Problems

2025 Paper 3 Q5

D: 1500.0 B: 1500.0

vectors position vectors tetrahedron barycentric coordinates geometric proof coplanar points centroid linear combinations

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 $k = 1 - (\alpha + \beta + \gamma)$.

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 $\overrightarrow{OL} = \frac{\alpha}{k+\alpha}\mathbf{a}$.
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?
State what the condition $\alpha + \beta + \gamma < 1$ tells you about the position of $P$ relative to the tetrahedron $OABC$.

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2024 Paper 2 Q4

D: 1500.0 B: 1500.0

vectors angle bisector dot product three-dimensional geometry proof collinearity

In this question, if $O$, $C$ and $D$ are non-collinear points in three dimensional space, we will call the non-zero vector $\mathbf{v}$ a \emph{bisecting vector} for angle $COD$ if $\mathbf{v}$ lies in the plane $COD$, the angle between $\mathbf{v}$ and $\overrightarrow{OC}$ is equal to the angle between $\mathbf{v}$ and $\overrightarrow{OD}$, and both angles are less than $90^\circ$.

Let $O$, $X$ and $Y$ be non-collinear points in three-dimensional space, and define $\mathbf{x} = \overrightarrow{OX}$ and $\mathbf{y} = \overrightarrow{OY}$. Let $\mathbf{b} = |\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}$.
1. Show that $\mathbf{b}$ is a bisecting vector for angle $XOY$. Explain, using a diagram, why any other bisecting vector for angle $XOY$ is a positive multiple of $\mathbf{b}$.
2. Find the value of $\lambda$ such that the point $B$, defined by $\overrightarrow{OB} = \lambda\mathbf{b}$, lies on the line $XY$. Find also the ratio in which the point $B$ divides $XY$.
3. Show, in the case when $OB$ is perpendicular to $XY$, that the triangle $XOY$ is isosceles.
Let $O$, $P$, $Q$ and $R$ be points in three-dimensional space, no three of which are collinear. A bisecting vector is chosen for each of the angles $POQ$, $QOR$ and $ROP$. Show that the three angles between them are either all acute, all obtuse or all right angles.

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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.

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 $\overrightarrow{OC} = \mathbf{c}$.

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 \[\mathbf{a}.(\mathbf{b}+\mathbf{c}) = |\mathbf{a}|^2.\]
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.
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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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 $\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$.

Show that $\mathbf{m}$ bisects the angle between $\mathbf{a}$ and $\mathbf{b}$.
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 \[ \cos\varphi = \frac{\cos\theta - \cos\alpha\cos\beta}{\sin\alpha\sin\beta}. \]
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:

$\,$ \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.
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*}
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 1 Q5

D: 1500.0 B: 1500.0

vectors quadrilateral parallelogram rhombus cube coplanarity centroid geometric proof scalar product optimisation

The four points $P$, $Q$, $R$ and $S$ are the vertices of a plane quadrilateral. What is the geometrical shape of $PQRS$ if $\vec{PQ} = \vec{SR}$? What is the geometrical shape of $PQRS$ if $\vec{PQ} = \vec{SR}$ and $|\vec{PQ}| = |\vec{PS}|$?
A cube with edges of unit length has opposite vertices at $(0,0,0)$ and $(1,1,1)$. The points $$P(p,0,0), \quad Q(1,q,0), \quad R(r,1,1) \quad \text{and} \quad S(0,s,1)$$ lie on edges of the cube. Given that the four points lie in the same plane, show that $$rq = (1-s)(1-p).$$
1. Show that $\vec{PQ} = \vec{SR}$ if and only if the centroid of the quadrilateral $PQRS$ is at the centre of the cube. Note: the centroid of the quadrilateral $PQRS$ is the point with position vector $$\frac{1}{4}(\vec{OP} + \vec{OQ} + \vec{OR} + \vec{OS}),$$ where $O$ is the origin.
2. Given that $\vec{PQ} = \vec{SR}$ and $|\vec{PQ}| = |\vec{PS}|$, express $q$, $r$ and $s$ in terms of $p$. Show that $$\cos PQR = \frac{4p-1}{5-4p+8p^2}.$$ Write down the values of $p$, $q$, $r$ and $s$ if $PQRS$ is a square, and show that the length of each side of this square is greater than $\frac{21}{20}$.

