Question

Find an equation of a plane that contains all the points that are equidistant from the given points:

(-5,1.-3), (2,-1,6)

Not even sure how to go about it.

Given solution is: 7x-2y+9z = 3

Answer 1

midpoint and normal

Answer 2

@pgpilot326 That doesn't seem to get me anywhere near the given solution O.o

Answer 3

there's a line going through the 2 points, right? hlaf way between is the midpoint. a plane that is normal to the line will be equidistant from both given points

Answer 4

Well, the answer I got doing that gave me something kinda, sorta close, just I guess there must've been a specific method that resulted in the solution given. The solution I'm given is

x-z = 0. Using the midpoint and the normal vector, I come up with
x+ 8y + z = 0

Answer 5

Hmm....hang on, lemme recheck that, I think I made a mistake.

Answer 6

Hmm...nvm, I'm just too unsure xD I simply found the normal vector by taking the cross-product of the two points.

Answer 7

Grr, damn it, the solution I posted was wrong, that was the solution for a different question @_@

Answer 8

Brain cramping today -_-

Answer 9

midpoint = <(-5+2)/2, (-1+1)/2 , (-3+6)/2> = (-3/2, 0 , 3/2)

Answer 10

Yeah, got that.

Answer 11

And used the cross-product of the two points to get <3, 24, 3>

Answer 12

Im just trying to see if I can find the correct given solution.

Answer 13

why cross at this point?

Answer 14

I thought you said I need the normal?

Answer 15

yeah, but what are you crossing?

Answer 16

The inital points given.

Answer 17

how will that work?

Answer 18

Dunno. I wasn't sure what else to cross to find the normal vector.