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March 29, 2024, 01:22:24 pm

Author Topic: Proving that the dot product between the normal vector of a plane and..  (Read 1675 times)  Share 

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#1procrastinator

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How can we show that the dot product between the unit normal vector to a plane and a position vector to any point on the plane always equals the same value?

I'm looking for a different argument/proof than this

therefore


(n is the normal vector, r and r_0 are the position vectors)

I suspect that there's a very nice geometric way of showing it for all cases but I'm dumb as mule and twice as ugly so I can't figure it out

nisha

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Re: Proving that the dot product between the normal vector of a plane and..
« Reply #1 on: November 23, 2012, 01:07:29 pm »
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show that cos(angle)=0.......
aka, show that the angle = 90 degrees.
« Last Edit: November 23, 2012, 01:19:37 pm by nisha »
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Re: Proving that the dot product between the normal vector of a plane and..
« Reply #2 on: November 23, 2012, 03:59:31 pm »
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Since the case of a plane passing through the origin is trivial, consider a plane that does not pass through the origin.

Let P be any point on the plane.

If there were a line in the direction of the unit normal vector passing through the origin, it would intersect the plane at one point, and it would intersect it at right angles. Call this point A.

It follows that OA is perpendicular to AP, so OAP is a right-angled triangle with a right-angle at A.

The cosine of the angle θ at O is equal to OA/OP. Therefore OPcos(θ) = OA.

Now, the normal vector is either in the same direction as the vector OA, or in the opposite direction. If it is in the same direction, then the dot product between the normal vector and the position vector OP is equal to |OP|cos(θ), which is equal to |OA| as we saw. If it is in the opposite direction, then the dot product is |OP|cos(π-θ) = -|OP|cos(θ) = -|OA|. In either case, the value is always the same for the same normal vector.
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kamil9876

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Re: Proving that the dot product between the normal vector of a plane and..
« Reply #3 on: November 23, 2012, 07:12:14 pm »
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How about figuring out what actually calculates. It is actually the distance from the origin to the plane with a sign (positive if the normal vector points towards the plane from the origin, negative if the normal vector points away from the origin). Proof: Think it terms of vector resolutes.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

#1procrastinator

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Re: Proving that the dot product between the normal vector of a plane and..
« Reply #4 on: November 25, 2012, 03:29:36 am »
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@nisha: I saw your original post and I don't remember exactly what you wrote but I remember feeling like it was the most obvious thing that didn't occur to me lol

@ClimbTooHigh:  I think my attempt below is similar to yours...

@kamil: I may've misunderstood you, but isn't the distance from the origin to the plane given by |r|?


This is my attempt, please point out any flaws, incorrect assumptions, circular reasoning etc.

Let P be a point on the plane such the position vector OP is parallel to the unit normal vector to the plane. We wish to show that the vector OQ, where Q is any point on the plane, dotted with the unit normal vector n is equal to |OP|

OQ.n=|OQ|*|n|*cos(A)

Since |OQ|=|OP|/cos(A), then

OQ.n=(|OP/cos(A))*|n|*cos(A)
OQ.n=|OP|*|n|
OQ.n=|OP|

Also, is the shortest distance to the plane from the origin is always the position vector that's parallel to the normal vector?

kamil9876

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Re: Proving that the dot product between the normal vector of a plane and..
« Reply #5 on: November 25, 2012, 02:49:25 pm »
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Quote
@kamil: I may've misunderstood you, but isn't the distance from the origin to the plane given by |r|?

By the distance from the plane to the origin we mean the i.e the length of the shortest segment from the origin to a point on the plane. We know that this is given by the segment from the origin to the plane that is perpendicular to the plane, hence it is a scalar multiple of i.e for some real . Using vector resolutes you can see that is this shortest vector from the origin to the plane (i.e perpendicular to the plane). So can't depend on .

edit: I didn't see this question of yours initially but I see that it might be helpful for understanding by argument above:

Quote
Also, is the shortest distance to the plane from the origin is always the position vector that's parallel to the normal vector?

Yes, in other words the shortest distance is given by the vector perpendicular to the plane. Proof: Suppose is a vector on the plane, then we can write with parralel to the plane and kn the vector on the plane parallel to . Then and so
« Last Edit: November 25, 2012, 03:15:48 pm by kamil9876 »
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#1procrastinator

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Re: Proving that the dot product between the normal vector of a plane and..
« Reply #6 on: November 26, 2012, 03:28:25 am »
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Quote
@kamil: I may've misunderstood you, but isn't the distance from the origin to the plane given by |r|?

By the distance from the plane to the origin we mean the i.e the length of the shortest segment from the origin to a point on the plane. We know that this is given by the segment from the origin to the plane that is perpendicular to the plane, hence it is a scalar multiple of i.e for some real . Using vector resolutes you can see that is this shortest vector from the origin to the plane (i.e perpendicular to the plane). So can't depend on .

edit: I didn't see this question of yours initially but I see that it might be helpful for understanding by argument above:

Quote
Also, is the shortest distance to the plane from the origin is always the position vector that's parallel to the normal vector?

Yes, in other words the shortest distance is given by the vector perpendicular to the plane. Proof: Suppose is a vector on the plane, then we can write with parralel to the plane and kn the vector on the plane parallel to . Then and so

So does that basically say for r^2 to be the smallest value, v^2 has to be 0 which means r^2 is parallel to n (sorry, I need everything spelled out lol)

In my book, I scribbled something down about the projection of n onto r being maximum when r is parallel to n l and since n.r is always the same, as you move further away from that point, the projection gets smaller and so r has to always get larger as a result.

Wasn't happy with it though cause it didn't looked like I'd put together a 'proper' proof plus I was using a result I was trying to prove.