Solution:

If $\vec{PQ} = \vec{SR}$ we have a parallelogram. $\vec{PQ} = \vec{SR}$ and $|\vec{PQ}| = |\vec{PS}|$ then we have a rhombus.
If the four points lie in a plane then $(\vec{RS} \times \vec{RP}) \cdot \vec{RQ} =0$, so \begin{align*} && 0 &=\left ( \begin{pmatrix}-r\\ s-1 \\ 0 \end{pmatrix} \times \begin{pmatrix}p-r\\ -1 \\ -1 \end{pmatrix}\right) \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ && &= \begin{pmatrix}1-s \\ -r \\r+(p-r)(1-s) \end{pmatrix} \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ &&&= (1-s)(1-r)-r(q-1)-r-(p-r)(1-s) \\ &&&=(1-s)(1-r-p+r)-rq \\ \Rightarrow && rq &= (1-s)(1-p) \end{align*}
1. $\,$ \begin{align*} && \vec{PQ} &= \vec{SR} \\ \Leftrightarrow && \begin{pmatrix}1-p\\q \\ 0 \end{pmatrix} &= \begin{pmatrix}r\\1-s \\ 0 \end{pmatrix} \\ \Leftrightarrow && 1-p = r & \quad ; \quad q = 1-s\\ \Leftrightarrow && 1= r+p & \quad ; \quad 1 = q+s\\ \end{align*} The centroid is $\frac14 (p+1+r, q+s+1, 2)$ which is clearly $\frac12(1,1,1)$ iff those equations are true.
2. $\,$ \begin{align*} && |\vec{PQ}| &= |\vec{PS}| \\ \Leftrightarrow && (1-p)^2+q^2+ 0^2 &= p^2+s^2+1)\\ \Leftrightarrow && 1-2p+p^2+q^2 &= p^2 + s^2 + 1 \\ \Leftrightarrow && -2p+q^2 &= s^2 \end{align*} From the previous equations we have $r = 1-p$, and $-2p+(1-s)^2 = s^2 \Rightarrow -2p + 1 -2s = 0 \Rightarrow s = \frac12 - p$ and $q = \frac12 + p$ \begin{align*} && \cos PQR &= \frac{\vec{QP}\cdot \vec{QR}}{|\vec{QP}||\vec{QR}|} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -q \\ 0 \end{pmatrix} \cdot \begin{pmatrix}r-1\\ 1-q \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+q^2}\sqrt{(r-1)^2+(1-q)^2+1^2}} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -\frac12-p \\ 0 \end{pmatrix} \cdot \begin{pmatrix}-p\\ \frac12-p \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+(-\frac12-p)^2}\sqrt{p^2+(\frac12-p)^2+1^2}} \\ &&&= \frac{ p-p^2-\frac14+p^2}{\sqrt{p^2-2p+1+\frac14+p+p^2}\sqrt{p^2+\frac14-p+p^2+1}} \\ &&&= \frac{4p-1}{\sqrt{8p^2-4p+5}\sqrt{8p^2-4p+5}}\\ &&&= \frac{4p-1}{8p^2-4p+5}\\ \end{align*} For $PQRS$ to be a square $\cos PQR = 0$, ie $p = \frac14$ and so $(p,q,r,s) = (\frac14, \frac34, \frac34, \frac14)$ and $|PQ| = \sqrt{(1-p)^2+q^2} = \sqrt{\left ( \frac34 \right)^2 + \left ( \frac34 \right)^2 } = \frac{3\sqrt{2}}4$, notice that $\left ( \frac{21}{20} \right)^2 = \frac{441}{400} < \frac{9}{8}$ ($441 < 450$) therefore the sides are at least as long as $\frac{21}{20}$

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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

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.
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:

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.
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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2019 Paper 3 Q8

D: 1500.0 B: 1500.0

trigonometry vectors pyramid scalar product perpendicular cross product geometric proof inequality 3D geometry

A pyramid has a horizontal rectangular base $ABCD$ and its vertex $V$ is vertically above the centre of the base. The acute angle between the face $AVB$ and the base is $\alpha$, the acute angle between the face $BVC$ and the base is $\beta$ and the obtuse angle between the faces $AVB$ and $BVC$ is $\pi - \theta$.

