\(A, B, C\) are points of three-dimensional space with coordinates \((1, 0, 3), (-1, -4, 7)\) and \((3, 3, -3)\) respectively.
Find
- 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 \(\angle BAC = 162.2^\circ\).
- 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?
