Review question

# Which point $P$ is on both this line and this plane? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6006

## Solution

The line $l$ has the equation $\mathbf{r}= \begin{pmatrix}5\\ 0\\ 5\end{pmatrix}+ \lambda \begin{pmatrix}2\\ 1\\ 0\end{pmatrix}, \hspace{1cm} \lambda \in \mathbb{R}.$

1. 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.

1. 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 $\begin{pmatrix}1\\ -2\\ 5\end{pmatrix}$.

1. 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 $\mathbf{r}.\begin{pmatrix}1\\ -2\\ -1\end{pmatrix}=0.$

1. 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 $\overrightarrow{OP}=\begin{pmatrix}5\\ 0\\ 5\end{pmatrix} - \begin{pmatrix}2\\ 1\\ 0\end{pmatrix}=\begin{pmatrix}3\\ -1\\ 5\end{pmatrix}.$