Vectors

Vectors in two dimensions (addition, scalar multiplication, equation of a line), scalar product

Showing 1-25 of 31 problems
2025 Paper 3 Q5
D: 1500.0 B: 1500.0

Three points, \(A\), \(B\) and \(C\), lie in a horizontal plane, but are not collinear. The point \(O\) lies above the plane. Let \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{c}\). \(P\) is a point with \(\overrightarrow{OP} = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c}\), where \(\alpha\), \(\beta\) and \(\gamma\) are all positive and \(\alpha + \beta + \gamma < 1\). Let \(k = 1 - (\alpha + \beta + \gamma)\).

  1. The point \(L\) is on \(OA\), the point \(X\) is on \(BC\) and \(LX\) passes through \(P\). Determine \(\overrightarrow{OX}\) in terms of \(\beta\), \(\gamma\), \(\mathbf{b}\) and \(\mathbf{c}\) and show that \(\overrightarrow{OL} = \frac{\alpha}{k+\alpha}\mathbf{a}\).
  2. Let \(M\) and \(Y\) be the unique pair of points on \(OB\) and \(CA\) respectively such that \(MY\) passes through \(P\), and let \(N\) and \(Z\) be the unique pair of points on \(OC\) and \(AB\) respectively such that \(NZ\) passes through \(P\). Show that the plane \(LMN\) is also horizontal if and only if \(OP\) intersects plane \(ABC\) at the point \(G\), where \(\overrightarrow{OG} = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\). Where do points \(X\), \(Y\) and \(Z\) lie in this case?
  3. State what the condition \(\alpha + \beta + \gamma < 1\) tells you about the position of \(P\) relative to the tetrahedron \(OABC\).

2019 Paper 2 Q7
D: 1500.0 B: 1500.0

  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.

Show 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 \(\frac{2}{\sqrt{3}}\)
2015 Paper 1 Q6
D: 1484.0 B: 1500.0

The vertices of a plane quadrilateral are labelled \(A\), \(B\), \(A'\) and \(B'\), in clockwise order. A point \(O\) lies in the same plane and within the quadrilateral. The angles \(AOB\) and \(A'OB'\) are right angles, and \(OA=OB\) and \(OA'=OB'\). Use position vectors relative to \(O\) to show that the midpoints of \(AB\), \(BA'\), \(A'B'\) and \(B'A\) are the vertices of a square. Given that the lengths of \(OA\) and \(OA'\) are fixed (and the conditions of the first paragraph still hold), find the value of angle \(BOA'\) for which the area of the square is greatest.

Show Solution
Let \(O\) be the origin, and let \(\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}'\) be the four points. The conditions give us \begin{align*} && \mathbf{a} \cdot \mathbf{b} &= 0 \\ && |\mathbf{a}| &= |\mathbf{b}| \\ && \mathbf{a}' \cdot \mathbf{b}' &= 0 \\ && |\mathbf{a}'| &= |\mathbf{b}'| \\ \end{align*} So \begin{align*} \text{midpoint }AB \text{ to midpoint } BA' &= (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}'))\cdot (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}')) \\ &= \frac12(\mathbf{a}-\mathbf{a}')\cdot \frac12(\mathbf{a} - \mathbf{a}') \\ \text{midpoint }BA' \text{ to midpoint } A'B' &= (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}')) \cdot (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}'))\\ &= \frac12(\mathbf{b}-\mathbf{b}')\cdot \frac12(\mathbf{b} - \mathbf{b}') \\ &= \frac14 (|\mathbf{b}|^2 + |\mathbf{b}'|^2 - 2\mathbf{b}\cdot\mathbf{b}')\\ &= \frac14(|\mathbf{a}|^2 + |\mathbf{a}'|^2 - 2\mathbf{b}\cdot\mathbf{b}') \\ \text{midpoint }A'B' \text{ to midpoint } B'A &= (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a})) \cdot (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a}))\\ &= \frac12(\mathbf{a}'-\mathbf{a})\cdot \frac12(\mathbf{a}' - \mathbf{a}) \\ \text{midpoint }B'A \text{ to midpoint } AB &= (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b})) \cdot (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b}))\\ &= \frac12(\mathbf{b}'-\mathbf{b})\cdot \frac12(\mathbf{b}' - \mathbf{b}) \\ \end{align*} So it's sufficient to prove \(\mathbf{a}\cdot \mathbf{a}' = \mathbf{b}\cdot \mathbf{b}'\) but this is clear from looking at a diagram for 1 second. Given the length of the square is what it is, we want to minimise \(\mathbf{b}\cdot \mathbf{b}'\) which is when they are vertically opposite each other, ie \(\angle BOA' = 90^\circ\)
2011 Paper 2 Q11
D: 1600.0 B: 1487.5

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 \(W\).

