# Vector given norm?

1. May 27, 2013

### Jbreezy

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

So this is part of a couple of questions.
Find the exact length of p, of OP, by considering a dot product with OP.(Hint OP will be orthogonal to the plane.) Hence find the position vector of P ( P is is the point on the plane closest to the origin F.Y.I)

2. Relevant equations

3. The attempt at a solution

So, I'm confused because what I did was find a vector orthogonal to the plane by using the cross product. I turned this into a unit vector. Then I did a dot product with one of the position vectors. The question describes a triangle in 3 space using position vectors another F.Y.I.
So now I have a norm correct? I have the orthogonal projection of position vector a onto my unit vector that is orthogonal to my plane.
So when the question says "Hence find the position vector of P" I'm lost because I do not know how to do this from where the problem has directed me. I feel like I can't find a vector from knowing the length. Thanks,
J

2. May 27, 2013

### Jbreezy

I just wondered does a unit vector of the point P suffice as a position vector? If that is the case then I would not have a problem.

3. May 27, 2013

### Jbreezy

OK so now I'm thinking that but I'm unsure. To get the position vector of P can I do p = (norm of p) times (the unit vector I was talking about above?)

This will give me p the position vector?

4. May 28, 2013

### Skins

P is the point on what plane ?

5. May 28, 2013

### Jbreezy

The plane of points A,B,C given by position vectors a,b,c.
Make more sense? Let me know.

The plane given by....
So, I'm confused because what I did was find a vector orthogonal to the plane by using the cross product.

Last edited: May 28, 2013
6. May 28, 2013

### HallsofIvy

You've written four posts but it is hard to understand what you are asking. You have three vectors, a, b, and c, that are the position vectors of four points, A, B, and C, right?

But your first post had "Find the exact length of p, of OP, by considering a dot product with OP.(Hint OP will be orthogonal to the plane.)" What happened to P? OP will be orthogonal to what plane? Was there a plane given in the problem that you didn't mention? And "a dot product with OP" will require another vector to take the dot product with.

Perhaps if you post the exact statement of the entire problem we can make more sense of it.

7. May 28, 2013

### Jbreezy

Points,A,B,C have position vectors <-1,3,2>,<4,-2,1> and <2,5,2> .P is the point on the plane containing , which is closest to the origin.
(a) Find a vector that is orthogonal to the plane containing the three points and find.
So, I just did a cross product so I got a vector I'm pretty sure this was correct

(b) Find the exact length,p , of OP by considering a dot product with OP . (Hint: will be orthogonal to the plane.) Hence find the position vector of P.
So I took the vector what I got from part a and made a unit vector and then I did a dot product with one of the position vectors. Now I have the magnitude of the position vector p.
My question is just a the unit vector I have found suffice as the position vector of P? I know it is but They say find the position vector of P. I have the unit vector of P. How would I get the position vector of p? Or can I just use the unit vector and say that is the position vector of P?

(c) Consider the position vector, , of a general point, on the plane with coordinates . Find first the vector equation involving and, using that, then find the Cartesian linear equation involving representing all points on the plane

For this one I will use the position vector p which is orthogonal to the plane and I will have something of the sorts of x= p +λ( b -a ) or whatever vector in the plane I want. Right?
Now coming back to question b. Can I use the unit vector of p in this equation?

Hope this makes sense now. Sorry for the fragmented questions before.

8. May 28, 2013

### HallsofIvy

There is something missing here. P is a point on the plane containing what?

Again, there is something missing. Find what?

9. May 28, 2013

### Jbreezy

Containing points A,B,C.
find v(hat).

I think because I copy and paste from pdf the eq and stuff didn't.
Good now? sorry?

10. May 28, 2013

### LCKurtz

You know the position vector to P is supposed to be p units long, which you have calculated. And you have a unit vector, which I will call $\hat P$ in the direction of OP. Then the position vector of P is $p\hat P$. You have to multiply the unit vector by the right length to make the position vector to the point P.

11. May 28, 2013

### Jbreezy

This vector is just the cross product of two vectors on the plane. This is the vector I turned into the unit vector and did the dot product with a position vector to find the length of p
So when you say multiply p times p hat how does that help?

Wait wait. OK I see. Then the vector eq I do x = p + λ( some vector on the plane say b-a).

So if I want Cartesian form I of the plane I can do <x,y,z> dot <p1, p2,p3> = <x1, y1, z1> dot <p1, p2,p3>
Where x1,y1, z1 is the points on the plane given from a vector on the plane say (b-a) = x1,y1,z1>
Is that proper? Thanks

Last edited: May 28, 2013
12. May 28, 2013

### LCKurtz

Why don't you show us your calculations? We can't tell what you really did. Did you cross two of the vectors A, B, C or something else.

