Question

Find an equation of the plane.
The plane passes through the points (5, 3, 1) and (5, 1, -5) and is perpendicular to the plane
8x + 9y + 2z = 11.

Answer 1

I found the vector of two points, crossed points w/ a,b,c from eq of plane. then I plugged into eq. of plane formula. what did I do wrong?

Answer 2

Check your answer by taking dot product of two perpendicular vectors.

Answer 3

I see a mistake in my calculation.

Answer 4

What is wrong in my method?

Answer 5

Show us your method.

Answer 6

(5,3,1) (5,1,-5) -> <0,-2,-6>
<0,-2,-6>X<8,9,2>=<50,-48,16>
<50,-48,16>.<8,9,2> = 0
<50,-48,16>=<25,-24,8>
25(x-5)^2-24(y-3)^2+8(z-1)=0
25x-125-24y-72+8z-8=0
25x-24y+8z=205

Answer 7

What does your answer say in book, apparently you are checking it?

Answer 8

chag (sorry to interrupt) did you ask about the calculus dream problem or was that someone else

Answer 9

Someone else.

Answer 10

k thanks

Answer 11

There is another method, I don't know if this is valid <a,b,c>dot V=0. it gives you
25x-27y+2z=0

Answer 12

Not correct

Answer 13

You asked what is wrong with your method. The mistake is when you did the equation formula you squared x and y; you shouldn't have squared it
25x-24y+8z=?
Plug in (5,3,1) and you get
25x-24y+8z=61