Question

The plane that passes through the point (2,3,-1) and is perpendicular to both 3x-2y+3z=27 and z-(2x+4y)=2 has _______ as its normal equation
I really need help I am stumped

Answer 1

Find the normal vectors of the planes and take the cross product. That will give you the normal vector of your plane <a, b, c>. Then take the dot product of the normal vector with the point, that will give you the constant d. Your plane is just ax + by + cz = d

Answer 2

I really need to get this done, could you go through this one with me so I can understand it?

Answer 3

Okay, do you know how to find the normal vectors of a plane?

Answer 4

Not really I am new to this and my teacher expects me to have it done for tomorrow I am lost :/

Answer 5

The normal vector of a plane is just the coefficients put in a vector.

Answer 6

So 3x-2y+3z=27 becomes <3, -2, 3>

Answer 7

so <3,-2,3> and <-2,-4,1> ?

Answer 8

Yes.

Answer 9

The cross product of those vectors will give you a vector perpendicular to them.

Answer 10

Which we can use as the normal vector for the plane we are trying to find.

Answer 11

alright.. hmmm

Answer 12

so <10,-9 , -16>?

Answer 13

Yeah, so what should our plane be?

Answer 14

Those are the coefficients.

Answer 15

I worked it out and got it ! Thanks a bunch!

Answer 16

Did you skip a step?

Answer 17

Did you use the point (2,3,-1) ?

Answer 18

I did the dot product and used those as coefficients

Answer 19

ok good.