# Equation of a plane

1. Sep 13, 2011

### arl146

1. The problem statement, all variables and given/known data
Find an equation of the plane for the plane that passes through the point (1, -1, 1) and contains the line with symmetric equations x=2y=3z.

2. Relevant equations
ax+by+cz=d
n=<a, b, c>
n $\bullet$ (r -r0) = 0

3. The attempt at a solution
I just have no idea where to start. How do I work through these kinds of problems?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 13, 2011

### ehild

There are two many parameters in your equation of the plane. It would be equivalent to write 1 instead of d.
Plug in (1,-1,1) for x,y,z into the equation of the plane. You get a relation among a,b,c.
The points of the line are also points of the plane. Replace y=x/2, z=x/3 in the equation. It has to be true for all x values. You can find the parameters from this condition.

ehild

3. Sep 13, 2011

### arl146

I got the equation of the plane right out of the book, im confused... and why would you write 1 instead of d ?

plugging the numbers in, youd get a-b+y=d but you said to put 1 for d, so a-b+y=1 ?

and replace y=x/2, z=x/3 in what equation

4. Sep 13, 2011

### ehild

The normal vector can have any length, it does not influence its direction, that it is normal to the plane.
In the equation ax+by+cz=d, a,b,c are the components of a normal vector. In vector form, it is rn=r0n where r0 is a vector of the plane. If you divide the equation by any number, the normal vector changes length, but the direction is the same. But you can not divide by d when it is zero. So you are right, it is better to keep d as you can not exclude that it is zero.

So you have an equation for a, b, c, d: a-b+c=d. You can express one of the parameters with the others: For example, c=d-a+b.

Replace the points of the line into the equation for the plane ax+by+cz=d. It will be

a x+b/2 x +c/3 x=d. Plug in the expression of c. You get an equation that must hold for all values of x. What you get when x =0? What you get when x=1?

ehild

5. Sep 13, 2011

### arl146

When x=0 don't you just get d=0 ? And when x=1 you have the same equation jut with all the x's as 1 I don't see how/why this helps

6. Sep 13, 2011

### arl146

Could i do this (if not why?) :

Use x=2y=3z and switch to parametric eqns and get x=t , y=(1/2)t , z=(1/3)t and so the direction vector is equal to <1, 1/2, 1/3> which is your a, b, and c. So use that for the eqn:
a(x-x0)+b(y-y0)+z(z-z0)=0
so then you'd get x+(1/2)y+(1/3)z=(5/6)

7. Sep 13, 2011

### HallsofIvy

Staff Emeritus
That doesn't work because the plane $a(x- x_0)+ b(y- y_0)+ c(z- z_0)= 0$ or, equivalently, $ax+ by+ cz= const$ has <a, b, c> as a normal vector. The plane you calculated has the given line perpendicular to the plane, not in the plane.

What I would do is this: taking t= 0 in the parametric equations x= 0, y= 0, z= 0 so (0, 0, 0) is a point on that line and so in the desired plane. Since we are given that (1,-1, 1) is also in the plane, the vector <1-0, -1- 0, 1- 0>= <1, -1, 1>, is in the plane. We now have two vectors, <1, -1, 1> and the direction vector of the given line, <1, 1/2, 1/3> in the plane. Their cross product is perpendicular to both and so perpendicular to the plane. You now have a point in the plane and a vector perpendicular to the plane.

8. Sep 13, 2011

### ehild

You have found d : d=0, so
a-b+c=0 and *
a x+b/2 x +c/3 x = 0.
The last equation holds for all x values. If you plug in x=1 you get
a+b/2+c/3=0**

Solve the system of equations * and **. One parameter is arbitrary, find the others in terms of that parameter.

ehild

9. Sep 15, 2011

### arl146

i used HallsofIvy's method:

so the answer i got is (5/6)(x-1) - (2/3)(y+1) + (3/2)(z-1) = 0

is that right ?

10. Sep 15, 2011

### ehild

Does it contain the line x=2y=3z?

ehild

11. Sep 15, 2011

### arl146

how do i check that?

12. Sep 15, 2011

### ehild

That line goes through the origin. So x=0, y=0, z=0 has to be a point of the plane. Is it?

ehild

13. Sep 15, 2011

### arl146

...no

14. Sep 15, 2011

### ehild

Check that vector product. You have only a sign error.

ehild

15. Sep 15, 2011

### arl146

oh, ok i see what i did. i did the plane's vector thing cross the dir vector of the line but its the other way around. how would i know that on a test?

16. Sep 15, 2011

### HallsofIvy

Staff Emeritus
No, that was not your error. Doing cross product "the other way around" gives a vector pointing in the opposite direction but still perpendicular to the plane.

That would be the same as multiplying the entire equation by -1: -a(x- 1)- b(y-1)- c(z-1)= 0 rather than a(x-1)+ b(x-1)+ c(x-1)= 0 but, in fact, they are the same equation.

Go ahead and do the cross product again. You have an error in the last component alone.

17. Sep 15, 2011

### ehild

The order does not matter when you get the normal vector. The negative of a normal vector is also a normal vector. There is one sign wrong in your equation: It should be
(5/6)(x-1) -(2/3)(y+1) - (3/2)(z-1) = 0.
And it would be nicer to write the equation in the standard form: ax+by+cz=d.

ehild

18. Sep 15, 2011

### arl146

Ohhhh I'm just dumb and did the cross product wrong. I got it now thanks!