Problems

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 \(OABC\).

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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Solution:

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The line \(AN\) is \(\mathbf{a} + \alpha (\mathbf{n}-\mathbf{a})\) \\ The line \(BM\) is \(\mathbf{b} + \beta (\mathbf{m} - \mathbf{b})\) The point \(OQ = OB + BQ = \mathbf{b} + \frac{1}{\mu+1} (\mathbf{m}-\mathbf{b})\) It is also \(OQ = OA + AQ = \mathbf{a} + \frac{1}{\nu+1} ( \mathbf{n} - \mathbf{a})\) \begin{align*} && \mathbf{q} &= \mathbf{a} + \frac{1}{\nu+1} ( n\mathbf{b} - \mathbf{a}) \\ && \mathbf{q} &= \mathbf{b} + \frac{1}{\mu+1} ( m\mathbf{a} - \mathbf{b}) \\ \Rightarrow && \frac{\nu}{\nu+1} &= \frac{m}{\mu+1} \\ \Rightarrow && m &= \frac{(1+\mu)\nu}{1+\nu} \\ \Rightarrow && \mathbf{m} &= \frac{(1+\mu)\nu}{1+\nu} \mathbf{a} \end{align*} By similar triangles (\(\triangle OAN \sim \triangle OML\), we can observe that \(\lambda = \mu \nu\). The significance of \(\mu \nu < 1\) \(L\) lies on the side \(OB\) and both \(M\) and \(N\) lie between \(O\) and \(A\) and \(B\) respectively.

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 \(T\).

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Solution:

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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 \(\frac{1}{OX} + \frac{1}{OA} = \frac{1}{m|\mathbf{a}|} + \frac{1}{|\mathbf{a}|} = \frac{m+1}{m|\mathbf{a}|} = \frac{1}{OT}\)

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.

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1. 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.
2. 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.
3. Find a condition on \(r_1\), \(r_2\) and \(r_3\) for \(Q\) to lie half-way between \(P\) and \(R\).

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.

1. 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.
2. 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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Solution:

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Notice that \(\mathbf{d} = \frac{1}{r+1} \mathbf{b}\) and \(\mathbf{e} = \frac{s}{s+1}\mathbf{a}\). We must also have that the line \(AD\) is \(\mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right)\) and \(BE\) is \(\mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right)\) at their point of intersection we must have \begin{align*} && \mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right) &= \mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\ [\mathbf{a}]: && 1 + \lambda &= -\frac{\mu s}{s+1} \\ [\mathbf{b}]: && -\frac{\lambda}{r+1} &= 1 + \mu \\ \Rightarrow && \lambda &= -\frac{1+s+\mu s}{s+1} \\ \Rightarrow && \mu &= \frac{1+s+\mu s}{(1+r)(1+s)} - 1 \\ \Rightarrow && (1+r+rs)\mu &= 1+s - 1 - r - s - rs \\ \Rightarrow && \mu &= -\frac{r+rs}{1+r+rs} \\ \Rightarrow && \mathbf{g} &= \mathbf{b} -\frac{r+rs}{1+r+rs}\left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\ &&&= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b} \end{align*} \item The line \(OG\) is \(\lambda \mathbf{g}\). The line \(AB\) is \(\mathbf{a} + \mu(\mathbf{b}-\mathbf{a})\), so we need \begin{align*} && \lambda \mathbf{g} &= \mathbf{a} + \mu(\mathbf{b}-\mathbf{a}) \\ [\mathbf{a}]: && \lambda \frac{rs}{1+r+rs} &= 1-\mu \\ [\mathbf{b}]: && \lambda \frac{1}{1+r+rs} &= \mu \\ \Rightarrow && \lambda \frac{1+rs}{1+r+rs} &= 1 \\ \Rightarrow && \lambda &= \frac{1+r+rs}{1+rs} \\ \Rightarrow && \mu &= \frac{1}{1+rs} \end{align*} Therefore the line is divided in the ratio \(rs : 1\), and therefore we have proven Ceva's Theorem.

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 \(Q\).

1. By considering the triangles \(P_1QP_4\) and \(P_2QP_3\) show that \((P_1Q)( QP_3) = (P_2Q) (QP_4)\,\).
2. 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}
3. 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 \(a_1a_3 (P_1P_3)^2 = a_2a_4 (P_2P_4)^2\,\).

