# Distance from a Point to a Plane

The distance *d* from a point *P*_{1}(*x*_{1}, *y*_{1}, *z*_{1}) to a plane with equation

is determined by the formula

Here:

*A*,*B*, and*C*are the coefficients of the normal vector to the plane.- (
*x*_{1},*y*_{1},*z*_{1}) are the coordinates of the point*P*_{1}. *D*is the constant term in the equation of the plane.

The numerator |*Ax*_{1} + *By*_{1} + *Cz*_{1} + *D*| represents the absolute value of the dot product between the normal vector of the plane and the vector formed by the coordinates of the point. The denominator \({\sqrt{A^2 + B^2 + C^2}}\) is the magnitude of the normal vector.

This formula essentially calculates the perpendicular distance from the point to the plane.

## Solved Problems

### Example 1.

Find the distance from point \(P\left({1,2,3}\right)\) to the plane that cuts off intercepts \(a = 1,\) \(b = 2,\) \(c = 3\) on the coordinate axes.

Solution.

The plane equation in intercept form is given by

Hence,

Let's convert this equation into general form:

Now we can easily find the distance from point *P* to the given plane:

### Example 2.

Find the equations of planes parallel to the plane \({2x + 2y - z + 3 = 0}\) such that the distance from point \(M\left({1,2,-1}\right)\) to the planes is \(d = 4.\)

Solution.

The equations of parallel planes differ only in the free term *D*. That is, these equations have the form

The distance from point \(M\left({x_1,y_1,z_1}\right)\) to the plane is determined by the formula

Let's substitute the known values *A*, *B*, *C*, and *x _{1}*,

*y*,

_{1}*z*:

_{1}So we got a simple equation

from which we can find the coefficient *D*. Two solutions are possible:

- \(7 + D = 3d,\) \(\Rightarrow D = 3d - 7 = 3\cdot4 - 7 = 5.\)
- \(-\left({7 + D}\right) = 3d,\) \(\Rightarrow D = -3d - 7 = -3\cdot4 - 7 = -19.\)

Therefore, the equations of the planes have the form

### Example 3.

The vertices of a tetrahedron have the coordinates \(A\left({3,0,0}\right),\) \(B\left({0,2,0}\right),\) \(C\left({0,0,5}\right),\) \(O\left({0,0,0}\right).\) Find the length of the height dropped from the origin to the face \(ABC.\)

Solution.

The length of the height *d* is equal to the distance from point *O* to the plane *p* passing through three points *A*, *B*, *C*.

Let's first write down the equation of the plane *p* through 3 points: *A*(*x*_{1}, *y*_{1}, *z*_{1}), *B*(*x*_{2}, *y*_{2}, *z*_{2}), *C*(*x*_{3}, *y*_{3}, *z*_{3}). It is written using the determinant as

Substitute the coordinates of points *A*, *B*, *C*:

Simplify:

Expand the determinant along the second column:

or

Now let's find the distance *d* from point *O*(*x*_{0}, *y*_{0}, *z*_{0}) = *O*(0, 0, 0) to plane *p* using the formula

After substitution we get

### Example 4.

Find the equation of a plane passing through the \(y-\)axis and equidistant from points \(P\left({2,7,3}\right)\) and \(Q\left({-1,1,0}\right).\)

Solution.

If a plane passes through the \(y-\)axis then its equation has the form

The distance from point \(P\left({2,7,3}\right)\) to the given plane is expressed by the formula

Similarly, we express the distance from point \(Q\left({-1,1,0}\right)\) to the plane. It is written as

According to the problem, \(d_P = d_Q.\) Hence

The following solutions are possible here

- \(2A + 3C = -A,\) \(\Rightarrow 3A + 3C = 0,\) \(\Rightarrow A = - C.\) Let \(C = 1,\) then \(A = -1.\)
- \(2A + 3C = -\left({-A}\right) = A,\) \(\Rightarrow A + 3C = 0,\) \(\Rightarrow A = - 3C.\) Let \(C = 1,\) then \(A = -3.\)

We get two equations of the plane:

### Example 5.

Find a point on the \(x-\)axis equidistant from point \(M\left({9,-2,2}\right)\) and from the plane *P* given by the equation \[3x - 6y + 2z - 3 = 0.\]

Solution.

Let our point on the \(x-\)axis have coordinates \(N\left({x_N,0,0}\right).\) The distance between two points *M* and *N* is

On the other hand, the distance from point *N* to plane *P* is expressed by the formula

By condition, \(d_{MN} = d_{NP}.\) Then we get an equation from which we can determine \(x_N:\)

Let's square both sides of the equation:

Simplify:

Divide both sides by 8:

Calculate the discriminant of the quadratic equation:

The roots of the quadratic equation are

We found that two points satisfy the given condition: