Distance between a point and plane

[SamitC]

Conventionally we have the following equation of a plane ax+by+cz=d where d = ax₀+by₀+cz₀. Where (x₀,y₀,z₀) is a known point on the plane.

Now if we try to find the distance between a point P and a plane we take any point on the plane Q (x,y,z) and find the vector from Q to P and project on the normal vector. Problem is... when we put the co-ordinates of point P to evaluate "d" in the equation of plane. In the equation of plane, (x₀,y₀,z₀) in "d" is of a point that is on the plane. But in this case P is outside the plane.
Am I missing out on something here?

For example please see: http://mathinsight.org/distance_point_plane_examples

 

[Mark44]

Conventionally we have the following equation of a plane ax+by+cz=d where d = ax₀+by₀+cz₀. Where (x₀,y₀,z₀) is a known point on the plane.

Now if we try to find the distance between a point P and a plane we take any point on the plane Q (x,y,z) and find the vector from Q to P and project on the normal vector. Problem is... when we put the co-ordinates of point P to evaluate "d" in the equation of plane. In the equation of plane, (x₀,y₀,z₀) in "d" is of a point that is on the plane. But in this case P is outside the plane.
Am I missing out on something here?

For example please see: http://mathinsight.org/distance_point_plane_examples

If P is a point that is not on the plane, you can't use it to find d, the constant in the plane equation.

 

[Buzz Bloom]

when we put the co-ordinates of point P to evaluate "d" in the equation of plane

Hi SamitC:

I don't understand what the above means. Why do you think P is used to "evaluate" d? Since the equation of the plane

ax+by+cz=d​

is known, a, b, c, and d must also be known. What do you mean by "evaluate", since "evaluate d" can't mean "solve for d". My guess is that the problem wants you to calculate the closest distance between P and plane. That means you need to find a point Q = (x,y,z) in the plane where the vector P-Q is normal to the plane. Then the length of P-Q is the desired distance.

Hope this helps.

Regards,
Buzz

 

[Mark44]

Hi Mark:
I don't understand what the above means. Why do you think P is used to "evaluate" d? Since the equation of the plane

ax+by+cz=d​

is known, a, b, c, and d must also be known. What do you mean by "evaluate", since "evaluate d" can't mean "solve for d". My guess is that the problem wants you to calculate the closest distance between P and plane. That means you need to find a point Q = (x,y,z) in the plane where the vector P-Q is normal to the plane. Then the length of P-Q is the desired distance.

I'm not the OP. Please direct your comments to SamitC, not me.

 

[Buzz Bloom]

Hi Mark:

I apologize for being careless in writing the addressee's name. Thanks for pointing my error out to me.

Regards,
Buzz

 

[Ssnow]

@SamitC in order to understand well your question if it possible to clarify what are the data of the problem that you have, because if you want to find the distance between a point and the plane, the essential data can be the point $P$ and the equation of the plane $ax+by+cz+d=0$

 

[SamitC]

Hi SamitC:

I don't understand what the above means. Why do you think P is used to "evaluate" d? Since the equation of the plane

ax+by+cz=d​

is known, a, b, c, and d must also be known. What do you mean by "evaluate", since "evaluate d" can't mean "solve for d". My guess is that the problem wants you to calculate the closest distance between P and plane. That means you need to find a point Q = (x,y,z) in the plane where the vector P-Q is normal to the plane. Then the length of P-Q is the desired distance.

Hope this helps.

Regards,
Buzz

Hi Buzz,
Thanks for your reply. Let me try to resolve it as per your direction.
Suppose the point P is (4, - 4, 3). Plane is 2x-2y+5x+8=0. Point Q = (a,b,c).
Vector QP = <4 - a, - 4 - b, 3 - c>.
Now this is a scalar multiple of the normal vector n = <2, - 2, 5> that we can get from the equation of the plane. We have so many unknowns. Still I do not know how to calculate the distance?