The edges $AB$ and $BC$ are parallel to the unit vectors $\mathbf{i}$ and $\mathbf{j}$, respectively, and the unit vector $\mathbf{k}$ is vertical. Find a unit vector that is perpendicular to the face $AVB$. Show that $$\cos \theta = \cos \alpha \cos \beta.$$
The edge $BV$ makes an angle $\phi$ with the base. Show that $$\cot^2 \phi = \cot^2 \alpha + \cot^2 \beta.$$ Show also that $$\cos^2 \phi = \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1 - \cos^2 \theta} \geq \frac{2 \cos \theta - 2 \cos^2 \theta}{1 - \cos^2 \theta}$$ and deduce that $\phi < \theta$.

Solution:

Let $A = (0,0,0)$ and then $B = 22b \mathbf{i}, D = 2d\mathbf{j}, C = 2b\mathbf{i}+2d\mathbf{j}$ and $V = b \mathbf{i} + d\mathbf{j} + h\mathbf{k}$ We also have \begin{align*} && \tan \alpha &= \frac{h}{d}\\ && \tan \beta &= \frac{d}{b} \\ && \vec{AV} \times \vec{VB} &= \begin{pmatrix} b \\ d \\ h \end{pmatrix} \times \begin{pmatrix} -b \\ d \\ h \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ -2bh \\ 2db \end{pmatrix} \\ &&&= 2b \begin{pmatrix} 0 \\ -d \tan \alpha \\ d \end{pmatrix} \\ &&&= k \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \end{align*} similarly for the vector perpendicular to the other face it must be $\begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix}$ Looking at the angle between these perpendicular (to find the angles between the faces we see: \begin{align*} \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \cdot \begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix} &= \cos \alpha \cos \beta \end{align*} But this is also $\pi -$ the angle between the planes, ie $\cos \theta = \cos \alpha \cos \beta$
$\,$ \begin{align*} && \cot^2 \phi &= \frac{b^2+d^2}{h^2} \\ && \cot^2 \alpha &= \frac{d^2}{h^2} \\ && \cot^2 \beta &= \frac{b^2}{h^2} \\ \Rightarrow && cot^2 \phi &= \cot^2 \beta+\cot^2 \alpha \end{align*} \begin{align*} && \cos^2 \phi &= \frac{b^2+d^2}{b^2+d^2+h^2} \\ && \cos^2 \alpha &= \frac{d^2}{d^2+h^2} \\ && \cos^2 \beta &= \frac{b^2}{b^2+h^2} \\ && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &= \frac{\frac{d^2}{d^2+h^2}+\frac{b^2}{b^2+h^2}-2\cdot \frac{d^2}{d^2+h^2} \cdot \frac{b^2}{b^2+h^2}}{1 - \frac{d^2}{d^2+h^2} \cdot\frac{b^2}{b^2+h^2}} \\ &&&= \frac{d^2(b^2+h^2)+b^2(d^2+h^2)-2d^2b^2}{(d^2+h^2)(b^2+h^2)-d^2b^2} \\ &&&= \frac{h^2(b^2+d^2)}{h^2(b^2+d^2+h^2)} \\ &&&= \frac{b^2+d^2}{b^2+d^2+h^2} \\ &&&= \cos^2\phi \end{align*} Also notice that \begin{align*} && \cos^2 \alpha + \cos^2 \beta &\underbrace{\geq}_{AM-GM} 2 \cos \alpha \cos \beta \\ &&&= 2 \cos \theta \\ \Rightarrow && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &\geq \frac{2 \cos \theta - 2\cos^2 \theta}{1-\cos^2 \theta} \\ &&&= \frac{2\cos \theta}{1+\cos \theta} = \cos \theta \frac{2}{1+\cos \theta} \\ &&&> \cos^2 \theta \\ \Rightarrow && \phi &< \theta \end{align*}

