Scalar Equation of a Plane Example 1

Uploaded by TheIntegralCALC on 07.12.2010

Hi everyone. Welcome back to Today we're going to be doing an example about
how to find the scalar equation of a plane. The example that we have gives us the normal
vector n <2, -9, 8> and the point P (-1, 4, -2).
So given those two pieces of information, the normal vector n and the point P, we can
find the scalar equation of the plane. When you’re finding the scalar equation of the
plane, the formula that you're always going to be using is this.
a(x-x_0)+ b(y-y_0)+ c(z-z_0)=0 And in this formula, the normal vector n is represented
by the coefficeints in a, b and c so the normal vector n, these three coordinates here are
a, b and c like our 2, negative 9 and 8. n=
And x sub 0, y0 and z0 are our three coordinates on the point P.
P=(x_0,y_0,z_0) So at this point, it's going to be plug and
play. We’re going to go ahead and drop these coordinates from the normal vector and the
coordinates from the point P into our equation, simplify and solve.
So let's go ahead and get started. Like I said, we're just going to plug in 2 for a,
negative 9 for b, and 8 for c. 2(x-x_0)+ {-9}(y-y_0)+ 8(z-z_0)=0
And then we're going to plug in negative 1, 4 and negative 2 for x0, y0 and z0.
2(x-{-1})+ {-9}(y-4)+ 8(z-{-2})=0 We’re going to leave x, y and z, competely
alone and those variables will remain in our equation. So when we plug in, we leave those
numbers exactly as they are. Then to simplify, I'm just going to go ahead and just remove
double negatives like this. 2(x+1)+ {-9}(y-4)+ 8(z+2)=0
Change this plus sign to a negative. 2(x+1)-9(y-4)+8(z+2)=0
And then in the next step, we'll go ahead and distribute the coefficients on these here.
You could leave the equation in this form but we can simplify it further if we multiply
it out because we'll have the constants here; the positive 2, positive 36 and positive 16,
that will combine into one constant. 2x+2-9y-36+ 8z+16=0
Once we do that, we have an x, y, z term; we're going to arrange that in that order
in front of our equation with the constants at the end.
2x-9y+8z+54=0 So with all that said and done, we end up
with 2x minus 9y plus 8z plus 54 equals 0. And that is our scalar equation of the plane
given this normal vector n and the point P. I hope that helped you guys and I'll see you
in the next video. Bye.