I didn't say multply p by p. Read it again. And now you are using p like it was a vector instead of a length.

It is impossible to figure out what you are doing. When you reply to this post, start with what you are given and show us the calculations you are doing with your given vectors. Give us a string of equations ending up with your answer.

13. May 28, 2013

### Jbreezy

ok give me a sec

14. May 28, 2013

### Jbreezy

OK, Proper.

Points,A,B,C have position vectors<-1,3,2>,<4,-2,1> and <2,5,2>. P is the point on the plane containing A,B,C, which is closest to the origin.
(a) Find a vector v that is orthogonal to the plane containing the three points and find v(hat).
(b) Find the exact length,p , OP of by considering a dot product with OP . (Hint:OP will be orthogonal to the plane.) Hence find the position vector of p.
(c) Consider the position vector,x , of a general point, X on the plane with coordinates x.y.z. Find first the vector equation involving x and, using that, then find the Cartesian linear equation involving x,y,z representing all points on the plane.

Now,
a.)
To get v
(c -a) = <2,5,2> - < -1,3,2> = < 3,2,0>
(b-a) = <4,-2,1> - < -1,3,2> = < 5,-5,-1>
Now, (c-a) cross (b -a) = < -2, 3, -25>
To get v(hat) = v/|v|
So, |v| = √((-2)^2 + (3)^2 + (-25)^2) = √638

so v(hat) = <1/√(638) <-2,3,-25>

b.) To get the length p.
We have a (dot) v(hat) = < -1, 3, 2> dot <1/√(638) <-2,3,-25> = |-39/√638| = 39/√638 ( I take abs because it is a length right ?)
To get the position vector of the point P
pv(hat) = p
So , (39/√638 )(<1/√(638) <-2,3,-25>) = <-78/638, 117/638, -975/ 638>

c.) For the vector equation
x = p + λ(b-a) ....b-a = < 5,-5,3> (I did b-a because it lies in the plane this is right?)
So , x = <-78/638, 117/638, -975/ 638> + λ< 5,-5,3>

To get Cartesian form
<x,y,z> dot < p1,p2,p3> = (b-a) dot <p1,p2,p3>
<x,y,z> dot <-78/638, 117/638, -975/ 638> = < 5,-5,3> dot <-78/638, 117/638, -975/ 638>
I got

(-78/638)x + (117/638)y - (975/638)z = -1950/319
So If I multiply the entire thing by -1 and by 638 to clear the fractions I get something of the form

78x - 117y + 975x = 3900
Is this correct way to do this problem. Thanks

15. May 28, 2013

### LCKurtz

Much better post!

Looks good through part b. For the vector equation you want $\vec x - A$, which is in the plane, to be perpendicular to the normal:$$(\langle x,y,z\rangle - \langle -1,3,2\rangle) \cdot \vec p = 0$$
Once you simplify the correct vector equation you should have it.

16. May 28, 2013

### Jbreezy

I thought that the vector b -a was in the plane so what is the issue? Since it is in the plane and and the normal is perpendicular it should be OK right? Why is it wrong?
So part c was proper how I went about it accept we have a disagreement about the vector equation. But my steps were OK in c right ?
Thanks.

17. May 28, 2013

### Jbreezy

I check min with dot product and it doesn't check How can this be//? If that vector is in the plane how can it not be orthogonal to p? can you explain?

18. May 28, 2013

### LCKurtz

The word is "except". No, your steps may be correct but they are not the right calculation. To get $\vec x = \langle x,y,z\rangle$ in the plane you need $\vec x - A$ perpendicular to the normal. You can check what you get and what you get using my equation. Just plug your three points into your final answer for the plane and see if they work, then try the other way.

19. May 28, 2013

### Jbreezy

OK, I understand your equation but I'm having a hard time understanding why mine is wrong.

I thought that the points A,B,C lie in a plane. So if I use one of those I described by their position vectors I will have a vector in the plane and orthogonal to p so I could use (b-a) or whatever. I did a dot product of (b-a) with p and it is not orthogonal. So how come?
Also in this form x = p + λ( what vector is in the plane here? )

I don't want to use x because it is a general point what vector should I use because like I said the ones I thought were in the plane are not.

Thanks

20. May 28, 2013

### Jbreezy

I did it your way I got

78x - 117y + 975x = 1521

So clearly my problem is that I have trouble with finding the vector that is in the plane to use. Can someone help me with my understanding of this. Please.

So this is not right.
x = p + λ(b-a)

but what do I use in this case and why is it not in the plane?