2007 Paper 1 Q5
D: 1500.0 B: 1484.0

Note: a regular octahedron is a polyhedron with eight faces each of which is an equilateral triangle.

  1. Show that the angle between any two faces of a regular octahedron is \(\arccos \left( -{\frac1 3} \right)\).
  2. Find the ratio of the volume of a regular octahedron to the volume of the cube whose vertices are the centres of the faces of the octahedron.

Show Solution
  1. Suppose the vertices are \((\pm1, 0,0), (0,\pm1,0), (0,0,\pm1)\), then clearly this is an octahedron. We can measure the angle between the faces, by looking at vectors in the same plane and also in two of the faces: \(\langle \frac12, \frac12, - 1\rangle\) and \(\langle \frac12, \frac12, 1\rangle\), then by considering the dot product: \begin{align*} && \cos \theta &= \frac{\frac14+\frac14-1}{\frac14+\frac14+1} \\ &&&= \frac{-2}{6} = -\frac13 \end{align*}
  2. The volume of our octahedron is \(2 \cdot \frac13 \cdot \underbrace{\sqrt{2}^2}_{\text{base}} \cdot \underbrace{1}_{\text{height}} = \frac43\). The centre of two touching faces are \(\langle \frac13, \frac13, \frac13 \rangle\) and \(\langle \frac13, \frac13, -\frac13 \rangle\) and so the length of the side of the cube is \(\frac23\) and so the volume of the cube is \(\frac8{27}\). Therefore the ratio is \(\frac{2}{9}\)
2006 Paper 2 Q6
D: 1600.0 B: 1516.0

By considering a suitable scalar product, prove that \[ (ax+by+cz)^2 \le (a^2+b^2+c^2)(x^2+y^2+z^2) \] for any real numbers \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\). Deduce a necessary and sufficient condition on \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\) for the following equation to hold: \[ (ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,. \]

  1. Show that \((x+2y+2z)^2 \le 9(x^2+y^2+z^2)\) for all real numbers \(x\), \(y\) and \(z\).
  2. Find real numbers \(p\), \(q\) and \(r\) that satisfy both \[ p^2+4q^2+9r^2 = 729 \text{ and } 8p+8q+3r = 243\,. \]

Show Solution
Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)
  1. By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
  2. Since \begin{align*} && (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\ \Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\ &&3^6 \cdot 3^4 &\geq 3^{10} \end{align*} Therefore we must be in the equality case, ie \(p = 8\lambda, 2q = 4\lambda, 3r = \lambda\) as well as \(64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3\) so we have \[ (p,q,r) = \left (24, 6, 1 \right) \]
2003 Paper 2 Q5
D: 1600.0 B: 1500.0

The position vectors of the points \(A\,\), \(B\,\) and \(P\) with respect to an origin \(O\) are \(a{\bf i}\,\), \(b{\bf j}\,\) and \(l{\bf i}+m{\bf j}+n{\bf k}\,\), respectively, where \(a\), \(b\), and \(n\) are all non-zero. The points \(E\), \(F\), \(G\) and \(H\) are the midpoints of \(OA\), \(BP\), \(OB\) and \(AP\), respectively. Show that the lines \(EF\) and \(GH\) intersect. Let \(D\) be the point with position vector \(d{\bf k}\), where \(d\) is non-zero, and let \(S\) be the point of intersection of \(EF\) and \(GH.\) The point \(T\) is such that the mid-point of \(DT\) is \(S\). Find the position vector of \(T\) and hence find \(d\) in terms of \(n\) if \(T\) lies in the plane \(OAB\).

Show Solution
\(E = \langle \frac{a}{2}, 0,0 \rangle, F = \langle \frac{l}{2}, \frac{m+b}{2}, \frac{n}{2} \rangle, G = \langle 0, \frac{b}{2}, 0 \rangle, H = \langle \frac{a+l}{2}, \frac{m}{2}, \frac{n}{2} \rangle\) Note that the midpoint of \(EF\) and \(GH\) are both $\langle \frac{a+l}{4}, \frac{m+b}{4}, \frac{n}{4} \rangle$, so clearly they must intersect at this point. The vector we just found is \(S\), and \(\mathbf{t} = \mathbf{d} + 2(\mathbf{s}-\mathbf{d}) = 2\mathbf{s} - \mathbf{d}\). Therefore \(T = \langle \frac{a+l}{2}, \frac{m+b}{2}, \frac{n-2d}{2} \rangle\). If \(T\) lies in the plane \(OAB\) then \(n - 2d = 0\) ie \(d = \frac{n}{2}\)
2002 Paper 2 Q7
D: 1600.0 B: 1484.0

