\bf k\) are the usual mutually perpendicular unit vectors with \(\bf j\) parallel to \(OA\) and \(\bf k\) vertically upwards.
  2. Find expressions in vector form for the forces acting on \(P\).
  3. Show that \(U=\sqrt6 V\) and find \(T\), \(U\) and \(V\) in terms of \(W\).

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2007 Paper 2 Q11
D: 1600.0 B: 1500.0
projectiles 3D projectile motion vectors distance minimisation bearing optimisation unit vectors relative motion

{\sl In this question take the acceleration due to gravity to be \(10\,{\rm m \,s}^{-2}\) and neglect air resistance.} The point \(O\) lies in a horizontal field. The point \(B\) lies \(50\,\)m east of \(O\). A particle is projected from \(B\) at speed \(25\,{\rm m\,s}^{-1}\) at an angle \(\arctan \frac12\) above the horizontal and in a direction that makes an angle \(60^\circ\) with \(OB\); it passes to the north of \(O\).

  1. Taking unit vectors \(\mathbf i\), \(\mathbf j\) and \(\mathbf k\) in the directions east, north and vertically upwards, respectively, find the position vector of the particle relative to \(O\) at time \(t\)~seconds after the particle was projected, and show that its distance from \(O\) is \[ 5(t^2- \sqrt5 t +10)\, {\rm m}. \] When this distance is shortest, the particle is at point \(P\). Find the position vector of \(P\) and its horizontal bearing from \(O\).
  2. Show that the particle reaches its maximum height at \(P\).
  3. When the particle is at \(P\), a marksman fires a bullet from \(O\) directly at \(P\). The initial speed of the bullet is \(350\,{\rm m\,s}^{-1}\). Ignoring the effect of gravity on the bullet show that, when it passes through \(P\), the distance between \(P\) and the particle is approximately~\(3\,\)m.

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