# Thread: Equation of the Plane

1. ## Equation of the Plane

I have to find the eq. of the plane which contains the following lines:

x = 1 - t
y = 2 + 3t
z = 2t

and

x = 1 - s
y = 5
z = 4 - 2s

So we have the vectors <-1, 3, 2> and <-1, 0, -2>

I found the cross product which is <-6, -4, 3> which is my normal vector.

Now what?

I know a_1(x - x_1) + a_2(x - x_2) + a_3(x - x_3) = 0; I have my a_1, ..., a_3, but how do I get the points? Apparently the point is (1, 2, 0), but I don't see how to get that.

2. Hello, Ideasman!

You may kick yourself . . .

I have to find the eq. of the plane which contains the following lines:

. . $\begin{Bmatrix}x \:= \:1 - t \\
y \:= \:2 + 3t \\ z = 2t\end{Bmatrix}$
.and . $\begin{Bmatrix}x \:= \:1 - s \\ y \:= \:5 \\ z \:= \:4 - 2s\end{Bmatrix}$

So we have the vectors $\langle-1, 3, 2\rangle$ and $\langle-1, 0, -2\rangle$ . . . Yes!

I found the cross product which is $\langle-6, -4, 3\rangle$ which is my normal vector. . . . Right!

Now what?

I know $a_1(x - x_1) + a_2(x - x_2) + a_3(x - x_3) \:= \:0$
I have my $a_1,\,a_2,\,a_3$, but how do I get the points?
Apparently the point is $(1, 2, 0)$, but I don't see how to get that.

Use any point on either line.

For example, in the first line, let $t =0$ . . . got it?

3. Of course!

Got it, thanks Soroban.

4. Originally Posted by Ideasman
I have to find the eq. of the plane which contains the following lines:

x = 1 - t
y = 2 + 3t
z = 2t

and

x = 1 - s
y = 5
z = 4 - 2s

So we have the vectors <-1, 3, 2> and <-1, 0, -2>

I found the cross product which is <-6, -4, 3> which is my normal vector.

Now what?

I know a_1(x - x_1) + a_2(x - x_2) + a_3(x - x_3) = 0; I have my a_1, ..., a_3, but how do I get the points? Apparently the point is (1, 2, 0), but I don't see how to get that.
You have two (non-linearly dependent) vectors, which lie in the plane if it
exists (1,3,2)' and (-1,0,-2)'. Then in vector form any point in plane can be writen:

(x,y,z)'=(a,b,c)' + lambda(1,3,2)' + mu(-1,0,-2),

where (a,b,c)' is any point in the plane, so we can use (1,2,0), so our plane
in parametric form is:

(x,y,z)'=(1,2,0)' + lambda(1,3,2)' + mu(-1,0,-2).

So from the y-component we have:

lambda=(y-2)/3.

From the z-component we have:

mu=lambda-z/2=(y-2)/3 - z/2.

Hence, from the x-component we have:

x=1+lambda-mu=1+(y-2)/3-(y-2)/3+z/2=1+z/2

or:

2x-z=1.

RonL