In 3-dimensional space, the lines \(m_1\) and \(m_2\) pass through the origin and have directions \(\bf i + j\) and \(\bf i +k \), respectively. Find the directions of the two lines \(m_3\) and \(m_4\) that pass through the origin and make angles of \(\pi/4\) with both \(m_1\) and \(m_2\). Find also the cosine of the acute angle between \(m_3\) and \(m_4\). The points \(A\) and \(B\) lie on \(m_1\) and \(m_2\) respectively, and are each at distance \(\lambda \surd2\) units from~\(O\). The points \(P\) and \(Q\) lie on \(m_3\) and \(m_4\) respectively, and are each at distance \(1\) unit from~\(O\). If all the coordinates (with respect to axes \(\bf i\), \(\bf j\) and \(\bf k\)) of \(A\), \(B\), \(P\) and \(Q\) are non-negative, prove that:

  1. there are only two values of \(\lambda\) for which \(AQ\) is perpendicular to \(BP\,\);
  2. there are no non-zero values of \(\lambda\) for which \(AQ\) and \(BP\) intersect.

2001 Paper 1 Q9
D: 1500.0 B: 1484.0

A ship sails at \(20\) kilometres/hour in a straight line which is, at its closest, 1 kilometre from a port. A tug-boat with maximum speed 12 kilometres/hour leaves the port and intercepts the ship, leaving the port at the latest possible time for which the interception is still possible. How far does the tug-boat travel?

Show Solution
The position of the ship is \(\mathbf{s} = \binom{20t}{1}\). Suppose the interception is at \(T\), then the ship leaves at \(T-\frac1{12}\underbrace{\sqrt{400T^2+1}}_{\text{distance to intercept}}\). We wish to maximise this, ie \begin{align*} && \frac{\d}{\d T} \left ( T - \frac1{12}\sqrt{400T^2+1}\right) &= 1 - \frac{1}{12} \cdot \frac12 \cdot 400 \cdot 2T \cdot \left (400T^2+1 \right)^{-1/2} \\ &&&= 1 - \frac{100}3 T(400T^2+1)^{-1/2} \\ \Rightarrow && \frac{T}{\sqrt{400T^2+1}} &= \frac{3}{100} \\ \Rightarrow && \frac{T^2}{400T^2+1} &= \frac{9}{10000} \\ \Rightarrow && 10000T^2 &= 3600T^2+9 \\ \Rightarrow && 6400T^2 &= 9 \\ \Rightarrow && T &= \pm \frac{3}{80} \quad \text{(T > 0)} \end{align*} Therefore the distance is \(\sqrt{400 \frac{9}{6400} + 1} = \sqrt{\frac{9}{16}+1} = \frac{5}{4} = 1.25 \text{ km}\)
2001 Paper 2 Q3
D: 1600.0 B: 1500.0

The cuboid \(ABCDEFGH\) is such \(AE\), \(BF\), \(CG\), \(DH\) are perpendicular to the opposite faces \(ABCD\) and \(EFGH\), and \(AB =2, BC=1, AE={\lambda}\). Show that if \(\alpha\) is the acute angle between the diagonals \(AG\) and \(BH\) then $$\cos {\alpha} = |\frac {3-{\lambda}^2} {5+{\lambda}^2} |$$ Let \(R\) be the ratio of the volume of the cuboid to its surface area. Show that \(R<\frac{1}{3}\) for all possible values of \(\lambda\). Prove that, if \(R\ge \frac{1}{4}\), then \(\alpha \le \arccos \frac{1}{9}\).

Show Solution
TikZ diagram
Set \(A\) to be the origin, then \(B = \langle 2, 0, 0 \rangle, G = \langle 2, 1, \lambda \rangle, H = \langle 0, 1, \lambda \rangle\), in particular \begin{align*} && AG&= \langle 2, 1, \lambda \rangle \\ && BH &= \langle -2, 1, \lambda \rangle \\ \Rightarrow && \cos \alpha &= |\frac{-4+1+\lambda^2}{\sqrt{2^2+1^2+\lambda^2}\sqrt{(-2)^2+1^2+\lambda^2}}| \\ &&&= |\frac{-3+\lambda^2}{5+\lambda^2}| \end{align*} \begin{align*} && \text{Volume} &= 2\lambda \\ && \text{Surface area} &= 2\cdot2\lambda + 2\cdot\lambda + 2\cdot2 \\ \Rightarrow && R&= \frac{\lambda}{3\lambda + 2} < \frac{1}{3} \\ && \frac14 &\leq R \\ \Rightarrow && 3\lambda +2 &\leq 4\lambda \\ \Rightarrow &&2 & \leq \lambda \end{align*} Then \(\frac{\lambda^2-3}{5+\lambda^2}\) is increasing as \(\lambda\) increases, in particularly the smallest value is \(\frac{1}{9}\).
2001 Paper 3 Q6
D: 1700.0 B: 1488.0

