Question

find a vector equation and parametric equations through the point (1,0,6) and perpendicular to the plane x + 3y + z =5

Answer 1

is the point on a line? thru the plane?

Answer 2

ummm what?

Answer 3

if so, the normal of the plance is the perp to it; strip off your coeefs for the line vector:

<1,3,1>

would be the vector that is perpenducular to the plane

Answer 4

since 1,0,6 is not on the plane im going to assume its on the line perp to it (the perp line is called a normal)

Answer 5

okay I'm with you so far

Answer 6

x = Px + 1t
y = Py + 3t
z = Pz + 1t


given the point (1,0,6) this makes

x = 1 + 1t
y = 0 + 3t
z = 6+ 1t

Answer 7

okay but why are the points (1,3,1) perpendicular to the plane?

Answer 8

its a result of how the equation of a plane is defined

Answer 9

the perp to the plane is a vector that dot producted to every line in the plane

Answer 10

so given a point (Px,Py,Pz) in space: we can construct any vector from it by taking any random point (x,y,z) and subtracting the 2 of them:

(x-Px, y-Py, z-Pz) is every vector that comes out from our point into space

Answer 11

to elimante all the vectors that are not in a plane; we need to establish a perp vector to it so that when we dot it to these vectors we get 0, so all vectors of the form:

(x-Px,y-Py,z-Pz) dot (Nx,Ny,Nz) = 0

Nx(x-Px) +Ny(y-Py) +Nz(z-Pz) = 0

is the equation of the plane

Answer 12

okay makes sense!

Answer 13

good :)

Answer 14

you're way better at explaining things than my book...

Answer 15

thats becasue im crazy lol

Answer 16

lol, does normal mean perpendicular?

Answer 17

yes.

Answer 18

Actually, yes in this context.
The word normal is used for other things in maths elsewhere.