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2018 Paper 2 Q7

D: 1600.0 B: 1500.0

vectors geometric proof triangle position vectors ratio parallel lines intersection point cevians

The points $O$, $A$ and $B$ are the vertices of an acute-angled triangle. The points $M$ and $N$ lie on the sides $OA$ and $OB$ respectively, and the lines $AN$ and $BM$ intersect at $Q$. The position vector of $A$ with respect to $O$ is $\bf a$, and the position vectors of the other points are labelled similarly. Given that $\vert MQ \vert = \mu \vert QB\vert $, and that $\vert NQ \vert = \nu \vert QA\vert $, where $\mu$ and $\nu$ are positive and $\mu \nu <1$, show that \[ {\bf m} = \frac {(1+\mu)\nu}{1+\nu} \, {\bf a} \,. \] The point $L$ lies on the side $OB$, and $\vert OL \vert = \lambda \vert OB \vert \,$. Given that $ML$ is parallel to $AN$, express $\lambda$ in terms of $\mu$ and $\nu$. What is the geometrical significance of the condition $\mu\nu<1\,$?

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2017 Paper 2 Q8

D: 1600.0 B: 1500.0

vectors geometric proof perpendicular triangle altitudes orthocentre scalar product line equations

All vectors in this question lie in the same plane. The vertices of the non-right-angled triangle $ABC$ have position vectors $\bf a$, $\bf b$ and $\bf c$, respectively. The non-zero vectors $\bf u$ and $\bf v$ are perpendicular to $BC$ and $CA$, respectively. Write down the vector equation of the line through $A$ perpendicular to $BC$, in terms of $\bf u$, $\bf a$ and a parameter $\lambda $. The line through $A$ perpendicular to $BC$ intersects the line through $B$ perpendicular to $CA$ at $P$. Find the position vector of $P$ in terms of $\bf a$, $\bf b$, $\bf c$ and $\bf u$. Hence show that the line $CP$ is perpendicular to the line $AB$.

Solution: The line through $A$ perpendicular to $BC$ is $\mathbf{a} + \lambda\mathbf{u}$. The line through $B$ perpendicular to $CA$ is $\mathbf{b} + \mu \mathbf{v}$. They intersect when $\mathbf{a} + \lambda\mathbf{u} = \mathbf{b} + \mu \mathbf{v}$. Since $\mathbf{v}$ is perpendicular to $CA$, we must have \begin{align*} && \mathbf{a} + \lambda\mathbf{u} &= \mathbf{b} + \mu \mathbf{v} \\ \Rightarrow && \mathbf{a}\cdot(\mathbf{c}-\mathbf{a}) + \lambda\mathbf{u}\cdot(\mathbf{c}-\mathbf{a}) &= \mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) + \mu \mathbf{v}\cdot(\mathbf{c}-\mathbf{a}) \\ \\ \Rightarrow && \lambda &= \frac{\mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) -\mathbf{a}\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u}\cdot(\mathbf{c}-\mathbf{a})} \\ &&&= \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \end{align*} Therefore the point is $\mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u}$. The line $CP$ is $\mathbf{c} + \nu \left (\mathbf{p} - \mathbf{c} \right)$, to check this is perpendicular with $AB$ we should dot $\mathbf{p}-\mathbf{c}$ with $\mathbf{a}-\mathbf{b}$, ie \begin{align*} && (\mathbf{p}-\mathbf{c}) \cdot (\mathbf{a}-\mathbf{b}) &= \left ( \mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} - \mathbf{c}\right) \cdot ( \mathbf{a}-\mathbf{b}) \\ &&&= \left ( \mathbf{a}- \mathbf{c} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} \right) \cdot ( \mathbf{a}-\mathbf{c}+(\mathbf{c}-\mathbf{b})) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})}\mathbf{u} \cdot (\mathbf{a}-\mathbf{c}) + \\ &&&\quad (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) + \lambda \underbrace{\mathbf{u} \cdot (\mathbf{c}-\mathbf{b})}_{=0} \\ &&&=(\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) -(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})+ (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}+\mathbf{b}-\mathbf{a}+\mathbf{c}-\mathbf{b}) \\ &&&= 0 \end{align*} as required.

