Question

find the equation of the plane through point (-2,8,10) and perpendicular to line x=1+T, y=2T, z=4-3T

Answer 1

is the answer: x+2y-3z=16?

Answer 2

our perp line gives us our normal to the plane so strip out the vector in it

Answer 3

<1,2,-3>

now apply this to any random point (x,y,z) - (-2,8,10)

1(x--2) +2(y-8) -3(z-10) = 0

x+2 +2y -16 -3z +30 = 0

x +2y -3z +16 = 0

x +2y -3z = -16

dbl chk the math to be sure, i tend to forget how to add and subtract lol

Answer 4

yay i'm figuring out how to do this!!

Answer 5

pretty soon youll be like; I cant believe I didnt know this stuff !!! :)

Answer 6

i figure in a few weeks we will be going over this in calc 3

Answer 7

okay so for the plane through the origin and points (x,y,z) and (a,b,c,) i would do do (a,b,c)x(x,y,z) to find the normal right?

Answer 8

and then use the normal to write an equation

Answer 9

yes, since the origin is one of our points; the other 2 points are also the vectors; so cross them and get the normal

Answer 10

then use the normal attached to any of the 3 points (0,0,0) being the most useful id assume

Answer 11

will my answer be a dot product?

Answer 12

the cross of 2 vectors is a vector

Answer 13

of the normal and (x,y,z)

Answer 14

your answer will be a dot product of the normal and every vector in the plane:

<x-0, y-0, z-0> would define all vector in the plane

Answer 15

okay awesome thats what i got!