$l_1: \mathbf{r} =\begin{pmatrix} 3 \\ -1 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \lambda \in \mathbb{R} \\ l_2: \mathbf{r} =\begin{pmatrix} 1 \\ -1 \\ -4 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}, \mu \in \mathbb{R}$.

1. Find, in vector form, the equation of the plane containing points $A,C$ and $D$.

$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA} = 2(\mathbf{i}+2\mathbf{j}-3\mathbf{k})$.

Similarly, $\overrightarrow{AD} = 2 \mathbf{i}-\mathbf{j}-\mathbf{k}$.

Hence $(\mathbf{i}+2\mathbf{j}-3\mathbf{k})$ and $(2 \mathbf{i}-\mathbf{j}-\mathbf{k})$ are two direction vectors of the plane which contain point $A(3,-1,6)$.

Equation of plane: $\mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}, \lambda, \mu \in \mathbb{R}$.

Since the plane contains $l_1$, it contains $A(3,-1,6)$ and the direction vector $\mathbf{d}_1 = (\mathbf{i}+2\mathbf{j}-3\mathbf{k})$.

Since it also contains $D$, $\mathbf{d}_2 = \overrightarrow{AD} = 2 \mathbf{i}-\mathbf{j}-\mathbf{k}$ is another direction vector.

$\mathbf{d}_1 \times \mathbf{d}_2 = -5 ( \mathbf{i} + \mathbf{j} + \mathbf{k})$.

Hence $\mathbf{n} = \mathbf{i} + \mathbf{j} + \mathbf{k}$ is a normal vector of the plane.

$\mathbf{a} \cdot \mathbf{n} = 3 -1 = 6 = 8$.

Hence equation of the plane $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 8$

Since the plane contains $l_1$, it contains the point $A$ and a direction vector $\mathbf{i}+2\mathbf{j}-3\mathbf{k}$.

Since it is parallel to $l_2$, the direction vector of $l_2$, $2 \mathbf{i}-\mathbf{j}-\mathbf{k}$ is also a direction vector of the plane.

By taking cross products similar to question 2, we can find that $\mathbf{n} = \mathbf{i} + \mathbf{j} + \mathbf{k}$ is a normal vector of the plane.

$\mathbf{a} \cdot \mathbf{n} = 8$ so the equation of the plane is $\mathbf{r}\cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) = 8$.

Converting to cartesian form gives us $x+y+z=8$.

### Vector equation of a plane

We can extend a line into a plane by adding a second, non-parallel direction vector.

Consider a plane $\pi$ that contains a point $A$ with position vector $\mathbf{a}$ and two non-parallel direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$.
The vector equation of a plane is given by

$$ \pi: \mathbf{r} = \mathbf{a} + \lambda \mathbf{d}_1 + \mu \mathbf{d}_2, \lambda, \mu \in \mathbb{R}$$

Unfortunately, this equation of a plane is often not useful for calculations or analysis. This is partly because there are infinite number of direction vectors in
a plane.

Instead, the most useful vector of a plane is what we term the ** normal vector **, a vector that is perpendicular to a plane. This vector is unique (up to a scalar).
We calculate the normal vector by taking the
vector/cross product
of two direction vectors.

### Equation of a plane in scalar product form

For example, if a plane $\pi$ passes through $A(1,2,3)$ and is has normal vector $(-\mathbf{i}+\mathbf{j}-3\mathbf{k})$, we calculate

$\mathbf{a} \cdot \mathbf{n} = -1 + 2 - 9 = -8$

and write the equation of the plane as
$$ \pi: \mathbf{r} \cdot \begin{pmatrix} -1 \\ 1 \\ -3 \end{pmatrix} = -8$$

Let $R$ be a general point on the plane $\pi$ with position vector $\mathbf{r}$. It is given that $\pi$ has normal vector $\mathbf{n}$ and passes through point $A$ with position vector $\mathbf{a}$.

$\overrightarrow{AR}$ is a direction vector of the plane and is thus perpendicular to $\mathbf{n}$.

Hence $\overrightarrow{AR}\cdot \mathbf{n}=0$.

$(\mathbf{r}-\mathbf{a}) \cdot \mathbf{n} = 0 \\
\mathbf{r}\cdot \mathbf{n} - \mathbf{a} \cdot \mathbf{n} = 0 \\
\mathbf{r}\cdot \mathbf{n} = \mathbf{a}\cdot \mathbf{n}$.

### Cartesian equation of a plane

Consider the plane with equation (in scalar product form) $$\pi: \mathbf{r} \cdot \begin{pmatrix} n_1 \\ n_2 \\ n_3 \end{pmatrix} = k$$

If we let $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, we get the cartesian equation of the plane:

If we use the plane from the earlier example, its cartesian equation is $-x+y-3z=-8$.

### Finding a point on a plane given in scalar product or cartesian form

To solve questions involving lines, the two most important objects are often the point that lies on the line and the direction vector of the line. These two are conveniently located in the vector equation of a line.

To solve questions planes, the two most important objects are often the point that lies on the plane and the normal vector of the plane. Because of the significance of the normal vector, we typically like to have our equations in scalar product form or the cartesian form. These two forms allow us to see what the normal vector is. However, these forms do not give rise to a point immediately.

For example, consider a plane with equation $2x-3y+5z = 6$. Immediately we can deduce that the normal vector is $2\mathbf{i}-3\mathbf{j}+5\mathbf{k}$. We do not, however, know the coordinates of any point on the plane. We need to find 3 numbers $x,y$ and $z$ that satisfy the equation. For example, $(2,1,1)$ will satisfy the equation. Unfortunately, such a point will be quite hard to think of.

The trick is to use a lot of 0s. If we let $y,z=0$, then $x$ must be 3 to satisfy the equation. Hence $(3,0,0)$ lies on the plane. Alternatively, $(0,-2,0)$ and $(0,0,\frac{6}{5})$ also lie on the plane.