Find eqn of plane thru (2,2,1), (-1,1,-1), perp to 2x-3y+z=3

[MarkSA]

Hello,

So I'm back for another semester of fun times in Calculus... this time Calc 3. We're starting out with vectors, planes, and 3d space and i'm having some real trouble understanding and visualizing some of this.

Q: Find the equation of the plane that passes through the points (2,2,1) and (-1,1,-1) and is perpendicular to the plane 2x-3y+z=3.

Here is what I thought. Since i'm looking for a plane perpendicular to the one given, couldn't I find the normal vector perpendicular to the 2x-3y+z=3 plane. I thought this would be <2,-3,1>. Now I have the normal vector, perpendicular to my plane, and I also am given a point in my plane. (Actually, two points.)

I thought using the scalar equation for the plane,
a(x - x1) + b(y - y1) + c(z - z1) = 0
I could just plug in say, (2,2,1) for (x1,y1,z1) and <2,-3,1> for <a,b,c>.

But I know i've gone wrong somewhere. First because I didn't use the other point, and second because if I plug the other point into the plane equation, I end up with a different equation. Any ideas? Thanks!

 

[pka]

Re: Find the equation of a plane

Find the vector \(\displaystyle <2,2,1>-<-1,1,-1>\).
Now find the cross product of that vector with \(\displaystyle <2,-3,1>\).
That is the normal of the required plane.

 

[MarkSA]

Re: Find the equation of a plane

thanks. can you verify if I am understanding what's going on here correctly?

<2,-3,1> is the normal vector for the 2x-3y+z=3 plane, let's call this plane A. this vector is perpendicular to plane A.
I have two points on another plane B: (2,2,1) and (-1,1,-1). Position vector for these points is <3,1,2>. I need the normal vector for this plane B to find the equation for the plane B...

The cross product gives me a vector that is perpendicular to both vectors being crossed. So the cross product of the position vector with the normal vector of plane A (i'm thinking of them as two lines with the resultant vector connecting them) is the normal vector, and since it's perpendicular to both of those vectors it must be perpendicular to plane B. With one point in plane B and the normal vector of the plane B I can find the equation of the plane.

Is this the correct line of thinking here? I end up with -7x - y + 9z = -7 for my answer.

 

[galactus]

Re: Find the equation of a plane

I think you are in error, Mark. But close. Unless I made a mistake.

\(\displaystyle P_{1}=[2,2,1], \;\ P_{2}=[-1,1,-1]\)

\(\displaystyle n=[2,-3,1]\) is normal to the given plane.

\(\displaystyle \overrightarrow{P_{1}P_{2}}=[-3,-1,-2]\) is parallel to the line.

Take the cross product of the normal and \(\displaystyle \overrightarrow{P_{1}P_{2}}\):

\(\displaystyle n\times \overrightarrow{P_{1}P_{2}}\)=? is normal to the plane whose equation is.............

Now, can you get it?.

 

[pka]

Re: Find the equation of a plane

Mark has a mistake in the z-component of the normal.
\(\displaystyle \left\langle {2, - 3,1} \right\rangle \times \left\langle {3,1,2} \right\rangle = \left\langle { - 7, - 1,11} \right\rangle\)
Note the 11.