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

The next quiz will be next week, and will cover the material until the end of Monday's class

Help Centers Monday-Friday 2:30-6:30 in MC 106 start on Monday, January 20.

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.

**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 material nicely.

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 points $P = (1, 1, 0)$, $Q = (0, 1, 2)$ and $R = (-1, 2, 1)$.
Solution on board.

But we get stuck on the normal form, which motivates:

**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.