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2016 Paper 1 Q6

D: 1500.0 B: 1484.7

vectors position vectors quadrilateral parallel lines intersection ratio geometric proof collinearity

The sides $OA$ and $CB$ of the quadrilateral $OABC$ are parallel. The point $X$ lies on $OA$, between $O$ and $A$. The position vectors of $A$, $B$, $C$ and $X$ relative to the origin $O$ are $\bf a$, $\bf b$, $\bf c$ and $\bf x$, respectively. Explain why $\bf c$ and $\bf x$ can be written in the form \[ {\bf c} = k {\bf a} + {\bf b} \text{ and } {\bf x} = m {\bf a}\,, \] where $k$ and $m$ are scalars, and state the range of values that each of $k$ and $m$ can take. The lines $OB$ and $AC$ intersect at $D$, the lines $XD$ and $BC$ intersect at $Y$ and the lines $OY$ and $AB$ intersect at $Z$. Show that the position vector of $Z$ relative to $O$ can be written as \[ \frac{ {\bf b} + mk {\bf a}}{mk+1}\,. \] The lines $DZ$ and $OA$ intersect at $T$. Show that \[ OT \times OA = OX\times TA \text{ and } \frac 1 {OT} = \frac 1 {OX} + \frac 1 {OA} \,, \] where, for example, $OT$ denotes the length of the line joining $O$ and $T$.

Solution:

Notice that $\mathbf{x} = m\mathbf{a}$ since $OX$ is parallel to $OA$ and $0 < m < 1$ since $X$ lies between them. $\overline{OC} = \overline{OB} + \overline{BC} = \mathbf{b} + k\mathbf{a}$ since $BC$ is parallel to $OA$, $k$ can take any value. The line $OB$ is $\lambda \mathbf{b}$, the line $AC$ is $\mathbf{a} + \mu (\mathbf{c}-\mathbf{a}) = \mu \mathbf{b} +(1+ \mu(k-1)) \mathbf{a}$ Therefore they meet when $\mu = \lambda$ and $(1+\mu(k-1)) = 0$, ie $\mu = \frac{1}{1-k}$ so $D$ is $\frac{1}{1-k} \mathbf{b}$ The line $XD$ is $m\mathbf{a} + \nu ( \frac{1}{1-k} \mathbf{b} - m \mathbf{a}) $ and $BC$ is $\mathbf{b} + \eta \mathbf{a}$ so they meet when $\nu = 1-k$ and $\eta = m-(1-k)m = km$. Therefore $Y = \mathbf{b} + km \mathbf{a}$ Therefore the line $OY$ is $\alpha(\mathbf{b} + km \mathbf{a})$ and AB is $\mathbf{a} + \beta(\mathbf{b} - \mathbf{a})$ so they intersect when $\alpha = \beta$ and $\alpha km = (1-\alpha) \Rightarrow \alpha = \frac{1}{1+km}$. Therefore $Z = \mathbf{a} + \frac{1}{1+km} (\mathbf{b} - \mathbf{a}) = \frac{\mathbf{b}+km\mathbf{a}}{1+km}$ The lines $DZ$ and $OA$ are $\frac{1}{1-k} \mathbf{b} + \gamma \left ( \frac{1}{1-k} \mathbf{b} - \frac{\mathbf{b}+km\mathbf{a}}{1+km} \right)$ and $\delta \mathbf{a}$. Therefore they intersect when $\frac{1}{1-k} + \gamma \left (\frac{1}{1-k} - \frac{1}{1+km} \right) = 0 \Rightarrow \gamma = \frac{(1-k)(1+km)}{(k-1)k(m+1)} = -\frac{1+km}{k(m+1)}$ and $\delta = -\gamma \frac{km}{1+km} = \frac{m}{m+1}$. Therefore $OT = \frac{m}{m+1} |\mathbf{a}|, OA = |\mathbf{a}|, OX = m|\mathbf{a}|, TA = \frac{1}{m+1}|\mathbf{a}|$, Therefore $OT \times OA = OX \times TA$. Also $\frac{1}{OX} + \frac{1}{OA} = \frac{1}{m|\mathbf{a}|} + \frac{1}{|\mathbf{a}|} = \frac{m+1}{m|\mathbf{a}|} = \frac{1}{OT}$