The plane \[ {x \over a} + {y \over b} +{z \over c} = 1 \] meets the co-ordinate axes at the points \(A\), \(B\) and \(C\,\). The point \(M\) has coordinates \(\left( \frac12 a, \frac12 b, \frac 12 c \right)\) and \(O\) is the origin. Show that \(OM\) meets the plane at the centroid \(\left( \frac13 a, \frac13 b, \frac 13 c \right)\) of triangle \(ABC\). Show also that the perpendiculars to the plane from \(O\) and from \(M\) meet the plane at the orthocentre and at the circumcentre of triangle \(ABC\) respectively. Hence prove that the centroid of a triangle lies on the line segment joining its orthocentre and circumcentre, and that it divides this line segment in the ratio \(2 : 1\,\). [The orthocentre of a triangle is the point at which the three altitudes intersect; the circumcentre of a triangle is the point equidistant from the three vertices.]

Show Solution
The line \(OM\) is \(\lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}\), then we need \(1 = \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix} = 3 \lambda \Rightarrow \lambda = \frac13\). Therefore \(OM\) meets the plane at the centroid. The orthocentre is the point \(\mathbf{h}\) such that \((\mathbf{a}-\mathbf{b}) \cdot (\mathbf{c} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} a \\ -b \\ 0 \end{pmatrix} \cdot \begin{pmatrix} -p \\ -q \\ c-r \end{pmatrix} \Leftrightarrow ap-bq = 0\) \((\mathbf{b}-\mathbf{c}) \cdot (\mathbf{a} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} 0 \\ b \\ -c \end{pmatrix} \cdot \begin{pmatrix} a-p \\ -q \\ -r \end{pmatrix} \Leftrightarrow bq-cr = 0\) \((\mathbf{c}-\mathbf{a}) \cdot (\mathbf{b} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} -a \\ 0 \\ c \end{pmatrix} \cdot \begin{pmatrix} -p \\ b-q \\ -r \end{pmatrix} \Leftrightarrow cr-ap = 0\) ie \(ap = bq = cr\) but this is clearly on the line \(\lambda \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix}\) therefore the orthocentre is on the perpendicular from \(O\) \(M-A = \begin{pmatrix} -a/2 \\ b/2 \\ c/2 \end{pmatrix}\) so \(|M-A|=|M-B|=|M-C|\) Also by pythagoras the point of intersection satisfies \(|M-P|^2 + |P-A|^2 = |M-A|^2\) so \(|P-A|^2 = |P-B|^2 = |P-C|^2\), therefore \(P\) is the circumcentre. Since all these points are in the same plane and \(OGM\) is a line, we have the points are in a line. Similar triangles gives the desired ratio
2001 Paper 3 Q10
D: 1700.0 B: 1516.0

Three ships \(A\), \(B\) and \(C\) move with velocities \({\bf v}_1\), \({\bf v}_2\) and \(\bf u\) respectively. The velocities of \(A\) and \(B\) relative to \(C\) are equal in magnitude and perpendicular. Write down conditions that \(\bf u\), \({\bf v}_1\) and \({\bf v}_2\) must satisfy and show that \[ \left| {\bf u} -{\textstyle\frac12} \l {\bf v}_1 + {\bf v}_2 \r \right|^2 = \left|{\textstyle\frac12} \l {\bf v}_1 - {\bf v}_2 \r \right|^2 \] and \[ \l {\bf u} -{\textstyle\frac12} \l {\bf v}_1 + {\bf v}_2 \r \r \cdot \l {\bf v}_1 - {\bf v}_2 \r = 0 \;. \] Explain why these equations determine, for given \({\bf v}_1\) and \({\bf v}_2\), two possible velocities for \(C\,\), provided \({\bf v}_1 \ne {\bf v}_2 \,\). If \({\bf v}_1\) and \({\bf v}_2\) are equal in magnitude and perpendicular, show that if \({\bf u} \ne {\bf 0}\) then \({\bf u} = {\bf v}_1 + {\bf v}_2\,\).

