equation of a plane (again)

[mathstresser]

Find the equation of the plane that passes through the point (-1,2,1)and contains the line of intersection of the planes x+y-z=2 and 2x-y+3z=1.

I tried to find the line of intersection, but I don't know how with 3 variables and only two equations. And I don't know how to change the equations from the form they are in now.

 

[soroban]

Hello, mathstresser!

Find the equation of the plane that passes through the point (-1,2,1)
and contains the line of intersection of the planes x+y-z=2 and 2x-y+3z=1.

I tried to find the line of intersection, but I don't know how with 3 variables and only two equations.

There are a number of ways to solve this problem.

Here's just one of them. \(\displaystyle \;\)Maybe others will suggest the alternatives.

To determine a plane, we need three points.
(I assume you know how to find the equation of a plane through 3 given points.)

We already have one point: \(\displaystyle (-1,2,1)\) . . . so we need two more points.

We'll find any two points on the line of intersection.

Try to solve the system: \(\displaystyle \;\begin{array}{cc}[1]\;x\,+\,y\,-\,z\:=\:2 \\ [2]\;2x\,-\,y\,+\,3x\:=\:1\end{array}\)

Add the two equations: \(\displaystyle \,3x\,+\,2x\:=\:3\;\;\Rightarrow\;\;x\:=\:1\,-\,\frac{2}{3}z\;\;\) [3]

Substitute into [1]: \(\displaystyle \,\left(1\,-\,\frac{2}{3}z\right)\,+\,y\,-\,z\:=\:2\;\;\Rightarrow\;\;y\:=\:1\,+\,\frac{5}{3}z\;\;\) [4]

Let \(\displaystyle z\,=\,3.\;\) Then [3] and [4] give us: \(\displaystyle \,x\,=\,-1,\;y\,=\,6\)
\(\displaystyle \;\;\)We have a second point: \(\displaystyle \,(-1,6,3)\)

Let \(\displaystyle z\,=\,-3.\;\) Then [3] and [4] give us: \(\displaystyle \,x\,=\,3,\;y\,=\,-3\)
\(\displaystyle \;\;\)We have a third point: \(\displaystyle \,(3,-4,-3)\)

Now we can finish the problem . . .

 

[mathstresser]

Okay, just so I don't do something completely wrong, I thought I'd ask first...

With the three points, do I do a matrix?

-1 2 1
-1 6 3
3 -4 3

that would give me

-1(18-(-12))-2(-3-9)+1(4-18)
= -20

But that doesn't look right, so I don't really know what to do next.

 

[pka]

This is the way I would do this problem.
The normal to the first plane is \(\displaystyle N_1 = \left\langle {1,1, - 1} \right\rangle\) ant the second normal is \(\displaystyle N_2 = \left\langle {2, - 1, 3} \right\rangle\). If \(\displaystyle P = ( - 1,2,1)\) is given point and \(\displaystyle Q = ( 1, 1, 0)\) is a point in the intersection of the two planes then the vector \(\displaystyle \vec{PQ} \times \left( {N_1 \times N_2 } \right)\) is the normal vector of your plane.