I am not able to use the "8" in the equation of the plane anywhere. Can you pls. help me with this confusion.
Regards,

 

[SamitC]

@SamitC in order to understand well your question if it possible to clarify what are the data of the problem that you have, because if you want to find the distance between a point and the plane, the essential data can be the point $P$ and the equation of the plane $ax+by+cz+d=0$

Hi Ssnow,
Thanks for your reply. Let us take the example: I have a plane 2x-2y+5x+8=0 and point (4,-4,3). We need to find the distance.
I have worked it out to whatever extent i could... I have uploaded in picture format. I know I am missing something but can't figure out. I am not using the "8" in the equation of the plane. That holds vital information. But I am not sure how to make use of it.

Can you help me find the distance in this case?
Regards,

 

[SamitC]

If P is a point that is not on the plane, you can't use it to find d, the constant in the plane equation.

 

[SamitC]

I'm not the OP. Please direct your comments to SamitC, not me.

 

[LCKurtz]

Hi Ssnow,
Thanks for your reply. Let us take the example: I have a plane 2x-2y+5z+8=0 and point (4,-4,3). We need to find the distance.
I have worked it out to whatever extent i could... I have uploaded in picture format. I know I am missing something but can't figure out. I am not using the "8" in the equation of the plane. That holds vital information. But I am not sure how to make use of it.

Can you help me find the distance in this case?
Regards,

I corrected your typo. Let's call the point $P(4,-4,3)$. This is just some point, not having anything to do with the plane. You want to know how far from the plane it is. In your example, you now look for a point on the plane. This is always real easy. You can see $Q(0,4,0)$ works by inspection and the 8 is used at that step. Now, as your OP example suggests, you just calculate $\vec V = P-Q$ to get a vector across from the plane to the point $P$. So in this example you have $\vec V = \langle 4,-8,3\rangle$. You already have a normal vector $\vec N = \langle 2,-2,5\rangle$ which you can make a unit vector $\hat N$ by dividing by its length $\sqrt {33}$. Then the distance is just the magnitude of the component of $\vec V$ in the direction of $\hat N$:$$
d = \left|\vec V \cdot \hat N\right| $$Using the absolute values means you don't have to worry about which side of the plane the point $P$ is on or which side the normal $\vec N$ points.

To convince your self that this all works, do the problem again with a different point chosen for $Q$ and compare.

 

[Ssnow]

Another procedure in order to find the distance is the following (less or more is the same of the previous). You can start to write the line in parametric form from $P$ that is perpendicular to the plane, this is $r= (4+2t,-4-2t,3+5t)$. You intersect this with the plane and you will find the point in the intersection $Q$. After you can obtain the distance using the distance formula between two points $P$ and $Q$.

 

[SamitC]

Conventionally we have the following equation of a plane ax+by+cz=d where d = ax₀+by₀+cz₀. Where (x₀,y₀,z₀) is a known point on the plane.

Now if we try to find the distance between a point P and a plane we take any point on the plane Q (x,y,z) and find the vector from Q to P and project on the normal vector. Problem is... when we put the co-ordinates of point P to evaluate "d" in the equation of plane. In the equation of plane, (x₀,y₀,z₀) in "d" is of a point that is on the plane. But in this case P is outside the plane.
Am I missing out on something here?

For example please see: http://mathinsight.org/distance_point_plane_examples

Now I am clear with this. Not sure why I did not understand this earlier.
In the above equation of the plane we have information about a specific point on the plane which is d. We need to utilize this information by using this point in calculating the distance. Earlier I was taking Q as any point. Instead, Q should be this specific point that is in "d".

