# Math 1600A Lecture 6, Section 002

## Announcements:

More texts, solutions manuals and packages coming soon, possibly today.

Read Section 1.4 (just the part on code vectors) and Section 2.1 for next class. Work through recommended homework questions. Scans of all of sections 1.1, 1.2 and 1.3 are available from the course home page.

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 have started.

Lecture notes (this page) available from course web page by clicking on our class times.

## Partial review of last lecture:

### Lines in $\R^2$ and $\R^3$

The vector form of the equation for a line $\ell$ is: $$\vx = \vp + t \vd ,$$ where $\vp$ is the position vector of a chosen point on the line, $\vd$ is a vector parallel to the line, and $t \in \R$.

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}

### Lines in $\R^2$

For a line in $\R^2$, there are additional ways to describe a line.

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.

### Lines in $\R^3$

There are also normal and general forms of equations for a line in $\R^3$, which I won't review here.

### Planes in $\R^3$

Normal form: $$\vn \cdot (\vx - \vp) = 0 \quad\text{or}\quad \vn \cdot \vx = \vn \cdot \vp .$$ When expanded into components, it gives the general form: $$a x + b y + c z = d,$$ where $\vn = [a, b, c]$ and $d = \vn \cdot \vp$.

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 point $P = (1, 2, 0)$ and has normal vector $\vn = [2, 1, -1]$.

Solution: For $\vp = [1,2,0]$ and $\vn = [2, 1, -1]$, the normal form is $$\vn \cdot \vx = \vn \cdot \vp = 4 .$$ 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 $\vv = \vec{PR}$ (see below). If $\vu$ and $\vv$ are parallel, you need to try again.

## New material

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 whiteboard.

### Cross products (Exploration after Section 1.3)

Given vectors $\vu$ and $\vv$ in $\R^3$, we would like a way to produce a new vector that is orthgonal to both $\vu$ and $\vv$. The cross product does this.

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 whiteboard.

Note: This only works in $\R^3$!

Can now finish the previous example.

### Distance from a point to a line (back to Section 1.3)

Recall the formula for the projection of $\vv$ onto $\vu$: $$\proj_{\vu}(\vv) = \left( \frac{\vu \cdot \vv}{\vu \cdot \vu} \right) \vu .$$ Applets for exploring projections: javascript (doesn't work under firefox), or Java.

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 whiteboard, 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.

### Distance from a point to a plane

Example: Find the distance from the point $B = (1,3,6)$ to the plane $\mathcal{P}$ whose equation is $2x + y - z = 2$. Solution on whiteboard, leading to $$d(B, \mathcal{P}) = \| \proj_\vn(\vv) \| = \frac{|\vn \cdot \vv|}{\|\vn\|} = |-3|/\sqrt{6} = 3/\sqrt{6} .$$