# Vector equation of a plane containing this line.

1. Jun 27, 2012

### manmachine

1. The problem statement, all variables and given/known data

What is the vector equation of a plane that contains the line y=(3,2,1)+x(1,0,2) and is parallel to the z-axis?

2. Relevant equations

Equation of a plane: Ax + By + Cz + D = 0

3. The attempt at a solution

Ok so i substitute the values into the plane,
A(1)+B(0)+C(2)+D=0

But then I am left with 3 variables. Can you please offer some help?

Kind Regards,

2. Jun 27, 2012

### tiny-tim

welcome to pf!

hi manmachine! welcome to pf!

sorry, but i don't understand what you're doing at all

the easiest way of finding the equation is probably first to find the normal to the plane …

what will that be?

(and i think the question probably means the parametric equation v = (3,2,1) + t(1,0,2))

3. Jun 27, 2012

### manmachine

Re: welcome to pf!

Yes but I don't know what the plane is. I know that the normal is the coefficients if the plane. Please help.

4. Jun 27, 2012

### HallsofIvy

Staff Emeritus
Why do you substitute those values? Those would be the components of a vector in the plane, not the coordinates of any point in the plane. (3, 2, 1) is a point in the plane. Setting x= 3, y= 2, z= 1, we have 3A+2B+ C+ D= 0.

Also, it is not a good idea to use "x" and "y" in that equation for the plane. In that vector equation "x" is a parameter, not the x- coordinate and "y" is a vector. Better to use, say, r= (3, 2, 1)+ t(1, 0, 2).

In that form, you can always divide through by any non=zero coefficient so you really have only two variables left. To eliminate one, use the fact that the plane is parallel to the z-axis. That means that the normal to the plane is also normal to the z-axis. That is, <0, 0, 1>.<A, B, C>= C= 0.

5. Jun 27, 2012

### manmachine

And yes, but how can I find the normal if don't have to vector equations IAXBI?

6. Jun 27, 2012

### tiny-tim

hi manmachine!
yes, but you do know the directions of two lines that the normal is perpendicular to, don't you?

7. Jun 27, 2012

### manmachine

There is only one line Tim! I can't use cross product :(

8. Jun 27, 2012

### tiny-tim

z-axis?

9. Jun 27, 2012

### manmachine

oh fudge :P (bangs head on wall)
So it is 0,1,0 by cross product right?

10. Jun 27, 2012

### tiny-tim

he he!
uhh?

11. Jun 27, 2012

### manmachine

Isn't that the normal to plane?

12. Jun 27, 2012

### tiny-tim

(0,1,0) is the y-axis, and is normal to the xz-plane

13. Jun 27, 2012

### manmachine

If I take the cross product of the z-axis (being 0,0,1) and our vector (1,0,2) this is what i get.

14. Jun 27, 2012

### tiny-tim

oh i see!!

ok, now you know the normal to the plane, and you know at least one point on the plane,

so the vector equation of the plane is … ?

15. Jun 27, 2012

### manmachine

Am i right :)?

Take coefficents
0a+1b+0c+D=0
(0)(3)+(1)(1)+D=0
d=-1

This equation of our is plane is this:
b+-1=0 ?

16. Jun 27, 2012

### tiny-tim

shouldn't there be a 2 in there somewhere?
what happened to x y and z ?

don't make it so complicated

you know it's vertical, and parallel to the xz-plane, so it must be y = constant

ok, what is obviously that constant??

17. Jun 27, 2012

### manmachine

I have no idea :p

18. Jun 27, 2012

### manmachine

What do I do next . I am sorry I just dont get it

19. Jun 27, 2012

### HallsofIvy

Staff Emeritus
This is very peculiar. You say that you know that any plane can be written as Ax+ By+ Cz+ D= 0. It has already been pointed out that you an divide through by any number, say, D, to reduce to three unknown constants, A'x+ B'y+ C'z+ 1= 0 (with A'
= A/D etc.). If the line given by the vector formula r= (3,2,1)+t(1,0,2) or parametric equations x= 3+ t, y= 2, z= 1+ 2t. In particular, taking t= 0, (3, 2, 1) is a point in the plane so we must have 3A'+ 2B'+ C'+ 1= 0.

Also, taking t= 1, (3+ 1, 2+ 0, 1+ 2)= (4, 2, 3) is also a point in the plane. so we must have 4A'+ 2B'+ 3C'+ 1= 0.

The last condition is that the plane is parallel to the z-axis. That means that its normal vector, <A, B, C> or <A', B', C'> is normal to <0, 0, 1>. Their dot product gives C= C'= 0 and so we have the two equations -3A'- 2B'- 1= 0 and 4A'+ 2B'+ 1= 0.

By the way, I find it simpler to use the fact that if a plane has normal vector <A, B, C> and contains point $(x_0, y_0, z_0)$, then the plane has equation $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.

20. Jun 27, 2012

### tiny-tim

you know that the plane is y = constant

you also know that (3,2,1) lies on the plane

soooo, y must be … ?