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 \(OABC\).
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 \(\mathbf{a}_1 \cdot \mathbf{a}_2 = \mathbf{a}_3 \cdot \mathbf{a}_4\).
Given that the four points lie in a plane, determine the shape of the quadrilateral with vertices \(A_1\), \(A_2\), \(A_3\) and \(A_4\).
Given instead that the four points are the vertices of a regular tetrahedron, find the length of the sides of this tetrahedron.
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*}
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
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 \(\frac{2}{\sqrt{3}}\)
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.
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 \(\angle BOA' = 90^\circ\)
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.
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.
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.
Find expressions in vector form for the forces acting on \(P\).
Show that \(U=\sqrt6 V\) and find \(T\), \(U\) and \(V\) in terms of \(W\).
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*}
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 \(\frac{2}{9}\)
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:
\[
(ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,.
\]
Show that \((x+2y+2z)^2 \le 9(x^2+y^2+z^2)\) for all real numbers \(x\), \(y\) and \(z\).
Find real numbers \(p\), \(q\) and \(r\) that satisfy both
\[
p^2+4q^2+9r^2 = 729
\text{ and }
8p+8q+3r = 243\,.
\]
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 \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)
By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
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
\[ (p,q,r) = \left (24, 6, 1 \right) \]
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 \(OAB\).
\(E = \langle \frac{a}{2}, 0,0 \rangle, F = \langle \frac{l}{2}, \frac{m+b}{2}, \frac{n}{2} \rangle, G = \langle 0, \frac{b}{2}, 0 \rangle, H = \langle \frac{a+l}{2}, \frac{m}{2}, \frac{n}{2} \rangle\)
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 \(d = \frac{n}{2}\)
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:
there are only two values of \(\lambda\) for which \(AQ\) is perpendicular
to \(BP\,\);
there are no non-zero values of \(\lambda\) for which \(AQ\) and \(BP\)
intersect.
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?
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 \(\alpha \le \arccos \frac{1}{9}\).
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.]
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
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 \({\bf u} = {\bf v}_1 + {\bf v}_2\,\).
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 \(R\).
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.
Find the volume of a regular octahedron whose edges are of unit length.
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.}]
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 \(PR\).)
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 \(c\).
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
\[\frac{b}{V}\left(\frac{c}{a}+\frac{a}{c}\right)
.
\]
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 \(OQ.\)
Show that \(\overrightarrow{AQ}\) is perpendicular to \(\overrightarrow{BC}\)
and deduce that the three altitudes of \(\triangle ABC\) are concurrent.
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.{]}
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
\[
\frac{k^{2}-2}{2k(k^{2}-1)}\sqrt{4k^{2}-2}.
\]
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.
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
\[
t+\frac{\sqrt{2}wd}{v^{2}-w^{2}}\ \text{hours.}
\]
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.
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.]
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 \((3,-2,4)\)
\(|\overrightarrow{AY}| = \sqrt{3^2+(-6)^2+6^2} = 3\sqrt{1+4+4} = 9\)
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.
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:
there are constants \(\alpha\) and \(\beta\), with \(\alpha>0\), such
that for any \(i\) and \(j\),
\[
\mathbf{a}_{i}\cdot\mathbf{a}_{j}=\begin{cases}
\alpha^{2} & \mbox{ when }i=j\\
\beta & \mbox{ when }i\neq j,
\end{cases}
\]
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 \(\beta = -\frac14\), \(\cos \phi = \frac{-1/4}{3/4} = -\frac13\). \(\sin \theta = \frac{\sqrt{15}}{4}\), \(\sin \phi = \frac{2\sqrt{2}}{3}\)
\(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{12} \\ \frac{\sqrt{30}}{6} \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{12} \\ -\frac{\sqrt{30}}{6} \end{pmatrix}, \right \}\)
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.