SUNY Geneseo Department of Mathematics

Planes

Thursday, February 13

Math 223 01
Spring 2020
Prof. Doug Baldwin

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Planes

Second half of section 11.5.

Equations

Yesterday we considered finding the plane that contains points ( 1, 0, 0 ), ( 0, 1, 0 ), and ( 0, 0, 1 ). By taking the cross product of two vectors between pairs of these points, we found a vector of coefficients for the plane equation, namely 〈 1, 1, 1, 〉. How come this vector gives the coefficients, and what does the equation actually look like?

Why: v × u is perpendicular to the plane, i.e., it’s a normal to the plane. The plane equations were built around the idea of the dot product between the normal and vectors to any point in the plane being 0:

Vector from P to Q dotted with normal must be 0

The complete plane equation comes from using components of the normal as coefficients, and components of any point in the plane as x0, y0, z0:

Coefficients and offsets for plane equation come from normal and any point in plane

How about the plane perpendicular to r(t) = (1,2,3) + t ⟨ 2, -2, 1 ⟩ and containing point ( 2, 1, 5 )?

Here you have the normal vector already, from the line’s direction vector. Combine that and the point to get the plane equation:

Plane with line passing through at right angle

Using the Equations

Find one new point in the second plane above.

Just find any x, y, and z, by any means you like, that satisfy the plane equation.

Picking 2 variables arbitrarily and solving for the 3rd is the easiest way to do this:

Plug x equals 0, y equals 0 into plane equation to find z equals 7

Line-plane intersections are useful, among other places, for “ray tracing” in computer graphics: mathematically trace backwards along rays of light reaching a virtual eye in order to find out where they came from and therefore what color they are.

Ray from eye towards tree

For example, suppose light arrived along the line r(t) = ( 0, 3, 1 ) + t ⟨ 1, -1, 1 ⟩ , and the plane x + y + z = 1 represents a hillside. Where (if at all) does the line intersect the plane?

Plug the parametric equations for x, y, and z into the equation for the plane, and solve for t. Then put that t back into the equation of the line to find out the actual point:

Ray towards plane, parametric functions from line plugged into plane equation

Problem Set

See handout for details

Next

Vector-valued functions

Read...

in section 12.1

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