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2015 Paper 1 Q6

D: 1484.0 B: 1500.0

vectors geometric proof quadrilateral midpoints square right angle optimisation position vectors

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.

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 $\angle BOA' = 90^\circ$

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2015 Paper 2 Q8

D: 1600.0 B: 1500.0

vectors circles tangent collinearity position vectors geometric proof ratio external tangent

\noindent

The diagram above shows two non-overlapping circles $C_1$ and $C_2$ of different sizes. The lines $L$ and $L'$ are the two common tangents to $C_1$ and $C_2$ such that the two circles lie on the same side of each of the tangents. The lines $L$ and $L'$ intersect at the point $P$ which is called the focus of $C_1$ and $C_2$.

Let $\mathbf{x}_1$ and $\mathbf{x}_2$ be the position vectors of the centres of $C_1$ and $C_2$, respectively. Show that the position vector of $P$ is \[ \frac{r_1 \mathbf{x}_2- r_2 \mathbf{x}_1}{r_1-r_2} \,, \] where $r_1$ and $r_2$ are the radii of $C_1$ and $C_2$, respectively.
The circle $C_3$ does not overlap either $C_1$ or $C_2$ and its radius, $r_3$, satisfies $r_1 \ne r_3 \ne r_2$. The focus of $C_1$ and $C_3$ is $Q$, and the focus of $C_2$ and $C_3$ is $R$. Show that $P$, $Q$ and $R$ lie on the same straight line.
Find a condition on $r_1$, $r_2$ and $r_3$ for $Q$ to lie half-way between $P$ and $R$.

Solution:

Notice that $P$ lies on $C_1C_2$, and that the triangles formed from $C_iPT_i$ where $T_i$ are the tangent points are similar, with ratios $\frac{r_1}{r_2}$. Therefore $\frac{C_1P}{r_1} = \frac{C_2P}{r_2}$, and hence $\frac{C_1P}{C_1C_2} = \frac{C_1P}{C_1P-C_2P} = \frac{1}{1-\frac{r_2}{r_1}} = \frac{r_1}{r_1-r_2}$ So we have $\mathbf{p} = \mathbf{x_1} + (\mathbf{x}_2 - \mathbf{x}_1)\cdot\frac{r_1}{r_1-r_2} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1-r_2}$
Suppose $\mathbf{x}_3 = \binom{\alpha}{\beta}$ in the basis of $\{ \mathbf{x}_1, \mathbf{x}_2 \}$, then we can see that \begin{align*} && \mathbf{p} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} \\ && \mathbf{q} &= \frac{r_1(\alpha \mathbf{x}_1 +\beta \mathbf{x}_2) - r_3\mathbf{x}_1}{r_1-r_3} \\ &&&= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ && \mathbf{r} &=\frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ && \mathbf{p}-\mathbf{q} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} - \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ &&&= \frac{1}{(r_1-r_2)(r_1-r_3)} \binom{(r_1-r_3)(-r_2)-(r_1-r_2)(r_1\alpha-r_3)}{(r_1-r_3)r_1 - (r_1-r_2)r_1\beta} \\ &&&= \frac{r_1}{(r_1-r_2)(r_1-r_3)} \binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \\ && \mathbf{q} - \mathbf{r} &= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} - \frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(r_2-r_3)(r_1\alpha-r_3) - (r_1-r_3)r_2\alpha)}{(r_2-r_3)r_1\beta - (r_1-r_3)(r_2\beta - r_3)} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(-r_2r_3+r_3^2) - \alpha(r_1r_3-r_3r_2)}{r_3(r_1-r_3)-\beta(r_1-r_2)} \\ &&&= \frac{r_3}{(r_1-r_3)(r_2-r_3)}\binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \end{align*} Therefore they are clearly parallel, and hence lie on a line.
$Q$ is halfway between $P$ and $R$ if \begin{align*} && \frac{r_1}{(r_1-r_2)(r_1-r_3)} &= \frac{r_3}{(r_1-r_3)(r_2-r_3)} \\ \Leftrightarrow && r_1(r_2-r_3) &= r_3(r_1-r_2) \\ \Leftrightarrow && r_1r_2 - r_1r_3 &= r_1r_3 - r_2r_3 \\ \Leftrightarrow && r_2 &= \frac{2r_1r_3}{r_1+r_3} \end{align*}

