Two non-parallel lines in 3-dimensional space are given by \(\mathbf{r}=\mathbf{p}_{1}+t_{1}\mathbf{m}_{1}\)
and \(\mathbf{r}=\mathbf{p}_{2}+t_{2}\mathbf{m}_{2}\) respectively,
where \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are unit vectors. Explain
by means of a sketch why the shortest distance between the two lines
is
\[
\frac{\left|(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}{\left|(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}.
\]
Find the shortest distance between the lines in the case
\[
\mathbf{p}_{1}=(2,1,-1)\qquad\mathbf{p}_{2}=(1,0,-2)\qquad\mathbf{m}_{1}=\tfrac{1}{5}(4,3,0)\qquad\mathbf{m}_{2}=\tfrac{1}{\sqrt{10}}(0,-3,1).
\]
Two aircraft, \(A_{1}\) and \(A_{2},\) are flying in the directions
given by the unit vectors \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\)
at constant speeds \(v_{1}\) and \(v_{2}.\) At time \(t=0\) they pass
the points \(\mathbf{p}_{1}\) and \(\mathbf{p}_{2}\), respectively.
If \(d\) is the shortest distance between the two aircraft during the
flight, show that
\[
d^{2}=\frac{\left|\mathbf{p}_{1}-\mathbf{p}_{2}\right|^{2}\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}-[(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2})]^{2}}{\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}}.
\]
Suppose that \(v_{1}\) is fixed. The pilot of \(A_{2}\) has chosen \(v_{2}\)
so that \(A_{2}\) comes as close as possible to \(A_{1}.\) How close
is that, if \(\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\)
are as in (i)?
Let \(\mathbf{a},\mathbf{b}\) and \(\mathbf{c}\) be the position vectors of points \(A,B\) and \(C\) in three-dimensional space. Suppose that \(A,B,C\) and the origin \(O\) are not all in the same plane. Describe
the locus of the point whose position vector \(\mathbf{r}\) is given by
\[
\mathbf{r}=(1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c},
\]
where \(\lambda\) and \(\mu\) are scalar parameters. By writing this equation in the form \(\mathbf{r}\cdot\mathbf{n}=p\) for a suitable
vector \(\mathbf{n}\) and scalar \(p\), show that
\[
-(\lambda+\mu)\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})+\lambda\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})+\mu\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})=0
\]
for all scalars \(\lambda,\mu.\)
Deduce that
\[
\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})=\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}).
\]
Say briefly what happens if \(A,B,C\) and \(O\) are all in the same plane.
\(\mathbf{r}=(1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a})+\mu(\mathbf{c}-\mathbf{a})\)
Therefore it is the plane through \(\mathbf{a}\) with direction vectors \(\mathbf{b}-\mathbf{a}\) and \(\mathbf{c}-\mathbf{a}\), ie it is the plane through \(\mathbf{a},\mathbf{b},\mathbf{c}\).
The normal to this plane will be \((\mathbf{b}-\mathbf{a} ) \times (\mathbf{c}-\mathbf{a}) = \mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a}\), so we must have:
\begin{align*}
&& \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) &= \mathbf{a} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\
&&&= \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c})
\end{align*}
Therefore,
\begin{align*}
&& \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c}) &= \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\
&&&= \left ( (1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} \right)\cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\
&&&= (1-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\
\Rightarrow && 0 &= (-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\
&&&= -(\lambda+ \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})+\lambda \mathbf{b}\cdot(\mathbf{c} \times \mathbf{a})+\mu \mathbf{c}\cdot(\mathbf{a} \times \mathbf{b}) \\
\end{align*}
The result follows from setting \(\mu = 0, \lambda = 1\) and \(\mu = 1, \lambda = 0\).
If they all lie in the same plane then the plane described is through the origin, and those values are all the same, but equal to \(0\).
The points \(A,B\) and \(C\) lie on the surface of the ground, which is an inclined plane. The point \(B\) is 100m due north of \(A,\) and \(C\) is 60m due east of \(B\). The vertical displacements from \(A\) to \(B,\) and from \(B\) to \(C\), are each 5m downwards. A plane coal seam lies below the surface and is to be located by making vertical bore-holes at \(A,B\) and \(C\). The bore-holes strike the coal seam at 95m, 45m and 76m below \(A,B\) and \(C\) respectively. Show that the coal seam is inclined at \(\cos^{-1}(\frac{4}{5})\) to the horizontal.
The coal seam comes to the surface along a line. Find the bearing of this line.
Set up a coordinate system so that \(x\) is E-W, \(y\) is N-S and \(z\) is the vertical direction. Also assume \(B\) is the origin, then, \(A = \begin{pmatrix} 0 \\ -100 \\ 5\end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}, C= \begin{pmatrix} 60 \\ 0\\ -5\end{pmatrix},\).
The coal seam has points: \(\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}, \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix},\)
Therefore we can find the normal to the coal seam:
\begin{align*}
\mathbf{n} &= \left (\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \times \left ( \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \\
&= \begin{pmatrix} 0 \\ - 100 \\ -45\end{pmatrix} \times \begin{pmatrix} 60 \\ 0 \\ -36\end{pmatrix} \\
&= \begin{pmatrix} 3600 \\ -60 \cdot 45 \\ 60 \cdot 100 \end{pmatrix} \\
&= 300\begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix}
\end{align*}
To measure the incline \(\theta\) to the horizontal we can take a dot with \(\hat{\mathbf{k}}\), to see:
\begin{align*}
\cos \theta &= \frac{20}{\sqrt{12^2+(-9)^2+20^2} \sqrt{1^2+0^2+0^2}} \\
&= \frac{20}{25} \\
&= \frac{4}{5}
\end{align*}
Therefore the angle is \(\cos^{-1} \tfrac 45\)
The equation of the seam is \(12x - 9y + 20z = -900\).
