Uploaded by TheIntegralCALC on 07.12.2010

Transcript:

Hi everyone. Welcome back to integralcalc.com. 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.

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.