# Finding the equation of a Plane

• Oct 20th 2009, 10:47 AM
craig
Finding the equation of a Plane
Hi, another vector question I'm afraid ;)

I have the 4 lines:

$BC = \begin{pmatrix} -2 \\ -4 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 6 \\ -3 \end{pmatrix}$, $AC = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} + \sigma\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}$.

$l2 = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} + \beta\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}$, $AD = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} + \alpha\begin{pmatrix} -3 \\ 0 \\ -3 \end{pmatrix}$.

I need to show that all 4 lines lie in a single plane and fine the Cartesian Equation of the plane.

The Cartesian equation of a plane is in the form $Ax+By+Cz=d$, so from this I can set up the following equations:

$-2A-4B+C=d$

$3A+2C=d$

$A+2B-2C=d$

From these I can get all $A,B,C$ in terms of $d$.

$A=d$, $B=C=-d$

Putting these values back into the original Cartesian Equation I got:

$dx-dy-dz = d$, which can be simplified to $x-y-z=1$.

Is this correct so far or have I gone horribly wrong?

Thanks
• Oct 21st 2009, 11:40 AM
craig
Over 20 views and none of you have any ideas ;)

Anyone got any comments at all?
• Oct 21st 2009, 12:27 PM
Plato
You are correct so far.
With each line there is a point and a direction vector
Test each line with with respect to the plane $x-y-z=1$.
Does the point satisify that equation?
Is the direction vector perpendicular to $<1,-1,-1>?$

If the answer is yes in all cases you are done.
• Oct 22nd 2009, 05:23 AM
craig
Quote:

Originally Posted by Plato
You are correct so far.
With each line there is a point and a direction vector
Test each line with with respect to the plane $x-y-z=1$.
Does the point satisify that equation?
Is the direction vector perpendicular to $<1,-1,-1>?$

If the answer is yes in all cases you are done.

Yes and yes.

I was pretty sure that I'd gone about it in the right way but just wanted to make sure.