Problems

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 \(90^\circ\).

1. 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.
2. 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.

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

1. Prove that a tetrahedron is isosceles if and only if all four faces have the same perimeter.

2. 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 \[\mathbf{a}.(\mathbf{b}+\mathbf{c}) = |\mathbf{a}|^2.\]
3. 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.
4. 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?

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

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

The plane \(\Pi\) has equation \(\mathbf{r} \cdot \mathbf{n} = 0\) where \(\mathbf{n}\) is a unit vector. Let \(P\) be a point with position vector \(\mathbf{x}\) which does not lie on the plane \(\Pi\). Show that the point \(Q\) with position vector \(\mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\mathbf{n}\) lies on \(\Pi\) and that \(PQ\) is perpendicular to \(\Pi\).

1. Let transformation \(T\) be a reflection in the plane \(ax+by+cz=0\), where \(a^2+b^2+c^2=1\). Show that the image of \(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) under \(T\) is \(\begin{pmatrix} b^2+c^2-a^2 \\ -2ab \\ -2ac \end{pmatrix}\), and find the images of \(\mathbf{j}\) and \(\mathbf{k}\) under \(T\). Write down the matrix \(\mathbf{M}\) which represents transformation \(T\).
2. The matrix \[ \begin{pmatrix} 0.64 & 0.48 & 0.6 \\ 0.48 & 0.36 & -0.8 \\ 0.6 & -0.8 & 0 \end{pmatrix} \] represents a reflection in a plane. Find the cartesian equation of the plane.
3. The matrix \(\mathbf{N}\) represents a rotation through angle \(\pi\) about the line through the origin parallel to \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), where \(a^2+b^2+c^2=1\). Find the matrix \(\mathbf{N}\).
4. Identify the single transformation which is represented by the matrix \(\mathbf{NM}\).

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*}

The points \(A\) and \(B\) have position vectors \(\bf a \) and \(\bf b\) with respect to an origin \(O\), and \(O\), \(A\)~and~\(B\) are non-collinear. The point \(C\), with position vector \(\bf c\), is the reflection of \(B\) in the line through \(O\) and \(A\). Show that \(\bf c\) can be written in the form \[ \bf c = \lambda \bf a -\bf b \] where \(\displaystyle \lambda = \frac{2\,{\bf a .b}}{{\bf a.a}}\). The point \(D\), with position vector \(\bf d\), is the reflection of \(C\) in the line through \(O\) and \(B\). Show that \(\bf d\) can be written in the form \[ \bf d = \mu\bf b - \lambda \bf a \] for some scalar \(\mu\) to be determined. Given that \(A\), \(B\) and \(D\) are collinear, find the relationship between \(\lambda\) and \(\mu\). In the case \(\lambda = -\frac12\), determine the cosine of \(\angle AOB\) and describe the relative positions of \(A\), \(B\) and \(D\).

The points \(P\), \(Q\) and \(R\) lie on a sphere of unit radius centred at the origin, \(O\), which is fixed. Initially, \(P\) is at \(P_0(1, 0, 0)\), \(Q\) is at \(Q_0(0, 1, 0)\) and \(R\) is at \(R_0(0, 0, 1)\).

1. The sphere is then rotated about the \(z\)-axis, so that the line \(OP\) turns directly towards the positive \(y\)-axis through an angle \(\phi\). The position of \(P\) after this rotation is denoted by \(P_1\). Write down the coordinates of \(P_1\).
2. The sphere is now rotated about the line in the \(x\)-\(y\) plane perpendicular to \(OP_1\), so that the line \(OP\) turns directly towards the positive \(z\)-axis through an angle \(\lambda\). The position of \(P\) after this rotation is denoted by \(P_2\). Find the coordinates of \(P_2\). Find also the coordinates of the points \(Q_2\) and \(R_2\), which are the positions of \(Q\) and \(R\) after the two rotations.
3. The sphere is now rotated for a third time, so that \(P\) returns from \(P_2\) to its original position~\(P_0\). During the rotation, \(P\) remains in the plane containing \(P_0\), \(P_2\) and \(O\). Show that the angle of this rotation, \(\theta\), satisfies \[ \cos\theta = \cos\phi \cos\lambda\,, \] and find a vector in the direction of the axis about which this rotation takes place.

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 vectors \({\bf a}\) and \({\bf b}\) lie in the plane \(\Pi\,\). Given that \(\vert {\bf a} \vert= 1\) and \({\bf a}.{\bf b} = 3,\) find, in terms of \({\bf a}\) and \({\bf b}\,\), a vector \({\bf p}\) parallel to \({\bf a}\) and a vector \({\bf q}\) perpendicular to \({\bf a}\,\), both lying in the plane \(\Pi\,\), such that $${\bf p}+{\bf q}={\bf a}+{\bf b}\;.$$ The vector \({\bf c}\) is not parallel to the plane \(\Pi\) and is such that \({\bf a}.{\bf c} = -2\) and \({\bf b}.{\bf c} = 2\,\). Given that \(\vert {\bf b} \vert = 5\,\), find, in terms of \({\bf a}, {\bf b}\) and \({\bf c},\) vectors \({\bf P}\), \({\bf Q}\) and \({\bf R}\) such that \({\bf P}\) and \({\bf Q}\) are parallel to \({\bf p}\) and \({\bf q},\) respectively, \({\bf R}\) is perpendicular to the plane \(\Pi\) and $${\bf P} + {\bf Q} + {\bf R} = {\bf a}+{\bf b}+{\bf c}\;.$$

Let \(ABCD\) be a parallelogram. By using vectors, or otherwise, prove that:

1. \(AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}\);
2. \(|AC^{2}-BD^{2}|\) is 4 times the area of the rectangle whose sides are any side of the parallelogram and the projection of an adjacent side on that side.