# Vector And Scalar Equations Of A Plane

1. Jul 11, 2012

### Plutonium88

1. The problem statement, all variables and given/known data
Find vector and scalar equations that passes through point P(3,7,-1) and is perpindicular to the line of intersection of 2 planes. P1:x-y-2z+3=0 P2: 3x-2y+z+5=0

So initially i started by finding the line between the two planes.

N1 Does Not Equal N2 so they are not paralell (N=Normal)

(1)x-y-2z+3=0 (2)3x-2y+z+5=0

if i multiply equation 1 by 3 i get
(3) 3x-3y-6z+9=0

If i subtract equation 3 from equation 2 i get

-y-7z+4=0

y=4-7z (4)

If i sub (4) into equation 1.

x - (4-7z) -2z + 3 = 0
x -1 +5z=0
x= 1-5z

z=t

therefore.

z=t
y=4-7t
x=1-5t

Equation of line is R=(1,4,0) + t(-5,-7,1)

(Also i noticed, when i did the cross product of the normals of these two planes, you will also find this same direction vector (-5-7,1)

So Now it says Find scalar + Vector equations Given point P(3,7,-1) and PERPINDICULAR to the line of intersection...

The only other case i have dealt with similiar to this, is where you have to find the planes equation when the intersection is PARALELL. In that case you just used the direction vector you find, in the equation of the line, to solve for vector + scalar equations..

***
So what i am lead to beleive with what i learned from that problem, since the plane is perpendicular to the line of intersection, that means that the direction vector of the line is parallel to the normal of the plane***

So this is where im stuck, can some one please tell me if this thinking is correct.. If it is correct, here is what i would make my answer...

NormalPlane3 = Direction Vector Of Line = (-5,-7,1)

-5x-7y+z + d = 0
(plugin point P to solve for D)
d= 65

-5x-7y+z+65=0 --SCALAR EQUATION

Also in terms of solving for a vector equation... how can i find a Direction vector i am stuck on this part as ewll.

2. Jul 11, 2012

### chiro

Hey Plutonium88.

For the vector equation, is this in the form n . (r - r0) = 0? If this is the case you know the normal (for ax + by + cz + d = 0, normal is (a,b,c) or multiple thereof but not 0) and then for some arbitrary r0 you solve -d = n.r0 where -d = ap + bq + cr and you just plug in any values that satisfy that identity.

As for the perpendicularity, you know that if the two lines are not parallel and intersect then these form a plane if the intersection is at the point. The reason for this is that there are two dimensions because one line has one parameterization and the other has a different one resulting in two distinct parameterizations as opposed to one in the case of two lines being equal (having the same parameterization).

So if you think of taking the cross product of the direction vectors of the two lines from the point of intersection, then by the definition of the cross product you will get a new vector that is perpendicular to these two direction vectors and because they are independent you get a non-zero normal vector.

Because the cross product of two vectors is perpendicular to both vectors, the plane formed by these two lines will have a normal that is a scalar multiple of the one that is perpendicular to both lines.

3. Jul 11, 2012

### HallsofIvy

Staff Emeritus
Yes, that is correct.
Sure. The cross product is perpendicular to both normal vectors and so parallel to both planes. The only line that does that is the line of intersection.

Normally, the first equation for a plane you work with is $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$ where $(x_0, y_0, z_0)$ is a point in the plane and (A, B, C) is a vector perpendicular to the plane. That is exactly the situation you have here!

-5x-7y+z + d = 0
(plugin point P to solve for D)
d= 65

-5x-7y+z+65=0 --SCALAR EQUATION

Also in terms of solving for a vector equation... how can i find a Direction vector i am stuck on this part as ewll.[/QUOTE]
You don't need a direction vector. A plane does not have a "direction vector", it has a normal vector and you already have that.
chiro takes the "vector equation of the plane to be the equation asserting that the normal vector is perpendicular to any vector in the plane. You already have that the vector (-5, -7, 1) is perendicular to the plane and (3, 7, -1) is a point in the plane so for any point (x, y, z) in the plane, (x- 3, y- 7, z+ 1) is vector in the plane and we must have $(3\vec{i}+ 7\vec{j}- vec{i})\cdot ((x-3)\vec{i}+ (y- 7)\vec{j}+ (z+1)\vec{k})= 0$.

Myself, I would interpret the "vector equation" for a figure as the position vector of any point in it, $x\vec{j}+ y\vec{j}+ z\vec{k}$ with x, y, and z functions of the appropriate parameters. For a plane this is $\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$ where u and v are two parameters (because any surface is two dimensional) and x, y, and z are linear functions of u and v (because a plane is flat).
The simplest thing to do is solve for z= 65+ 5x+ 7y and use u= x and v= y as parameters:
$$\vec{r}(u,v)= x\vec{i}+ y\vec{j}+ z\vec{k}= u\vec{i}+ v\vec{j}+ (65+ 5u+ 7v)\vec{k}$$

You, of course, do it whichever way your teacher wants you to.

Last edited: Jul 11, 2012
4. Jul 11, 2012

### Plutonium88

Is this equation of the plane the same as writing the 'vector equation'

r= Po + ta +sb

Where Po is a point which the plane passes through, a and b are vectors. Do you know how ti is possible to obtain this form? (and from before, i thought a and b were direction vectors, and thats why it was called vector equation of a plane? I'm a bit confused now)

The other thing i wanted to ask, Can you tell me whether that scalar equation of the plane is correct or not that i currently have? Because from what you're telling me i have done everything correctly so far up to that point.

******The only other case i have dealt with similiar to this, is where you have to find the planes equation when the intersection is PARALELL. In that case you just used the direction vector you find, in the equation of the line, to solve for vector + scalar equations..*******

IE: lets say the line of intersection is r=(xo,yo,zo) + t(a,b,c)

If it was paralell to the other plane, the vector equation of the plane could be written as

R= Po + t(a,b,c) <---- and this is the vector eqn of the plane
where Po is the point the plane passes through.

but because it is perpindicular i don't understand how to go about it.

Last edited: Jul 11, 2012
5. Jul 11, 2012

### Plutonium88

Ax+By+Cz+D=0

-5x-7y+z+65=0

so using the interecepts i can find 3 different points.

A(13,0,0)
B(0,65/7,0)
C(0,0,-65)

AB=(-13,65/7,0)
AC(-13,0,-65)

r= A + tAB + sAC

R= (13,0,0) +t(-13,65/7,0) + s(-13,0,-65)

Is this a legit vector equation?

Last edited: Jul 11, 2012
6. Jul 11, 2012

### HallsofIvy

Staff Emeritus
Well, its an equations involving vectors and it is true for all points on the plane and only them. I'm not saying the dot product form is not valid. But neither of us can read the minds of the teacher!

If Po is the position vector of the point and a and b are independent vectors in the plane, yes, that's a valid vector equation describing the plane.

No, not if (a, b, c) is a vector parallel to the plane. In the form ax+ by+ cz+ d= 0 or, equivalently, a(x- x0)+ b(y- y0)+ c(z- z0)= 0, (a, b, c) is perpendicular to the plane.

Last edited: Jul 11, 2012
7. Jul 11, 2012

### Plutonium88

But because the direction vector is perpendicular to the plane, that means it is the normal vector in this case, is what i meant to say :(

( i was just using the parallel scenario, as a case to compare too)

And also, thank you muchly Halls Of Ivy, you are the greatest ∠3

(i also did what u are saying basically, i asked my teacher if the form of the vector equation using intercepts was acceptable, and he said yes, i just couldnt reach him to find out before...)