Question

The plane that passes through the point (5,2,5) and is perpendicular to both 5x+3y=15 and 2x+3y+5z=-19. What is its normal equation?

Answer 1

Okay, so we have the normal vectors <5, 3, 0> and <2, 3, 5> for the two planes to which we want to be perpendicular. Remember that if we take the cross product of two vectors, then we will get a vector which is perpendicular to those vectors. Does this make sense?

Answer 2

So using the cross product will get us the normal vector for the plane we want. Can you do that ConDawg?

Answer 3

yes

Answer 4

Let's start with that.

Answer 5

okay

Answer 6

Cross <5,3,0> and <2,3,5>?

Answer 7

Yeah.

Answer 8

i got <15,-25,9>

Answer 9

You can double check. If you do the dot product with the other two individually, you should get 0.

Answer 10

Anyway, we have some plane where 15x - 25y + 9z = c. Now we need to find c. How do you suppose we do that?

Answer 11

<15,-25,9> (dot) (5,2,5)??

Answer 12

:D thank you!

Answer 13

Well, you could just plug it into the equation and solve for c, but yeah that gets you the same result.

Answer 14

By the way, do you have a method of verifying your solution?

Answer 15

No I dont