Find Length of P in Vector Norm Given Plane

[Jbreezy]

Homework Statement

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)

Homework Equations

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

 

[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.

 

[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?

 

[Skins]

P is the point on what plane ?

 

[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.

 

[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.

 

[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.

 

[HallsofIvy]

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

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

, which is closest to the origin.
(a) Find a vector that is orthogonal to the plane containing the three points and find.

Again, there is something missing. Find what?

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.

 

[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?

 

[LCKurtz]

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?

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.

 

[Jbreezy]

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 P^ in the direction of OP. Then the position vector of P is pP^. You have to multiply the unit vector by the right length to make the position vector to the point P.

you know the position vector to P is supposed to be p units long, which you have calculated.

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

 

[LCKurtz]

This vector is just the cross product of two vectors on the plane.

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.

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).

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.

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

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.

 

[Jbreezy]

ok give me a sec

 

[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

 

[LCKurtz]

OK, Proper.

Much better post!

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>

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:$$
(\langle x,y,z\rangle - \langle -1,3,2\rangle) \cdot \vec p = 0$$

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

Once you simplify the correct vector equation you should have it.

 

[Jbreezy]

Looks good through part b. For the vector equation you want x⃗ −A, which is in the plane, to be perpendicular to the normal:
(⟨x,y,z⟩−⟨−1,3,2⟩)⋅p⃗ =0

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?

Once you simplify the correct vector equation you should have it.

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.

 

[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?

 

[LCKurtz]

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.

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.

 

[Jbreezy]

Looks good through part b. For the vector equation you want x⃗ −A, which is in the plane, to be perpendicular to the normal:

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

 

[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?

 

[haruspex]

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?

The problem with your (b-a) method is that it only added one degree of freedom, so gave a line, not a plane. You needed to consider e.g. c-a as well.

 

[Jbreezy]

Can you explain more? I thought that this was in the plane. I know it gave me a line. I thought that a vector equation of a line was when in 3 space and converted to Cartesian form was the equation of a plane because there are infinitely many points which satisfy the line.


Thanks

 

[Jbreezy]

What has to go here ? And how can I be sure it is in the plane?
x = p + λ(b-a)

I considered your b-a and c-a I tried to add them didn't work. I tried to subtract them didn't work. I'm lost.

 

[Skins]

If you have 3 points in soace you can define a plane containing those 3 points. So, if you have 3 position vectors , say.
$A = <a_{1}, a_{2}, a_{3}>$ and $B = <b_{1},b_{2},b_{3}>$ and $C = <c_{1},c_{2},c_{3}>$

Then the vectors B-A and C-A lie in the plane and are parallel to the plane and their cross product will be normal (perpendicular to the plane) i.e. N = (B-A) x (C-A) where N is the normal vector.

Now, if you take some arbitrary vector in the plane, say $X = <x,y,z>$ then the vector (X-A) is also in the plane. Now you have a normal vector and an arbitrary vector in the plane thus the ddot produce of (X-A) and N has to be zero.

$$N.(X-A) = 0$$

and this will give you the equation of the plane,

 

[LCKurtz]

I think one thing that is confusing jbreezy is the distinction between a point $(x,y,z)$ in the plane and the position vector $\vec X = \langle x,y,z\rangle$ to that point. A position vector is represented with its tail at the origin. Because the tail is at the origin, the coordinates of the point and the components of the position vector are the same numbers, but they aren't the same thing. The position vector is not "in" nor parallel to the plane. If $\vec A$ is the position vector to a point $(a_1,a_2,a_3)$ in the plane, then $\vec X - \vec A$ is a vector in (parallel to) the plane. That's what you have to dot into the normal and set to zero to be perpendicular.

 

[LCKurtz]

If you have 3 points in soace you can define a plane containing those 3 points. So, if you have 3 position vectors , say.
$A = <a_{1}, a_{2}, a_{3}>$ and $B = <b_{1},b_{2},b_{3}>$ and $C = <c_{1},c_{2},c_{3}>$

Then the vectors B-A and C-A lie in the plane and are parallel to the plane and their cross product will be normal (perpendicular to the plane) i.e. N = (B-A) x (C-A) where N is the normal vector.

Now, if you take some arbitrary vector in the plane, say $X = <x,y,z>$ then the vector (X-A) is also in the plane.

Be careful here. If $(x,y,z)$ is in the plane, the vector $\langle x,y,z\rangle$ generally isn't. The OP is already confused enough about that

 

[Skins]

Be careful here. If $(x,y,z)$ is in the plane, the vector $\langle x,y,z\rangle$ generally isn't. The OP is already confused enough about that

My apologies. I should have made clear that specified X = <x,y,z> is a position vector from the origin to a arbitrary point (x,y,z) in the plane. Vector X is not in the plane but the vector difference X-A is a vector in the plane and that vector dotted with the normal vector N to the plane = 0 , i.e. $N.(X-A) = 0$ which will give an equation of the plane in terms of x,y,z.

Sorry if I didn't make that point sharp enough in my original message.

 

[LCKurtz]

My apologies. I should have made clear that specified X = <x,y,z> is a position vector from the origin to a arbitrary point (x,y,z) in the plane.

No need to apologize. We all occasionally mis-state what we mean. But I'm not so sure about how well the OP understands it.

 

[haruspex]

I considered your b-a and c-a I tried to add them didn't work. I tried to subtract them didn't work. I'm lost.

Pls post your working for that.

 

[Jbreezy]

I got it I got it. Thanks thanks for the responses.:)