Finding the distance between a point and a plane using dot products

[haackeDc]

I know how I would be able to do this using projection, but am not so sure with dot products.

Do I dot the normal vector with an imaginary point and then figure something out from there?

If the normal is a= <a1,a2,a3>
and the random point is (p1,p2,p3)

If I dot them, I would get a1p1 + a2p2 + a3p3 = 0

And then solve for this point?

I don't see how to do it from here, could anyone help me?

 

[conquest]

(0, 0, 0) solves this!

I think though that you want to find the distance between any given point and any given plane. I think it should work if you shift the plane into the origin ( by setting the euqaiton for the plane to zero). Then shift your point accordingly. then you dot the vector going to the point into the unit normal vector to the plane.

 

[HallsofIvy]

If the plane is Ax+ By+ Cz= D and the point is $(x_0, y_0, z_0)$, you know that <A, B, C> is a normal vector to the plane and so $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$ is a line that passes through $(x_0, y_0, z_0)$ and is perpendicular to the plane. Replace x, y, and z in the equation of the plane with those to solve for the point where the normal line passes through the plane and find the distance from $(x_0, y_0, z_0)$ to that point.

 

[haackeDc]

If the plane is Ax+ By+ Cz= D and the point is $(x_0, y_0, z_0)$, you know that <A, B, C> is a normal vector to the plane and so $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$ is a line that passes through $(x_0, y_0, z_0)$ and is perpendicular to the plane. Replace x, y, and z in the equation of the plane with those to solve for the point where the normal line passes through the plane and find the distance from $(x_0, y_0, z_0)$ to that point.

Question resolved. Thanks for your help everyone.

 

[blue_raver22]

Hi there,

Thank you for your question. Finding the distance between a point and a plane using dot products is a common problem in mathematics and physics. The process involves using the properties of dot products and vectors to calculate the distance between the point and the plane.

First, let's review the properties of dot products. The dot product of two vectors, a and b, is defined as a · b = |a| |b| cosθ, where |a| and |b| are the magnitudes of the vectors and θ is the angle between them. This means that the dot product of two perpendicular vectors is 0, since cos90° = 0.

Now, let's consider a point P with coordinates (p1, p2, p3) and a plane with a normal vector a = <a1, a2, a3>. The distance between the point and the plane can be calculated as the length of the perpendicular line from the point to the plane. This perpendicular line is also known as the shortest distance between the point and the plane.

To find this distance, we can use the dot product. We know that the dot product of two perpendicular vectors is 0, so we can use this property to find the distance. Let's call the distance d. We can represent the vector from the point P to any point on the plane as r = <p1, p2, p3> - <x, y, z>, where <x, y, z> is any point on the plane. This vector is perpendicular to the normal vector a, so the dot product of r and a must be 0. This can be written as:

r · a = 0

Substituting the values of r and a, we get:

(p1 - x)a1 + (p2 - y)a2 + (p3 - z)a3 = 0

Solving for x, y, and z, we get:

x = p1 - d(a1/|a|)

y = p2 - d(a2/|a|)

z = p3 - d(a3/|a|)

where d is the distance between the point and the plane. We can use the Pythagorean theorem to find the value of d:

d^2 = (p1 - x)^2 + (p2 - y)^2 + (p3 - z)^2

Substituting the