Joe O'Bryant
Week#1
jro6766
1.2.6

Problem

Find both a parametric form and an equation form for the plane passing through (1,0,0), (1,1,1), and (0,0,0)

Solution

Let's find the parametric form first:

The parametric form of a plane is usually written as

where  is a point in the plane, and  and  are vectors tangent to the plane.  We use s and t as two independently varying parameters to sweep out all the points contained by the plane.

and  can be formed by subtracting the first vector from the other two vectors:

= <1,1,1> - <1,0,0> = <0,1,1>
= <0,0,0> - <1,0,0> = <-1,0,0>

Using <1,0,0> to represent , we obtain the following equation:

= <0,s,s> + <-t,0,0,> + <1,0,0> =>
= <1-t,s,s>

Using the results from finding the parametric form, we can now determine the equation form, which is usually written as

a*x + b*y + c*z = d (a,b,c, and d are constants)

Let = <a,b,c>.  We must find a vector n which is perpendicular to both and  .  Since the cross product of and  give rise to a vector perpendicular to both and , we can use=X 

The constant d can be determined by the following equation

d =*

This implies that d = <0,-1,1>*<1,0,0> = 0*1 - 1*0 + 1*0 = 0.

Since = <a,b,c> = <0,-1,1,>, and a*x + b*y + c*z = d, the equation form of the plane is: z - y = 0