Math 1410, Spring 2020

Planes and Distances

January 23, 2020

Sean Fitzpatrick
University of Lethbridge

Warm-up

Given \(\uu = \langle 2,0,-1\rangle\) and \(\vv =\langle 1,3,-1\rangle\text{,}\) find vectors \(\vec{a}\) and \(\vec{b}\) such that:

\(\vec{a}\) is parallel to \(\vec{u}\)
\(\vec{b}\) is orthogonal to \(\vec{u}\)
\(\vec{a}+\vec{b}=\vec{v}\)

Vector \(\vec{a}\) is called the projection of \(\vv\) onto \(\uu\text{.}\) Notation and formula:

\begin{equation*} \proj_{\uu}\vv = \left(\frac{\uu\dotp \vv}{\norm{\uu}^2}\right)\uu\text{.} \end{equation*}

Distance

Example:

Find the distance from the point \(P=(3,1,2)\) to the line
\begin{equation*} \la x,y,z\ra = \la 0, 2, -1\ra + t \la 2, -1, 2\ra\text{.} \end{equation*}
What is the point on the line that is closest to \(P\text{?}\)
Find the distance between the parallel lines
\begin{equation*} \vec{r}_1(t) = \la 2, -1, 3\ra + t\la 1, 2, 3\ra, \text{ and } \vec{r}_2(t) = \la -4, 1, 3\ra + t\la 2, 4, 6\ra\text{.} \end{equation*}

Planes in \(\R^3\)

Two ways to describe a plane:

A point, and two vectors parallel to the plane.
A point, and one vector perpendicular to the plane.

Second option is simpler.

Suppose \(\vec n = \la a,b,c\ra\) is perpendicular to the plane.
Suppose also \(P_0 = (x_0,y_0,z_0)\) and \(P=(x,y,z)\) are in the plane.
Then \(\orr{P_0P}\) is parallel to the plane, so \(\vec n \dotp \orr{P_0P}=0\text{.}\)

Finding equations

Find the equation of the plane:

Through \(P_0=(2,-1,3)\) and perpendicular to \(\vec n = \la 5, -3, 4\ra\)
Perpendicular to the line \(\vec{r}(t) = \la 3+2t, -5+3t, -2-9t\ra\text{,}\) and containing the origin.
Containing the points \(P=(1,0,4), Q=(2,-1,3), R=(5,2,4)\)
Containing the lines
\begin{align*} \vec{r}_1(s) \amp = \la 0, 1, 2\ra + s\la 4, -2, 1\ra\\ \vec{r}_2(t) \amp = \la -2, -3, 7 \ra + t\la 3, 1, -2\ra \end{align*}
(We found that these intersect at \(P_0=(4,-1,3)\) on Tuesday.)

Intersection: plane and a line

Suppose we have:

A plane \(ax+by+cz=d\)
A line \(\la x,y,z\ra = \la x_0+at, y_0+bt,z_0+ct\ra\)

How do we find what point (if any) they have in common?

Example:

Find the point of intersection of the plane

\begin{equation*} 2x-3y+4z=6 \end{equation*}

and the line

\begin{equation*} \la x,y,z\ra = \la 2,-1,0\ra + t\la 1,1,-2\ra\text{.} \end{equation*}

Intersection: two planes

Like parallel lines in \(\R^2\text{,}\) parallel planes in \(\R^3\) do not interesect, but any non-parallel planes do. What does the intersection look like? How do we find it?

Example:

Find the intersection of the planes \(x-2y+3z=2\) and \(3x-y-4z=8\text{.}\)

Two methods:

Using cross products
Solving a system

Distance: point to plane

Yes, there's a formula in the book.
Don't use it.
The formula lets you get an answer without understanding what's going on. (That is not a good thing.)

Setup is similar to point–to–line distance, but we project onto a normal vector, not a direction vector. Given a point \(P\) and a plane \(ax+by+cz=d\text{:}\)

Choose a point \(P_0\) on the plane.
Form the vector \(\orr{P_0P}\text{.}\)

This goes from the plane to the point, but probably not at a right angle.
Project \(\orr{P_0P}\) onto the normal vector \(\vec n = \la a,b,c\ra\text{.}\)

Result goes from plane to point, and is as short as possible.

Example

Example:

Find the distance from the point \(P=(1,4,1)\) to the plane \(2x-y+3z=4\text{.}\)

Also find the point \(Q\) on the plane that is closest to \(P\text{.}\)

Two possible methods for solving the problem:

Using projection
Using a normal line

Distance: parallel planes

Example:

Find the distance between the planes \(2x-3y+z=4\) and \(2x-3y+z=10\text{.}\)

Distance: skew lines

Example:

Find the distance between the skew lines

\begin{align*} \vec{r}_1(s) \amp = \la 3, 1, 1\ra + s\la 2, -1, 3\ra\\ \vec{r}_2(t) \amp = \la 1,0,-1\ra + t\la 1, 2, 1\ra\text{.} \end{align*}

Bonus opportunity: find the two points (one on each line) that have this minimum distance.