The lines \(l_1\) and \(l_2\) have equations \(\mathbf{r} = \mathbf{a} + s\mathbf{b}\) and \(\mathbf{r} = \mathbf{c} + t\mathbf{d}\), where \(s\) and \(t\) are real numbers and
\[\mathbf{a} = 5\mathbf{i}+ 6\mathbf{j}- \mathbf{k}, \quad\mathbf{b} = \mathbf{i} +\mathbf{j}+ \mathbf{k}, \quad\mathbf{c} = \mathbf{i}+ 13\mathbf{j} +7\mathbf{k}, \quad\mathbf{d} = \mathbf{i} + 6\mathbf{j} + 3\mathbf{k}.\]
Also, the point \(P\) on \(l_1\) and the point \(Q\) on \(l_2\) are such that \(PQ\) is perpendicular to both \(l_1\) and \(l_2.\) In any order:
show that \(PQ = \sqrt{38},\)
find the position vectors of \(P\) and \(Q,\)
We can write the position vector of \(P\) as \(\mathbf{p} = \begin{pmatrix}5+s \\ 6+s \\ -1+s\end{pmatrix}\), and the position vector of \(Q\) as \(\mathbf{q} = \begin{pmatrix}1+t \\ 13+6t\\ 7+3t\end{pmatrix}.\)
Thus \(\mathbf{q} - \mathbf{p} = \begin{pmatrix}t-s-4 \\ 6t-s+7 \\ 3t-s+8\end{pmatrix}\). This vector has to be perpendicular to both \(l_1\) and \(l_2\), and so using the scalar product, \[\begin{pmatrix}t-s-4 \\ 6t-s+7 \\ 3t-s+8\end{pmatrix}.\begin{pmatrix}1 \\ 1\\1\end{pmatrix}=0 \quad\text{and}\quad \begin{pmatrix}t-s-4 \\ 6t-s+7 \\ 3t-s+8\end{pmatrix}.\begin{pmatrix}1\\ 6 \\ 3\end{pmatrix}=0.\]
This gives us the two equations \(10t-3s+11=0\) and \(46t-10s+62=0\), which solve to give \(s = -3\), \(t = -2\).
Thus \(\mathbf{p} = \begin{pmatrix}2 \\ 3 \\ -4\end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix}-1 \\ 1\\ 1\end{pmatrix}\). We also have \(\mathbf{q}-\mathbf{p} = \begin{pmatrix}-3 \\ -2\\ 5\end{pmatrix}\), and so \(\big\vert\mathbf{q}-\mathbf{p} \big\vert = \sqrt{(-3)^2+(-2)^2+5^2} = \sqrt{38}\).
- obtain the Cartesian equation of the plane containing \(l_1\) and \(PQ\).
Using vector product we can find a vector perpendicular to two others, thus \[(\mathbf{q}- \mathbf{p}) \times \mathbf{b}=\mathbf{n}\] where \(\mathbf{n}\) is a vector perpendicular to both \(\mathbf{b}\) (the direction vector of \(l_1\)) and \(PQ\).
\[\begin{pmatrix}-3\\-2\\5\end{pmatrix} \times \begin{pmatrix}1 \\ 1\\1\end{pmatrix}=\begin{pmatrix}-2-5\\ -(-3-5)\\-3--2\end{pmatrix}=\begin{pmatrix}-7 \\ 8\\-1\end{pmatrix}\]
A normal vector immediately yields the vector equation of the plane \(\mathbf{r}.\mathbf{n}=d\) where \(d\) can be calculated from any known point on the plane.
So \(\mathbf{q}.\mathbf{n}= 7+8-1=d=14\).
Hence the vector equation of the plane is \[\begin{pmatrix}x\\y\\z\end{pmatrix} . \begin{pmatrix} -7 \\ 8\\-1\end{pmatrix}=14\] which gives rise to the Cartesian equation \(-7x+8y-z=14\) or \[7x-8y+z+14=0\]
Check; is \(\mathbf{a}\) on this plane? Substituting in the coordinates we have \(7(5)-8(6)+(-1)+14=0\), so yes.
As an alternative method, an equation of the plane containing both \(PQ\) and \(l_1\) is \[\mathbf{r} = \begin{pmatrix}x \\ y\\ z\end{pmatrix}= \begin{pmatrix}2\\ 3 \\ -4\end{pmatrix}+ \lambda\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix} + \mu\begin{pmatrix}3 \\ 2\\ -5\end{pmatrix}.\]
Thus \(x = 2+\lambda + 3\mu\), \(y = 3+\lambda + 2\mu\), \(z = -4+\lambda -5\mu\).
Subtracting the first two, \(x-y=-1+\mu\), and so \(\mu = x-y+1\).
Subtracting the second two, \(y-z= 7+7\mu\), and so \(y-z-7=7(x-y+1)\).
This becomes \(7x-8y+z+14=0\), which is the equation of the plane we seek.
