A straight line passes through the points $(1, 2, 3)$ and $(-3, -2, -1)$.
I have calculated the system of equations of this line to be $$ x = 1 - t,\, y = 2 - t, \, z = 3 - t $$
The question I have been asked to solve is to find the equation of the plane containing this line.
I'm not exactly sure how to go about this.
Thanks!
Since one form of the equation of a plane is $ax+by+cz+d = 0$, you want $0 =a(1-t)+b(2-t)+c(3-t)+d =a+b+c+d-t(a+2b+3c) $.
For this to be true for all $t$, you must have $a+b+c+d=0$ and $a+2b+3c=0$.
You can choose any two of these arbitrarily and determine the others in terms of those. For example, choose $a$ and $b$. Then $c = -(a+2b)/3$ and $d =-a-b-c =-a-b+(a+2b)/3 =-(2a+b)/3 $.
You might say "Hey! There are two free parameters here, but there should be only one, since all the planes must pass through the line."
This is resolved by noting that if all of $a, b, c, $ and $d$ are multiplied by any non-zero real, the resulting plane is the same. We can therefore adopt some normalization convention, such as $a^2+b^2+c^2+d^2 = 1$, to eliminate this possibility. This reduces the number of free parameters from two to one.