1998 Paper 2 Q8
D: 1600.0 B: 1487.4

Points \(\mathbf{A},\mathbf{B},\mathbf{C}\) in three dimensions have coordinate vectors \(\mathbf{a},\mathbf{b},\mathbf{c}\), respectively. Show that the lines joining the vertices of the triangle \(ABC\) to the mid-points of the opposite sides meet at a point \(R\). \(P\) is a point which is {\bf not} in the plane \(ABC\). Lines are drawn through the mid-points of \(BC\), \(CA\) and \(AB\) parallel to \(PA\), \(PB\) and \(PC\) respectively. Write down the vector equations of the lines and show by inspection that these lines meet at a common point \(Q\). Prove further that the line \(PQ\) meets the plane \(ABC\) at \(R\).

1998 Paper 3 Q6
D: 1700.0 B: 1516.0

  1. Show that four vertices of a cube, no two of which are adjacent, form the vertices of a regular tetrahedron. Hence, or otherwise, find the volume of a regular tetrahedron whose edges are of unit length.
  2. Find the volume of a regular octahedron whose edges are of unit length.
  3. Show that the centres of the faces of a cube form the vertices of a regular octahedron. Show that its volume is half that of the tetrahedron whose vertices are the vertices of the cube.
\noindent [{\em A regular tetrahedron (octahedron) has four (eight) faces, all equilateral triangles.}]

1997 Paper 1 Q5
D: 1500.0 B: 1484.0

Four rigid rods \(AB\), \(BC\), \(CD\) and \(DA\) are freely jointed together to form a quadrilateral in the plane. Show that if \(P\), \(Q\), \(R\), \(S\) are the mid-points of the sides \(AB\), \(BC\), \(CD\), \(DA\), respectively, then \[|AB|^{2}+|CD|^{2}+2|PR|^{2}=|AD|^{2}+|BC|^{2}+2|QS|^{2}.\] Deduce that \(|PR|^{2}-|QS|^{2}\) remains constant however the vertices move. (Here \(|PR|\) denotes the length of \(PR\).)

1997 Paper 1 Q9
D: 1500.0 B: 1484.0

A single stream of cars, each of width \(a\) and exactly in line, is passing along a straight road of breadth \(b\) with speed \(V\). The distance between the successive cars is \(c\).

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A chicken crosses the road in safety at a constant speed \(u\) in a straight line making an angle \(\theta\) with the direction of traffic. Show that \[u\geqslant \frac{Va}{c\sin\theta+a\cos\theta}.\] Show also that if the chicken chooses \(\theta\) and \(u\) so that it crosses the road at the least possible uniform speed, it crosses in time \[\frac{b}{V}\left(\frac{c}{a}+\frac{a}{c}\right) . \]

1995 Paper 1 Q7
D: 1500.0 B: 1516.0

Let \(A,B,C\) be three non-collinear points in the plane. Explain briefly why it is possible to choose an origin equidistant from the three points. Let \(O\) be such an origin, let \(G\) be the centroid of the triangle \(ABC,\) let \(Q\) be a point such that \(\overrightarrow{GQ}=2\overrightarrow{OG},\) and let \(N\) be the midpoint of \(OQ.\)

  1. Show that \(\overrightarrow{AQ}\) is perpendicular to \(\overrightarrow{BC}\) and deduce that the three altitudes of \(\triangle ABC\) are concurrent.
  2. Show that the midpoints of \(AQ,BQ\) and \(CQ\), and the midpoints of the sides of \(\triangle ABC\) are all equidistant from \(N\).
{[}The centroid of \(\triangle ABC\) is the point \(G\) such that \(\overrightarrow{OG}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}).\) The altitudes of the triangle are the lines through the vertices perpendicular to the opposite sides.{]}

1995 Paper 3 Q11
D: 1700.0 B: 1500.0

A ship is sailing due west at \(V\) knots while a plane, with an airspeed of \(kV\) knots, where \(k>\sqrt{2},\) patrols so that it is always to the north west of the ship. If the wind in the area is blowing from north to south at \(V\) knots and the pilot is instructed to return to the ship every thirty minutes, how long will her outward flight last? Assume that the maximum distance of the plane from the ship during the above patrol was \(d_{w}\) miles. If the air now becomes dead calm, and the pilot's orders are maintained, show that the ratio \(d_{w}/d_{c}\) of \(d_{w}\) to the new maximum distance, \(d_{c}\) miles, of the plane from the ship is \[ \frac{k^{2}-2}{2k(k^{2}-1)}\sqrt{4k^{2}-2}. \]