Three distinct points, \(X_1\), \(X_2\) and \(X_3\), with position vectors \({\bf x}_1\), \({\bf x}_2\) and \({\bf x}_3\) respectively, lie on a circle of radius 1 with its centre at the origin \(O\). The point \(G\) has position vector \(\frac13({\bf x}_1+{\bf x}_2+{\bf x}_3)\). The line through \(X_1\) and \(G\) meets the circle again at the point \(Y_1\) and the points \(Y_2\) and \(Y_3\) are defined correspondingly. Given that \(\overrightarrow{GY_1} =-\lambda_1\overrightarrow{GX_1}\), where \(\lambda_1\) is a positive scalar, show that \[ \overrightarrow{OY_1}= \tfrac13 \big( (1-2\lambda_1){\bf x}_1 +(1+\lambda_1)({\bf x}_2+{\bf x}_3)\big) \] and hence that \[ \lambda_1 = \frac {3-\alpha-\beta-\gamma} {3+\alpha -2\beta-2\gamma} \,,\] where \(\alpha = {\bf x}_2 \,.\, {\bf x}_3\), \(\beta = {\bf x}_3\,.\, {\bf x}_1\) and \(\gamma = {\bf x}_1\,.\, {\bf x}_2\). Deduce that $\dfrac {GX_1}{GY_1} + \dfrac {GX_2}{GY_2} + \dfrac {GX_3}{GY_3} =3 \,$.

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Solution:

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\begin{align*} && \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\ &&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 - \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ && 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\ \Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\ &&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\ && 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\ \Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\ &&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\ \Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)} \end{align*} as required. Since \(\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}\) we must have, \begin{align*} && \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} + \frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\ &&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\ &&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\ &&&= 3 \end{align*}

Relative to a fixed origin \(O\), the points \(A\) and \(B\) have position vectors \(\bf{a}\) and \(\bf{b}\), respectively. (The points \(O\), \(A\) and \(B\) are not collinear.) The point \(C\) has position vector \(\bf c\) given by \[ {\bf c} =\alpha {\bf a}+ \beta {\bf b}\,, \] where \(\alpha\) and \(\beta\) are positive constants with \(\alpha+\beta<1\,\). The lines \(OA\) and \(BC\) meet at the point \(P\) with position vector \(\bf p\) and the lines \(OB\) and \(AC\) meet at the point \(Q\) with position vector \(\bf q\). Show that \[ {\bf p} =\frac{\alpha {\bf a} }{1-\beta}\,, \] and write down \(\bf q\) in terms of \(\alpha,\ \beta\) and \(\bf {b}\). Show further that the point \(R\) with position vector \(\bf r\) given by \[ {\bf r} =\frac{\alpha {\bf a} + \beta {\bf b}}{\alpha + \beta}\,, \] lies on the lines \(OC\) and \(AB\). The lines \(OB\) and \(PR\) intersect at the point \(S\). Prove that $ \dfrac{OQ}{BQ} = \dfrac{OS}{BS}\,$.

The non-collinear points \(A\), \(B\) and \(C\) have position vectors \(\bf a\), \(\bf b\) and \(\bf c\), respectively. The points \(P\) and \(Q\) have position vectors \(\bf p\) and \(\bf q\), respectively, given by \[ {\bf p}= \lambda {\bf a} +(1-\lambda){\bf b} \text{ \ \ \ and \ \ \ } {\bf q}= \mu {\bf a} +(1-\mu){\bf c} \] where \(0<\lambda<1\) and \(\mu>1\). Draw a diagram showing \(A\), \(B\), \(C\), \(P\) and \(Q\). Given that \(CQ\times BP = AB\times AC\), find \(\mu\) in terms of \(\lambda\), and show that, for all values of \(\lambda\), the the line \(PQ\) passes through the fixed point \(D\), with position vector \({\bf d}\) given by \({\bf d= -a +b +c}\,\). What can be said about the quadrilateral \(ABDC\)?