The equation of the surface is \(5x + 3y + 60z = 0\)
We can compute the direction of the overlap again with a cross product:
\begin{align*}
\mathbf{d} &= \begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \times \begin{pmatrix} 5 \\ 3 \\ 60\end{pmatrix} \\
&= \begin{pmatrix} -600 \\ -620 \\ 81 \end{pmatrix}
\end{align*}
To get the bearing of this vector we just need to look at the \(x\) and \(y\) components, so it will be \(\tan^{-1} \frac{600}{620} = \tan^{-1} \frac{30}{31}\)
The surface \(S\) in 3-dimensional space is described by the equation
\[
\mathbf{a}\cdot\mathbf{r}+ar=a^{2},
\]
where \(\mathbf{r}\) is the position vector with respect to the origin \(O\), \(\mathbf{a}(\neq\mathbf{0})\) is the position vector of a fixed point, \(r=\left|\mathbf{r}\right|\) and \(a=\left|\mathbf{a}\right|.\)
Show, with the aid of a diagram, that \(S\) is the locus of points which are equidistant from the origin \(O\) and the plane \(\mathbf{r}\cdot\mathbf{a}=a^{2}.\)
The point \(P\), with position vector \(\mathbf{p},\) lies in \(S\), and the line joining \(P\) to \(O\) meets \(S\) again at \(Q\). Find the position vector of \(Q\).
The line through \(O\) orthogonal to \(\mathbf{p}\) and \(\mathbf{a}\) meets \(S\) at \(T\) and \(T'\). Show that the position vectors of \(T\) and \(T'\) are
\[
\pm\frac{1}{\sqrt{2ap-a^{2}}}\mathbf{a}\times\mathbf{p},
\]
where \(p=\left|\mathbf{p}\right|.\)
Show that the area of the triangle \(PQT\) is
\[
\frac{ap^{2}}{2p-a}.
\]
The plane is the same as the plane \((\mathbf{r} - \mathbf{a}) \cdot \mathbf{a} = 0\), ie the plane through \(\mathbf{a}\) whose normal is parallel to \(\mathbf{a}\)
The distance from \(\mathbf{r}\) to the plane therefore is \(\lambda\) where \(\mathbf{r}+\lambda \frac{1}{a}\mathbf{a}\) must be on the plane, ie \((\mathbf{r}+\frac{\lambda}{a} \mathbf{a} - \mathbf{a})\cdot \mathbf{a} = 0 \Rightarrow \lambda = \frac{a^2-\mathbf{a} \cdot \mathbf{r}}{a}\)
But if \(\mathbf{a} \cdot \mathbf{r} = a^2 - ar\) then \(\lambda = r\), ie the distance to the plane is the same as the distance to the origin.
\(\mathbf{q} = k \mathbf{p}\) and so \(\mathbf{a} \cdot k \mathbf{p} + a |k|p = a^2\) if \(k > 0\) we will find \(k = 1\) the position vector we already know about, therefore suppose \(k < 0\) so:
\begin{align*}
&& \mathbf{a} \cdot k \mathbf{p} - ka p &= a^2 \\
\Rightarrow && k(a^2-ap)-kap &= a^2 \\
\Rightarrow && k(a^2-2ap) &= a^2 \\
\Rightarrow && k &= \frac{a^2}{a^2-2ap}
\end{align*}
Therefore \(\mathbf{q} = \frac{a^2}{a^2-2ap} \mathbf{p}\)
The line through \(O\) orthogonal to \(\mathbf{p}\) and \(\mathbf{a}\) will be parallel to \(\mathbf{a} \times \mathbf{p}\). Therefore we should consider points of the from \(s \mathbf{a} \times \mathbf{p}\) on the surface \(S\).
\begin{align*}
&& s\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{p}) + sa^2p |\sin \theta| &= a^2
\end{align*}
The angle between \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{ap} = \frac{a^2-ap}{ap} \Rightarrow |\sin \theta| = \sqrt{1-\frac{(a-p)^2}{p^2}} = \frac{1}{p} \sqrt{2ap-a^2}\)
Therefore \(sa^2 \sqrt{2ap-a^2} = a^2 \Rightarrow s = \frac{1}{\sqrt{2ap-a^2}}\) and so the points are as required.
Noting that \(|\mathbf{p} \times \mathbf{t}| = |\frac{1}{p \sin \theta}\mathbf{p} \times (\mathbf{p} \times \mathbf{a}) | = |\frac{1}{p \sin \theta}p^2a \sin \theta | = pa\)
The area of triangle \(PQT\) is :
\begin{align*}
\frac12 | (\mathbf{p} - \mathbf{t}) \times (\mathbf{q} - \mathbf{t}) | &= \frac12 |\mathbf{p} \times \mathbf{q} - \mathbf{t} \times \mathbf{q} - \mathbf{p} \times \mathbf{t} - \mathbf{t} \times \mathbf{t}| \\
&= \frac12 |\mathbf{t} \times (\mathbf{p} - \mathbf{q})| \\
&= \frac12 \cdot (1 - \frac{a^2}{a^2-2ap})| \mathbf{t} \times \mathbf{p}| \\
&= \frac12 \frac{2ap}{a^2-2ap} \cdot ap \\
&= \frac{ap^2}{a^2-ap}
\end{align*}