**Read** Sections 2.0 and 2.1 for next class.
Work through recommended homework questions.

**Quiz** today and tomorrow in lab, covering 1.1, 1.2 and the code
vectors part of 1.4.

**Office hour:** today, 11:30-noon, MC103B.

**Help Centers:** Monday-Friday 2:30-6:30 in MC 106 starting today (Sept 17).

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

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 $\vp = \vec 0$.

**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.

**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 earlier).

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.

**Example:** Find all four forms of the equations for the plane in $\R^3$
which goes through the points $P = (1, 1, 0)$, $Q = (0, 1, 2)$ and $R = (-1, 2, 1)$.
Solution on board.

**Definition:** The **cross product** of $\vu$ and $\vv$ is the vector
$$\vu \times \vv := [ u_2 v_3 - u_3 v_2,\ u_3 v_1 - u_1 v_3,\ u_1 v_2 - u_2 v_1].$$

**Theorem:** $\vu \times \vv$ is orthogonal to both $\vu$ and $\vv$.
That is, $\vu \cdot ( \vu \times \vv ) = 0$ and $\vv \cdot ( \vu \times \vv ) = 0$.

Explain on board.

**Note: The cross product only makes sense in $\R^3$!**

Can now finish the previous example, on board.

**Theorem:** The cross product also has the following properties:

(a) $\vv \times \vu = - (\vu \times \vv)$ !

(b) $\vu \times \vec 0 = \vec 0$

(c) $\vu \times \vu = \vec 0$

(d) $\vu \times k \vv = k(\vu \times \vv) = (k \vu) \times \vv$

(e) $\vu \times k \vu = \vec 0$

(f) $\vu \times (\vv + \vw) = \vu \times \vv + \vu \times \vw$

This is exercise 5 from the cross product exploration, and I encourage you to check these properties.

**Example:** Find the distance from the point $B = (1,3,6)$ to the
line through $P = (1, 1, 0)$ in the direction $\vd = [0,-1,1]$.
Solution on board, leading to $d(B, \ell) = \| \vv - \proj_\vd(\vv) \| = 4\sqrt{2}$,
where $\vv = \vec{PB}$.

If the line was in $\R^2$ and had been described in normal form, one could instead compute $\| \proj_\vn(\vv) \|$, which saves one step.