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2014 Paper 1 Q7

D: 1516.0 B: 1500.0

vectors geometric proof triangle ratio intersection of lines position vectors section formula

In the triangle $OAB$, the point $D$ divides the side $BO$ in the ratio $r:1$ (so that $BD = rDO$), and the point $E$ divides the side $OA$ in the ratio $s:1$ (so that $OE =s EA$), where $r$ and $s$ are both positive.

The lines $AD$ and $BE$ intersect at $G$. Show that \[ \mathbf{g}= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b} \,, \] where $\mathbf{a}, \mathbf{b}$ and $\mathbf{g}$ are the position vectors with respect to $O$ of $A$, $B$ and $G$, respectively.
The line through $G$ and $O$ meets $AB$ at $F$. Given that $F$ divides $AB$ in the ratio $t:1$, find an expression for $t$ in terms of $r$ and $s$.

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2014 Paper 3 Q7

D: 1700.0 B: 1484.0

vectors cyclic quadrilateral similar triangles position vectors geometric proof linear dependence diagonals intersection circle theorems

The four distinct points $P_i$ ($i=1$, $2$, $3$, $4$) are the vertices, labelled anticlockwise, of a cyclic quadrilateral. The lines $P_1P_3$ and $P_2P_4$ intersect at $Q$.

By considering the triangles $P_1QP_4$ and $P_2QP_3$ show that $(P_1Q)( QP_3) = (P_2Q) (QP_4)\,$.
Let $\+p_i$ be the position vector of the point $P_i$ ($i=1$, $2$, $3$, $4$). Show that there exist numbers $a_i$, not all zero, such that \begin{equation} \sum\limits_{i=1}^4 a_i =0 \qquad\text{and}\qquad \sum\limits_{i=1}^4 a_i \+p_i ={\bf 0} \,. \tag{$*$} \end{equation}
Let $a_i$ ($i=1$,~$2$, $3$,~$4$) be any numbers, not all zero, that satisfy~$(*)$. Show that $a_1+a_3\ne 0$ and that the lines $P_1P_3$ and $P_2P_4$ intersect at the point with position vector \[ \frac{a_1 \+p_1 + a_3 \+p_3}{a_1+a_3} \,. \] Deduce that $a_1a_3 (P_1P_3)^2 = a_2a_4 (P_2P_4)^2\,$.

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2013 Paper 1 Q3

D: 1500.0 B: 1500.0

vectors binary operation associativity geometric proof sequences convex combination line segment division algebraic manipulation

For any two points $X$ and $Y$, with position vectors $\bf x$ and $\bf y$ respectively, $X*Y$ is defined to be the point with position vector $\lambda {\bf x}+ (1-\lambda){\bf y}$, where $\lambda$ is a fixed number.

