Question

Given the plane that passes through the points (3, -4, -6), (5, -9, 10) and (-5, -7, -2). Let A(x0,y0,z0) be a point on the plane with x0 = 0 and y0 = 3. Find z0

Answer 1

this we can cross to get a normal

Answer 2

lets set a point at origin to determine the vectors involved:

(3, -4, -6) (5, -9, 10) (-5, -7, -2)
-3 +4 +6 -3 +4 +6 -3 +4 +6
---------------------------------
(0, 0, 0) <2,-5, 16> <-8,-3,4>

Answer 3

do you see why id did that?

Answer 4

when we can measure the points from the origin, we can turn them into vectors

Answer 5

yeah just titing the triangle so that 1 side starts at the origin. what does that do tho

Answer 6

< 2,-5, 16>
<-8,-3, 4 >
-----------
x = -20+48 = 28
y = -128-8 = -136
z = -6-40 = -46

check the math on that

Answer 7

you can turn 2 of them into vectors well erm

Answer 8

? where are those numbers coming from

Answer 9

im crossing the 2 vectors i found from the points

Answer 10

a x b = normal to the plane

Answer 11

tell me where any of this doesnt make sense

Answer 12

nah i understand what you are doing graphically

but where does this come from
x = -20+48 = 28
y = -128-8 = -136
z = -6-40 = -46

Answer 13

i found my 2 vectors that reside in the plane:

< 2,-5, 16> <-8,-3, 4 > ; given 2 vectors in the plane, if you cross them you get the normal that is perp to the plane