Continue reading Section 1.3 and also the Exploration on cross products for next class. Work through recommended homework questions.

Quiz 1 this week in tutorials. See last lecture for how they run.

My office hour cancelled today.

Lecture notes (this page) available from course web page.

The **vector form** of the equation for $\ell$ is:
$$
\vx = \vp + t \vd
$$
where $\vp$ is the position vector of a point on the line,
$\vd$ is a vector parallel to the line, and $t \in \R$.
This is concise and works in $\R^2$ and $\R^3$.

If we expand the vector form into components, we get the
**parametric form** of the equations for $\ell$:
$$
\begin{aligned}
x &= p_1 + t d_1\\
y &= p_2 + t d_2\\
( z &= p_3 + t d_3 \quad \text{if we are in $\R^3$})
\end{aligned}
$$

The **normal form** of the equation for $\ell$ is:
$$
\vn \cdot (\vx - \vp) = 0 \quad\text{or}\quad \vn \cdot \vx = \vn \cdot \vp ,
$$
where $\vn$ is a vector that is *normal = perpendicular* to $\ell$.

If we write this out in components, with $\vn = [a, b]$, we get the
**general form** of the equation for $\ell$:
$$
a x + b y = c,
$$
where $c = \vn \cdot \vp$.
When $b \neq 0$, this can be rewritten as $y = m x + k$, where
$m = -a/b$ and $k=c/b$.

**Note:** All of these simplify when the line goes through the origin,
as then you can take $\vp = \vec 0$.

**Example:** Find all four forms of the equations for the line in $\R^2$
going through $A = [1,1]$ and $B = [3,2]$.

**Note:** None of these equations is *unique*, as $\vp$, $\vd$ and $\vn$
can all change. The general form is closest to being unique: it is unique
up to an overall scale factor.

**Question:** What are the pros and cons of the different ways of
describing a line?

When expanded into components, it gives the

**Note:** You can read off $\vn$ from the general form.
Two planes are parallel if and only if their normal vectors are parallel.

A plane can also be described in **vector form**.
You need to specify a point $\vp$ in the plane as well as
two vectors $\vu$ and $\vv$ which are parallel to the plane but not parallel to each other.
$$
\vx = \vp + s \vu + t \vv
$$
When expanded into components, this gives the **parametric equations** for a plane:
$$
\begin{aligned}
x &= p_1 + s u_1 + t v_1\\
y &= p_2 + s u_2 + t v_2\\
z &= p_3 + s u_3 + t v_3 .
\end{aligned}
$$
Table 1.3 in the text summarizes this nicely (except for the one typo mentioned above).

It may seem like there are lots of different forms, but really there are two: vector and normal, and these can be expanded into components to give the parametric and general forms.

**Example:** Find all four forms of the equations for the plane in $\R^3$
which goes through the point $P = (1, 2, 0)$ and has normal vector
$\vn = [2, 1, -1]$.

**Solution:** For $\vp = [1,2,0]$ and $\vn = [2, 1, -1]$,
we have $\vn \cdot \vp = 4$. So the normal form is
$$ \vn \cdot \vx = 4 .$$
The general form is
$$ 2 x + y - z = 4 .$$
To get the vector form, we need two vectors parallel to the plane, so
we need two vectors perpendicular to $\vn$.
Can get these by trial and error, for example, $\vu = [-1, 2, 0]$
and $\vv = [0, 1, 1]$. Then the vector form is
$$ \vx = \vp + s \vu + t \vv .$$
Expanding into components gives the parametric form:
$$
\begin{aligned}
x &= 1 - s\\
y &= 2 + 2 s + t \\
z &= \phantom{2 + 2 s + }\, t .
\end{aligned}
$$
You can also find parallel vectors by finding two other points $Q$ and $R$
in the plane and then taking $\vu = \vec{PQ}$ and $\vv = \vec{PR}$.
If $\vu$ and $\vv$ are parallel, you need to try again.

**True/false:**
The planes given by
$$ 2 x + 3 y + 4 z = 7 $$
and
$$ 4 x + 6 y + 8 z = 9 $$
are parallel.

If the 9 was changed to 14, the two planes would be **equal**,
but the answer would still be **true**, as a plane is parallel to
itself. The right hand side shifts the position of a plane, but
not its orientation.

**True/false:**
The lines given by
$$
\begin{aligned}
\vx &= \vp_1 + t \vd_1, & \vp_1 &= [1,2,3], & \vd_1 &= [2,0,-2] \\
\vx &= \vp_2 + t \vd_2, & \vp_2 &= [2,4,6], & \vd_2 &= [2,1,0] \\
\end{aligned}
$$
are parallel.