1994 Paper 3 Q3
D: 1700.0 B: 1516.0

Describe geometrically the possible intersections of a plane with a sphere. Let \(P_{1}\) and \(P_{2}\) be the planes with equations \begin{alignat*}{1} 3x-y-1 & =0,\\ x-y+1 & =0, \end{alignat*} respectively, and let \(S_{1}\) and \(S_{2}\) be the spheres with equations \begin{alignat*}{1} x^{2}+y^{2}+z^{2} & =7,\\ x^{2}+y^{2}+z^{2}-6y-4z+10 & =0, \end{alignat*} respectively. Let \(C_{1}\) be the intersection of \(P_{1}\) and \(S_{1},\) let \(C_{2}\) be the intersection of \(P_{2}\) and \(S_{2}\) and let \(L\) be the intersection of \(P_{1}\) and \(P_{2}.\) Find the points where \(L\) meets each of \(S_{1}\) and \(S_{2}.\) Determine, giving your reasons, whether the circles \(C_{1}\) and \(C_{2}\) are linked.

1994 Paper 3 Q10
D: 1700.0 B: 1502.0

The island of Gammaland is totally flat and subject to a constant wind of \(w\) kh\(^{-1},\) blowing from the West. Its southernmost shore stretches almost indefinitely, due east and west, from the coastal city of Alphabet. A novice pilot is making her first solo flight from Alphaport to the town of Betaville which lies north-east of Alphaport. Her instructor has given her the correct heading to reach Betaville, flying at the plane's recommended airspeed of \(v\) kh\(^{-1},\) where \(v>w.\) On reaching Betaport the pilot returns with the opposite heading to that of the outward flight and, so featureless is Gammaland, that she only realises her error as she crosses the coast with Alphaport nowhere in sight. Assuming that she then turns West along the coast, and that her outward flight took \(t\) hours, show that her return flight takes \[ \left(\frac{v+w}{v-w}\right)t\ \text{hours.} \] If Betaville is \(d\) kilometres from Alphaport, show that, with the correct heading, the return flight should have taken \[ t+\frac{\sqrt{2}wd}{v^{2}-w^{2}}\ \text{hours.} \]

1993 Paper 3 Q8
D: 1700.0 B: 1484.0

A square pyramid has its base vertices at the points \(A\) \((a,0,0)\), \(B\) \((0,a,0)\), \(C\) \((-a,0,0)\) and \(D\) \((0,-a,0)\), and its vertex at \(E\) \((0,0,a)\). The point \(P\) lies on \(AE\) with \(x\)-coordinate \(\lambda a\), where \(0<\lambda<1\), and the point \(Q\) lies on \(CE\) with \(x\)-coordinate \(-\mu a\), where \(0<\mu<1\). The plane \(BPQ\) cuts \(DE\) at \(R\) and the \(y\)-coordinate of \(R\) is \(-\gamma a\). Prove that $$ \gamma = {\lambda \mu \over \lambda + \mu - \lambda \mu}. $$ Show that the quadrilateral \(BPRQ\) cannot be a parallelogram.

1992 Paper 3 Q5
D: 1700.0 B: 1484.0

The tetrahedron \(ABCD\) has \(A\) at the point \((0,4,-2)\). It is symmetrical about the plane \(y+z=2,\) which passes through \(A\) and \(D\). The mid-point of \(BC\) is \(N\). The centre, \(Y\), of the sphere \(ABCD\) is at the point \((3,-2,4)\) and lies on \(AN\) such that \(\overrightarrow{AY}=3\overrightarrow{YN}.\) Show that \(BN=6\sqrt{2}\) and find the coordinates of \(B\) and \(C\). The angle \(AYD\) is \(\cos^{-1}\frac{1}{3}.\) Find the coordinates of \(D\). [There are two alternative answers for each point.]

