The line \(l\) has the equation \[\mathbf{r}= \begin{pmatrix}5\\ 0\\ 5\end{pmatrix}+ \lambda \begin{pmatrix}2\\ 1\\ 0\end{pmatrix}, \hspace{1cm} \lambda \in \mathbb{R}.\]
- Show that \(l\) lies in the plane whose equation is \[\mathbf{r}.\begin{pmatrix}-1\\ 2\\ 0\end{pmatrix}=-5.\]
We need to show that every vector \(\mathbf{r}\) that satisfies the line equation also satisfies the plane equation.
Substituting in, we find
\[\left[\begin{pmatrix}5\\ 0\\ 5\end{pmatrix}+ \lambda \begin{pmatrix}2\\ 1\\ 0\end{pmatrix} \right].\begin{pmatrix}-1\\ 2\\ 0\end{pmatrix}=\begin{pmatrix}5+2\lambda\\ \lambda\\ 5\end{pmatrix} .\begin{pmatrix}-1\\ 2\\ 0\end{pmatrix}= -5 -2\lambda+2\lambda=-5.\]
This is true for every \(\lambda\), so we’ve shown that the line does indeed lie in the given plane.
- Find the position vector of \(L\), the foot of the perpendicular from the origin \(O\) to \(l\).
Let \(\overrightarrow{OL} = \begin{pmatrix}5\\ 0\\ 5\end{pmatrix}+ \lambda \begin{pmatrix}2\\ 1\\ 0\end{pmatrix}.\) We need
\[\left[\begin{pmatrix}5\\ 0\\ 5\end{pmatrix}+ \lambda \begin{pmatrix}2\\ 1\\ 0\end{pmatrix} \right].\begin{pmatrix}2\\ 1\\ 0\end{pmatrix}=0\]
which is true if and only if \(10+4\lambda+\lambda =0,\) or \(\lambda = -2.\) Thus \(L\) is the point \(\begin{pmatrix}1\\ -2\\ 5\end{pmatrix}\).
- Find an equation of the plane containing \(O\) and \(l\).
We know that \(\overrightarrow{OL}\) is in the plane, and putting \(\lambda = 0\), we know also that \(\overrightarrow{OM}\) is in the plane, where \(\overrightarrow{OM}\) = \(\begin{pmatrix}5\\ 0\\ 5\end{pmatrix}\).
Any point in the plane can be expressed as a linear combination of these vectors. So an equation of the plane is \[\mathbf{r} = \alpha \begin{pmatrix} 5 \\ 0 \\ 5 \end{pmatrix}+ \beta \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix}.\]
Alternatively, we could write the equation as \(\mathbf{r}.\mathbf{n}=0\) where \(\mathbf{n}\) is the normal to the plane. Substituting in the two points \(L\) and \(M\) tells us that \(\mathbf{n}\) is a multiple of \(\begin{pmatrix}1\\ -2\\ -1\end{pmatrix}\), so the plane equation can be written as \[\mathbf{r}.\begin{pmatrix}1\\ -2\\ -1\end{pmatrix}=0.\]
- Find the position vector of the point \(P\) where \(l\) meets the plane \(\pi\) whose equation is \[\mathbf{r}.\begin{pmatrix}1\\ 2\\ 2\end{pmatrix} = 11.\]
Let \(\overrightarrow{OP} = \begin{pmatrix}5\\ 0\\ 5\end{pmatrix} + \lambda \begin{pmatrix}2\\ 1\\ 0\end{pmatrix}.\) If \(P\) is on the plane, then \(\overrightarrow{OP}.\begin{pmatrix}1\\ 2\\ 2\end{pmatrix} = 11,\) and so
\[\begin{align*} \left[ \begin{pmatrix}5\\ 0\\ 5\end{pmatrix}+ \lambda \begin{pmatrix}2\\ 1\\ 0\end{pmatrix} \right].\begin{pmatrix}1\\ 2\\ 2\end{pmatrix} = 11. \end{align*}\]This implies
\[\begin{align*} 15 + 4\lambda = 11, \end{align*}\]from which we conclude \(\lambda=-1\). Thus \[\overrightarrow{OP}=\begin{pmatrix}5\\ 0\\ 5\end{pmatrix} - \begin{pmatrix}2\\ 1\\ 0\end{pmatrix}=\begin{pmatrix}3\\ -1\\ 5\end{pmatrix}.\]