If $X$ and $Y$ are distinct, show that $X*Y$ and $Y*X$ are distinct unless $\lambda$ takes a certain value (which you should state).
Under what conditions are $(X*Y)*Z$ and $X*(Y*Z)\,$ distinct?
Show that, for any points $X$, $Y$ and $Z$, \[ (X*Y)*Z = (X*Z)*(Y*Z)\, \] and obtain the corresponding result for $X*(Y*Z)$.
The points $P_1$, $P_2$, $\ldots$ are defined by $ P_1 = X*Y$ and, for $n \ge2$, $P_n= P_{n-1}*Y\,.$ Given that $X$ and $Y$ are distinct and that $0<\lambda<1$, find the ratio in which $P_n$ divides the line segment $XY$.

Solution:

Suppose $X*Y = Y*X$, then \begin{align*} && X * Y &= \lambda \mathbf{x} + (1-\lambda) \mathbf{y} \\ && Y * X &= \lambda \mathbf{y} + (1-\lambda) \mathbf{x}\\ \Rightarrow && 0 &= (2\lambda - 1)(\mathbf{x} -\mathbf{y}) \end{align*} Therefore, either $\mathbf{x} = \mathbf{y}$ or $\lambda = \frac12$. Since we assumed $X,Y$ were distinct, $\mathbf{x} \neq \mathbf{y}$ and so $X*Y$ and $Y*X$ are distinct unless $\lambda = \frac12$
Suppose $(X*Y)*Z = X*(Y*Z)$ \begin{align*} &&(X*Y)*Z &= (\lambda \mathbf{x} + (1-\lambda) \mathbf{y}) * \mathbf{z} \\ &&&= (\lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)\mathbf{z}\\ &&X*(Y*Z) &=\mathbf{x}* (\lambda \mathbf{y} + (1-\lambda) \mathbf{z}) \\ &&&= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z}\\ \Rightarrow && 0 &= (\lambda^2 - \lambda)\mathbf{x} + ((1-\lambda) - (1-\lambda)^2)\mathbf{z} \\ &&&=(1-\lambda)(-\lambda \mathbf{x} +\lambda \mathbf{z}) \\ &&&= \lambda(1-\lambda)(\mathbf{z}-\mathbf{x}) \end{align*} Therefore they are distinct unless $\lambda = 1, 0$ or $\mathbf{x} = \mathbf{z}$.
Claim: $(X*Y)*Z = (X*Z)*(Y*Z)$ Proof: \begin{align*} && (X*Y)*Z &= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ && (X*Z)*(Y*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) * (\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) + (1-\lambda)(\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda) \mathbf{z} \end{align*} Claim: $X*(Y*Z) = (X*Y)*(X*Z)$ Proof: \begin{align*} X*(Y*Z) &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ (X*Y)*(X*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y})*(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda (\lambda \mathbf{x} + (1-\lambda)\mathbf{y}) + (1-\lambda)(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \end{align*}
$P_1 = X*Y$ divides the line segment into the ratio $\lambda:(1-\lambda)$. $P_n$ divides the line segment $P_{n-1}Y$ into the ratio $\lambda:(1-\lambda)$, therefore it divides the line segment $XY$ in the ratio $\lambda^n : 1- \lambda^n$ Alternatively, \begin{align*} P_1 &= \lambda \mathbf{x} + (1-\lambda)\mathbf{y} \\ P_2 &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y} )*\mathbf{y} \\ &= \lambda^2 \mathbf{x} + (1-\lambda^2) \mathbf{y} \end{align*} Suppose $P_k = \lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}$ then \begin{align*}P_{k+1} &= (\lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}) * \mathbf{y} \\ &= \lambda^{k+1}\mathbf{x} + \lambda(1-\lambda^k)\mathbf{y} + (1-\lambda)\mathbf{y}\\ & = \lambda^{k+1}\mathbf{x} + (1-\lambda^{k+1})\mathbf{y}\end{align*}

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