Show Solution
Since \(B\) and \(C\) are reflections of each other in the plane \(y+z=2\) (since that's what it means to be symmetrical), we must have that \(N\) also lies on the plane \(y+z=2\). Since \(\overrightarrow{AY}=3\overrightarrow{YN}.\) we must have \(\overrightarrow{AN}=\overrightarrow{AY}+\overrightarrow{YN} = \frac43\overrightarrow{AY} = \frac43\begin{pmatrix} 3\\-6\\6\end{pmatrix} = \begin{pmatrix} 4\\-8\\8\end{pmatrix}\) and \(N\) is the point \((4,-4,6)\) (which fortunately is on our plane as expected). \(Y\) is the point \((3,-2,4)\) \(|\overrightarrow{AY}| = \sqrt{3^2+(-6)^2+6^2} = 3\sqrt{1+4+4} = 9\)
TikZ diagram
Notice that \(BN^2 + 3^2 = 9^2 \Rightarrow BN^2 = 3\sqrt{3^2-1} = 6\sqrt2\). Therefore \(\overrightarrow{NB} = \pm 6\sqrt{2} \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}\) and \(\{ B, C\} =\{ (4, 2, 12), (4, -10, 0)\}\). Suppose \(D = (x,y,z)\) then \begin{align*} && \begin{pmatrix} -1 \\ 2 \\ -2\end{pmatrix} \cdot \begin{pmatrix} x- 3 \\ y+2 \\ z-4\end{pmatrix} &= 3 \cdot 9 \cdot \frac13 = 9\\ \Rightarrow && 9 &= 3-x+2(y+2)-2(z-4) \\ &&&= -x+2y-2z+15 \\ \Rightarrow && 6 &= x-2y+2z \\ && 2 &= x -4y \\ \\ \Rightarrow && 81 &= (4y+2-3)^2+(y+2)^2+(2-y-4)^2 \\ &&&= (4y-1)^2+2(y+2)^2 \\ &&&= 16y^2-8y+1+2y^2+8y+8 \\ &&&= 18y^2+9 \\ \Rightarrow && y^2 &= 2 \\ \Rightarrow && y &= \pm 2 \end{align*} Therefore \(\displaystyle D \in \left \{ (10, 2, 0), (-6, -2, 4) \right \}\)
1992 Paper 3 Q7
D: 1700.0 B: 1474.8

The points \(P\) and \(R\) lie on the sides \(AB\) and \(AD,\) respectively, of the parallelogram \(ABCD.\) The point \(Q\) is the fourth vertex of the parallelogram \(APQR.\) Prove that \(BR,CQ\) and \(DP\) meet in a point.

Show Solution
Let \(\overrightarrow{AX} = \mathbf{x}\) for all points, so: \begin{align*} \mathbf{p} &= p\mathbf{b}\\ \mathbf{r} &= r\mathbf{d}\\ \mathbf{q} &= \mathbf{p}+\mathbf{r} \\ &= p\mathbf{b} + r\mathbf{d} \end{align*} Therefore \begin{align*} BR: && \mathbf{b} + \lambda(\mathbf{r}-\mathbf{b}) \\ &&= (1-\lambda) \mathbf{b}+ \lambda r \mathbf{d} \\ CQ: && \mathbf{c} + \mu(\mathbf{q} - \mathbf{c}) \\ &&= \mathbf{b}+\mathbf{d} + \mu(p\mathbf{b}+r\mathbf{d} - (\mathbf{b}+\mathbf{d}) ) \\ &&= (1+\mu(p-1))\mathbf{b} + (1+\mu(r-1))\mathbf{d} \\ DP: && \mathbf{d} + \nu (\mathbf{p} - \mathbf{d}) \\ &&= \nu p\mathbf{b} +(1-\nu) \mathbf{d} \end{align*} So we need \(1-\nu = \lambda r, \nu p = 1-\lambda, \) so lets say \(1 = \nu + \lambda r, 1 = \lambda + \nu p \Rightarrow \lambda(pr-1) = p-1 \Rightarrow \lambda = \frac{p-1}{pr-1}\) so they intersect at \(\frac{rp-r}{pr-1} \mathbf{d} + \frac{pr-p}{pr-1}\mathbf{b}\). If we take \(\mu = -\frac{\lambda}{p-1} = 1-pr\) this is clearly also on \(CQ\) hence they all meet at a point
1991 Paper 1 Q5
D: 1500.0 B: 1484.0

A set of \(n\) distinct vectors \(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n},\) where \(n\geqslant2\), is called regular if it satisfies the following two conditions:

  1. there are constants \(\alpha\) and \(\beta\), with \(\alpha>0\), such that for any \(i\) and \(j\), \[ \mathbf{a}_{i}\cdot\mathbf{a}_{j}=\begin{cases} \alpha^{2} & \mbox{ when }i=j\\ \beta & \mbox{ when }i\neq j, \end{cases} \]
  2. the centroid of \(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}\) is the origin \(\mathbf{0}.\) {[}The centroid of vectors \(\mathbf{b}_{1},\mathbf{b}_{2},\ldots,\mathbf{b}_{m}\) is the vector \(\frac{1}{m}(\mathbf{b}_{1}+\mathbf{b}_{2}+\cdots+\mathbf{b}_{m}).\){]}
Prove that (i) and (ii) imply that \((n-1)\beta=-\alpha^{2}.\) If $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0 \end{pmatrix},\( where \)\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}$ is a regular set of vectors in 2-dimensional space, show that either \(n=2\) or \(n=3\), and in each case find the other vectors in the set. Hence, or otherwise, find all regular sets of vectors in 3-dimensional space for which $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\( and \)\mathbf{a}_{2}\( lies in the \)x\(-\)y$ plane.

