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.