Review question

# Can we express a point in this plane in terms of two parameters? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7187

## Solution

$A, B, C$ are points of three-dimensional space with coordinates $(1, 0, 3), (-1, -4, 7)$ and $(3, 3, -3)$ respectively.

Find

1. the magnitude of the angle $BAC$;

We are told that $\overrightarrow{OA}= \begin{pmatrix}1 \\ 0 \\ 3\end{pmatrix}, \overrightarrow{OB}= \begin{pmatrix}-1 \\ -4 \\ 7\end{pmatrix}, \overrightarrow{OC}= \begin{pmatrix}3 \\ 3 \\ -3\end{pmatrix}$ and so $\overrightarrow{AB} = \begin{pmatrix}-2 \\ -4 \\ 4\end{pmatrix}, \overrightarrow{AC} = \begin{pmatrix}2 \\ 3 \\ -6\end{pmatrix}.$

Now $\overrightarrow{AB}.\overrightarrow{AC} = \big \vert \overrightarrow{AB} \big \vert \big \vert \overrightarrow{AC} \big \vert \cos \angle BAC.$

We have $\overrightarrow{AB}.\overrightarrow{AC} = -4 -12 -24 = -40.$

Additionally $\big \vert \overrightarrow{AB} \big \vert = \sqrt{4+16+16} = 6,$ while $\big \vert \overrightarrow{AC} \big \vert = \sqrt{4+9+36} = 7.$

Thus $\cos \angle BAC = \dfrac{-40}{42}$, and $\angle BAC = 162.2^\circ$.

1. expressions for the coordinates of any point of the plane $ABC$ in terms of two parameters $s$ and $t$.

We have two non-parallel vectors lying in the plane, so any vector in the plane can be expressed as a combination of these two. The equation of the plane is $\mathbf{r} = \overrightarrow{OA}+s\overrightarrow{AB}+t\overrightarrow{AC},$ or $\mathbf{r} = \begin{pmatrix}1 \\ 0 \\ 3\end{pmatrix}+s\begin{pmatrix}-2 \\ -4 \\ 4\end{pmatrix}+t\begin{pmatrix}2 \\ 3 \\ -6\end{pmatrix}.$

We can simplify this to $\mathbf{r} = \begin{pmatrix}1 \\ 0 \\ 3\end{pmatrix}+s\begin{pmatrix}1 \\ 2 \\ -2\end{pmatrix}+t\begin{pmatrix}2 \\ 3 \\ -6\end{pmatrix},$ or $\mathbf{r} = \begin{pmatrix}1 +s+2t\\ 2s+3t \\ 3-2s-6t\end{pmatrix}.$

Is the vector form of the equation of the plane unique?