Show Solution
\begin{align*} && \mathbf{0} &= \sum_i \mathbf{a}_i \tag{ii} \\ \Rightarrow && 0 &= \mathbf{a}_i \cdot \mathbf{0} \\ &&&= \sum_j \mathbf{a}_i \cdot \mathbf{a}_j \\ &&&= (n-1)\beta + \alpha^2 \tag{i} \\ \Rightarrow && (n-1)\beta &= -\alpha^2 \end{align*} Suppose we have \(\mathbf{a}_j = \binom{x}{y}\), \(j \neq 1\) then \(x = \beta\). We also must have \(\beta^2 + y^2 = 1\), so there are at most two values for \(y\), ie two extra vectors. ie \(n = 2, 3\). If \(n = 2 \Rightarrow \mathbf{a}_2 = - \mathbf{a}_1\). If \(n = 3\) \begin{align*} && \mathbf{0} &= \binom{1}{0} + \binom{\beta}{y} + \binom{\beta}{-y} \\ \Rightarrow && \beta = -1/2 \\ \Rightarrow && y &= \pm \frac{\sqrt{3}}{2} \end{align*} Suppose $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\(, \)\mathbf{a}_{2}=\begin{pmatrix}\cos \theta \\ \sin \theta \\ 0 \end{pmatrix}$ (since we need \(\mathbf{a}_2 \cdot \mathbf{a}_2 = 1\)). \(\beta = \cos \theta\)). We can have \(\cos \theta = - 1\). Suppose we have \(\mathbf{a}_j =\begin{pmatrix}x\\ y \\ z \end{pmatrix}\), so \(x = \cos \theta\), and \(y^2 + z^2 = \sin^2 \theta\), so we can write it as: \(\mathbf{a}_j =\begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix}\). We must also have \(\beta = \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta\\ 0 \end{pmatrix} = \cos^2 \theta + \sin^2 \theta \cos \phi = \cos \theta\), so \(\cos \phi = \frac{\cos \theta - \cos^2\theta}{1-\cos^2 \theta} = \frac{\cos \theta}{1+\cos \theta}\). Therefore there is one value for \(\cos \phi\), so at most two values for \(\sin \phi\), Therefore we can have either \(2, 3,4\) or \(5\) different values in the set. \(n = 2\), we've already handled. If \(n = 3\), then \(\beta = -\frac12\), \(\cos \phi = -1\), so we can only have two different values for \(\sin \theta\), ie: \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ \frac{\sqrt{3}}{2} \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ -\frac{\sqrt{3}}{2} \\ 0 \end{pmatrix} \right \}\) Finally, if \(n = 4\), we have \(\beta = -\frac13\), \(\cos \phi = \frac{-1/3}{2/3} = -\frac12\). \(\sin \theta = \pm \frac{\sqrt{3}}{2}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) If \(n = 5\), then \(\beta = -\frac14\), \(\cos \phi = \frac{-1/4}{3/4} = -\frac13\). \(\sin \theta = \frac{\sqrt{15}}{4}\), \(\sin \phi = \frac{2\sqrt{2}}{3}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{12} \\ \frac{\sqrt{30}}{6} \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{12} \\ -\frac{\sqrt{30}}{6} \end{pmatrix}, \right \}\)
1991 Paper 2 Q10
D: 1600.0 B: 1500.0

A straight stick of length \(h\) stands vertically. On a sunny day, the stick casts a shadow on flat horizontal ground. In cartesian axes based on the centre of the Earth, the position of the Sun may be taken to be \(R(\cos\theta,\sin\theta,0)\) where \(\theta\) varies but \(R\) is constant. The positions of the base and tip of the stick are \(a(0,\cos\phi,\sin\phi)\) and \(b(0,\cos\phi,\sin\phi)\), respectively, where \(b-a=h\). Show that the displacement vector from the base of the stick to the tip of the shadow is \[ Rh(R\cos\phi\sin\theta-b)^{-1}\begin{pmatrix}-\cos\theta\\ -\sin^{2}\phi\sin\theta\\ \cos\phi\sin\phi\sin\theta \end{pmatrix}. \] {[}`Stands vertically' means that the centre of the Earth, the base of the stick and the tip of the stick are collinear, `horizontal' means perpendicular to the stick.