The points \(A\) and \(B\) have position vectors \(\bf a\) and \(\bf b\), respectively, relative to the origin \(O\). The points \(A\), \(B\) and \(O\) are not collinear. The point \(P\) lies on \(AB\) between \(A\) and \(B\) such that \[ AP : PB = (1-\lambda):\lambda\,. \] Write down the position vector of \(P\) in terms of \(\bf a\), \(\bf b\) and \(\lambda\). Given that \(OP\) bisects \(\angle AOB\), determine \(\lambda\) in terms of \(a\) and \(b\), where \(a=\vert \bf a\vert\) and $b=\vert \mathbf{b}\vert\(. The point \)Q\( also lies on \)AB\( between \)A\( and \)B\(, and is such that \)AP=BQ$. Prove that $$OQ^2-OP^2=(b-a)^2\,.$$

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Solution:

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\(\mathbf{p} = \lambda \mathbf{a} +(1- \lambda) \mathbf{b}\) Applying the Sine rule, we can see that: \begin{align*} && \frac{OA}{\sin \angle APO} = \frac{AP}{\sin \angle AOP} \\ && \frac{OB}{\sin \angle BPO} = \frac{BP}{\sin \angle BOP} \\ \end{align*} But \(\angle AOP = \angle BOP = \frac12 \angle AOP\) (since \(OP\) bisects the angle) and \(\angle APO = 180^{\circ} -\angle BPO\), so their sines are also equal. Therefore \begin{align*} && \frac{a}{AP} &= \frac{b}{BP} \\ \Rightarrow && \frac{a}{b} &= \frac{1-\lambda}{\lambda} \\ \Rightarrow && \lambda &= \frac{b}{a+b} \end{align*} Note that \(\mathbf{p} = \frac{b\mathbf{a} + a \mathbf{b}}{a+b} = \mathbf{a} + \frac{a}{a+b}(\mathbf{b}-\mathbf{a})\) and \(\mathbf{q} = \mathbf{b} +\frac{a}{a+b}(\mathbf{a}-\mathbf{b}) = \frac{a\mathbf{a} +b \mathbf{b}}{a+b}\) Therefore \begin{align*} && OQ^2 &= \frac{1}{(a+b)^2} \left (a\mathbf{a} + b \mathbf{b} \right)\cdot \left (a\mathbf{a} + b \mathbf{b} \right) \\ &&&= \frac{a^4+b^4+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ &&OP^2 &= \frac{1}{(a+b)^2} \left (b\mathbf{a} + a \mathbf{b} \right)\cdot \left (b\mathbf{a} + a \mathbf{b} \right) \\ &&&= \frac{2a^2b^2+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ \\ && OQ^2 - OP^2 &= \frac{a^4+b^4-2a^2b^2}{(a+b)^2} \\ &&&= \frac{(a^2-b^2)^2}{(a+b)^2} \\ &&&= (a-b)^2 \end{align*}

The points \(B\) and \(C\) have position vectors \(\mathbf{b}\) and \(\mathbf{c}\), respectively, relative to the origin \(A\), and \(A\), \(B\) and \(C\) are not collinear.

1. The point \(X\) has position vector \(s \mathbf{b}+t\mathbf{c}\). Describe the locus of \(X\) when \(s+t=1\).
2. The point \(P\) has position vector \(\beta \mathbf{b}+\gamma\mathbf{c}\), where \(\beta\) and \(\gamma\) are non-zero, and \(\beta+\gamma\ne1\). The line \(AP\) cuts the line \(BC\) at \(D\). Show that \(BD:DC=\gamma:\beta\).
3. The line \(BP\) cuts the line \(CA\) at \(E\), and the line \(CP\) cuts the line \(AB\) at \(F\). Show that \[ \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}=1\,. \]

Show that the line through the points with position vectors \(\bf x\) and \(\bf y\) has equation \[{\bf r} = (1-\alpha){\bf x} +\alpha {\bf y}\,, \] where \(\alpha\) is a scalar parameter. The sides \(OA\) and \(CB\) of a trapezium \(OABC\) are parallel, and \(OA>CB\). The point \(E\) on \(OA\) is such that \(OE : EA = 1:2\), and \(F\) is the midpoint of \(CB\). The point \(D\) is the intersection of \(OC\) produced and \(AB\) produced; the point \(G\) is the intersection of \(OB\) and \(EF\); and the point \(H\) is the intersection of \(DG\) produced and \(OA\). Let \(\bf a\) and \(\bf c\) be the position vectors of the points \(A\) and \(C\), respectively, with respect to the origin \(O\).

1. Show that \(B\) has position vector \(\lambda {\bf a} + {\bf c}\) for some scalar parameter \(\lambda\).
2. Find, in terms of \(\bf a\), \(\bf c\) and \(\lambda\) only, the position vectors of \(D\), \(E\), \(F\), \(G\) and \(H\). Determine the ratio \(OH:HA\).

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 \(\tfrac32 \pi\,\).

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 \(OAB\).