Equation form of a line in 3-D

• yvonnars
In summary, to find the equation form of the line perpendicular to the plane Q and passing through the origin, we can use the parametric form of a line to get the equation L=(x,y,z)=(0-s,0+3s,0+2s). By making s the subject for each variable, we can then set them all equal to each other to get the equation x+y+z=4s for the line L.

Homework Statement

Fine the equation form of the line L that is perpendicular to the plane Q with equation
-x+3y+2z=1 and passing through the origin.

Homework Equations

Parametric form of a line:
L=(0,0,0)+s(-1,3,2)
so x/-1=y/3=z/2

The Attempt at a Solution

x+y+z=4s
but then what is the value of s?

s is just a parameter, for different values of s you will get different points on the line.

Remember L is also (x,y,z) so get s in terms of each x,y and z and they will all be equal to one another.

rock.freak667 said:
s is just a parameter, for different values of s you will get different points on the line.

Remember L is also (x,y,z) so get s in terms of each x,y and z and they will all be equal to one another.

So will the equation form of this line just be x+y+z=4s?

yvonnars said:
So will the equation form of this line just be x+y+z=4s?

No, from L=(0,0,0)+s(-1,3,2) = (0-s,0+3s,0+2s)

L = (x,y,z) = (0-s,0+3s,0+2s)

x = 0-s and so on. Just make s the subject for each one and then put them all equal to each other.