Thank you all for your help.
Regards,
Samit

 

[SamitC]

I corrected your typo. Let's call the point $P(4,-4,3)$. This is just some point, not having anything to do with the plane. You want to know how far from the plane it is. In your example, you now look for a point on the plane. This is always real easy. You can see $Q(0,4,0)$ works by inspection and the 8 is used at that step. Now, as your OP example suggests, you just calculate $\vec V = P-Q$ to get a vector across from the plane to the point $P$. So in this example you have $\vec V = \langle 4,-8,3\rangle$. You already have a normal vector $\vec N = \langle 2,-2,5\rangle$ which you can make a unit vector $\hat N$ by dividing by its length $\sqrt {33}$. Then the distance is just the magnitude of the component of $\vec V$ in the direction of $\hat N$:$$
d = \left|\vec V \cdot \hat N\right| $$Using the absolute values means you don't have to worry about which side of the plane the point $P$ is on or which side the normal $\vec N$ points.

To convince your self that this all works, do the problem again with a different point chosen for $Q$ and compare.

Thank you very much. Once I linked Q to "d" I am clear now.
Regards

 

[HallsofIvy]

A crucial point in this and in similar problems is that "the distance from set A to set B" is defined as the shortest of all distances from a point in set A to a point in set B. In the case of a point and a line, it is easy to show that this is along the line through the given point perpendicular to the given plane. Here, in your example, the plane is given by 2x-2y+5x+8=0. A vector perpendicular to that plane is <2, -2, 5> and a line in that direction, through the point (4,-4,3) is x= 4+ 2t, y= -4- 2t, z= 3+ 5t. Determine where that line crosses the plane, by setting those formulas for x, y, z to get 4(4+ 2t)- 2(-4- 2t)+ 5(3+ 5t)= 0. Solve that for t, use that to determine the point and find the distance between that point and (4, -4, 3).

 

[LCKurtz]

Now I am clear with this. Not sure why I did not understand this earlier.
In the above equation of the plane we have information about a specific point on the plane which is d. We need to utilize this information by using this point in calculating the distance. Earlier I was taking Q as any point. Instead, Q should be this specific point that is in "d".

Thank you all for your help.
Regards,
Samit

I'm afraid that doesn't make any sense at all. "d" is not a specific point in the plane and it doesn't make sense to say "Q should be this specific point that is in "d"." When you have the equation of a plane given, such as $2x-2y+5z=-8$, it describes a plane in the sense that the set of points $(a,b,c)$ that satisfy the equation form a plane. To find a point in the plane, you just have to pick three numbers $a,b,c$ which satisfy the equation of the plane.

 

[Buzz Bloom]

Where (x0,y0,z0) is a known point on the plane.

Hi SamitC:

What is the known point (x0,y0,z0)?

ADDED

You said:

In the above equation of the plane we have information about a specific point on the plane which is d.​

d is NOT a point in the plane. (x0,y0,z0) is a point in the plain. d is the DOT product of (x0,y0,z0) and (a,b,c).

Do you know about DOT products?

Regards,
Buzz

 

[Buzz Bloom]

You already have a normal vector ⃗N=⟨2,−2,5⟩

Hi LCKurt:

I disagree that N=⟨2,−2,5⟩ is a vector normal to the plane. N is the vector from the origin (0,0,0) point and the N point (2,-2,5).
The point in the plane closest to N is Q= (x₀, y₀,z₀).

Regards,
Buzz

 

[LCKurtz]

Hi LCKurt:

I disagree that N=⟨2,−2,5⟩ is a vector normal to the plane. N is the vector from the origin (0,0,0) point and the N point (2,-2,5).
The point in the plane closest to N is Q= (x₀, y₀,z₀).

Regards,
Buzz

We were talking about the plane $2x-2y+5z=-8$. $\vec N =\langle 2,-2,5\rangle$ most certainly is perpendicular to that plane. You can also think of it as a position vector from the origin to the point $(2,-2,5)$ if you want to, but that point is irrelevant to the problem.

 

[Buzz Bloom]

Hi LCKurtz:

I apologize for my senior moment stupidity. Of course N is normal to the plane. I think my aberration was that for some reason I had it in my mind that the origin (0,0,0) was also in the plane.

Regards,
Buzz

 

[LCKurtz]

No need to apologize. It happens to everyone, not just us seniors. I have had my share of woops on this forum too.