Question

How do I calculate the distance of a point to an intersection of two planes ?

Can you please explain it in detail and give an example ?

Answer 1

Given; two planes of the form:

`a_1x+b_1y+c_1z=k_1`
`a_2x+b_2y+c_2z = k_2`

and a point `(x_0,y_0,z_0)`

You want to find the vector, `vecv`, of the line of intersection of the two planes. To find this vector, compute the cross product of the normal vectors for the two planes:

`vecv = (a_1hati+b_1hatj+c_1hatk)xx(a_2hati+b_2hatj+c_2hatk)`

I am going to assume that you know how to compute the cross product of two 3 dimensional vectors but, if not, the please open the link in a separate tab and follow the instructions. Let's assume that the resulting vector is:

`vecv = a_3hati+b_3hatj+c_3hatk`

In addition to being the vector of the line of intersection, it is the normal vector for the plane that must contain the given point, `(x_0,y_0,z_0)` and the point on the line, `(x_1,y_1,z_1)`, that is orthogonal to the given point.

To write the equation of this plane, use the normal vector components:

`a_3x+b_3y+c_3z= k_3` where the value of `k_3` is unknown.

To find the value of `k_3`, substitute the given point `(x_0,y_0,z_0)`

`k_3" = a_3x_0+b_3y_0+c_3z_0`

You, now know all of the values for the equation of the third plane that contains both points `(x_0,y_0,z_0)` and `(x_1,y_1,z_1)`:

`a_3x+b_3y+c_3z= k_3`

When you solve the 3 equations of the 3 planes:

`a_1x+b_1y+c_1z=k_1`
`a_2x+b_2y+c_2z = k_2`
`a_3x+b_3y+c_3z= k_3`

by any method that you like, you will obtain the point where the line of intersection of the two planes intersects the third plane `(x_1, y_1,z_1)`

Use the distance equation to find the distance between the given point and the point on the line of intersection that is orthogonal to the given point:

`d = sqrt((x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2)`