# Distance from a Point to a Plane

The distance d from a point P1(x1, y1, z1) to a plane with equation

$Ax + By + Cz + D = 0$

is determined by the formula

$d = \frac{\left|{Ax_1 + By_1 + Cz_1 + D}\right|}{\sqrt{A^2 + B^2 + C^2}}$

Here:

• A, B, and C are the coefficients of the normal vector to the plane.
• (x1, y1, z1) are the coordinates of the point P1.
• D is the constant term in the equation of the plane.

The numerator |Ax1 + By1 + Cz1 + 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

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1.$

Hence,

$\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1.$

Let's convert this equation into general form:

$6x + 3y + 2z - 6 = 0.$

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

$d = \frac{\left|{Ax_P + By_P + Cz_P + D}\right|}{\sqrt{A^2 + B^2 + C^2}} = \frac{\left|{6\cdot 1 + 3\cdot 2 + 2\cdot 3 - 6}\right|}{\sqrt{6^2 + 3^2 + 2^2}} = \frac{\left|{6 + 6 + \cancel{6} - \cancel{6}}\right|}{\sqrt{36 + 9 + 4}} = \frac{12}{7}.$

### 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

$2x + 2y - z + D = 0.$

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

$d = \frac{\left|{Ax_1 + By_1 + Cz_1 + D}\right|}{\sqrt{A^2 + B^2 + C^2}}.$

Let's substitute the known values A, B, C, and x1, y1, z1:

$d = \frac{\left|{2\cdot1 + 2\cdot2 - 1\cdot\left(-1\right) + D}\right|}{\sqrt{2^2 + 2^2 + \left(-1\right)^2}} = \frac{\left|{2 + 4 + 1 + D}\right|}{\sqrt{4 + 4 + 1}} = \frac{\left|{7 + D}\right|}{3}.$

So we got a simple equation

$\left|{7 + D}\right| = 3d,$

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

1. $$7 + D = 3d,$$ $$\Rightarrow D = 3d - 7 = 3\cdot4 - 7 = 5.$$
2. $$-\left({7 + D}\right) = 3d,$$ $$\Rightarrow D = -3d - 7 = -3\cdot4 - 7 = -19.$$

Therefore, the equations of the planes have the form

$2x + 2y - z + 5 = 0, \;2x + 2y - z - 19 = 0.$

### 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(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3). It is written using the determinant as

$\left| {\begin{array}{*{20}{l}} {{x - x_1}}&{{y - y_1}}&{{z - z_1}}\\ {{x_2 - x_1}}&{{y_2 - y_1}}&{{z_2 - z_1}}\\ {{x_3 - x_1}}&{{y_3 - y_1}}&{{z_3 - z_1}} \end{array}} \right| = 0.$

Substitute the coordinates of points A, B, C:

$\left| {\begin{array}{*{20}{c}} {{x - 3}}&{{y - 0}}&{{z - 0}}\\ {{0 - 3}}&{{2 - 0}}&{{0 - 0}}\\ {{0 - 3}}&{{0 - 0}}&{{5 - 0}} \end{array}} \right| = 0.$

Simplify:

$\left| {\begin{array}{*{20}{c}} {{x - 3}}&{{y}}&{{z}}\\ {{- 3}}&{{2}}&{{0}}\\ {{- 3}}&{{0}}&{{5}} \end{array}} \right| = 0.$

Expand the determinant along the second column:

$-y \left| {\begin{array}{*{20}{r}} {{3}}&{{0}}\\ {{-3}}&{{5}} \end{array}} \right| + 2\left| {\begin{array}{*{20}{c}} {{x-3}}&{{z}}\\ {{-3}}&{{5}} \end{array}} \right|= 0,$
$-y\left({-15-0}\right) + 2\left[{5\left({x-3}\right) + 3z}\right] = 0,$
$15y + 10x - 30 + 6z = 0,$

or

$10x + 15y + 6z - 30 = 0.$

Now let's find the distance d from point O(x0, y0, z0) = O(0, 0, 0) to plane p using the formula

$d = \frac{\left|{Ax_0 + By_0 + Cz_0 + D}\right|}{\sqrt{A^2 + B^2 + C^2}}.$

After substitution we get

$d = \frac{\left|{10\cdot0 + 15\cdot0 + 6\cdot0 - 30}\right|}{\sqrt{10^2 + 15^2 + 6^2}} = \frac{30}{\sqrt{100+225+36}} = \frac{30}{\sqrt{361}} = \frac{30}{19}.$

### 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

$Ax + Cz = 0.$

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

$d_P = \frac{\left|{A\cdot2 + B\cdot7 + C\cdot3 + D}\right|}{\sqrt{A^2 + B^2 + C^2}} = \frac{\left|{2A + 3C}\right|}{\sqrt{A^2 + C^2}}.$

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

$d_Q = \frac{\left|{A\cdot\left({-1}\right) + B\cdot1 + C\cdot0 + D}\right|}{\sqrt{A^2 + B^2 + C^2}} = \frac{\left|{-A}\right|}{\sqrt{A^2 + C^2}}.$

According to the problem, $$d_P = d_Q.$$ Hence

$\frac{\left|{2A + 3C}\right|}{\sqrt{A^2 + C^2}} = \frac{\left|{-A}\right|}{\sqrt{A^2 + C^2}}, \Rightarrow \left|{2A + 3C}\right| = \left|{-A}\right|.$

The following solutions are possible here

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

We get two equations of the plane:

$-x + z = 0\;\text{ or }\; x - z = 0,$
$-3x + z = 0\;\text{ or }\; 3x - z = 0.$

### 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

$d_{MN} = \sqrt{\left({x_N - 9}\right)^2 + \left({0 + 2}\right)^2 + \left({0 - 2}\right)^2} = \sqrt{\left({x_N - 9}\right)^2 + 8}.$

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

$d_{NP} = \frac{\left|{3\cdot x_N - 6\cdot0 + 2\cdot0 - 3}\right|}{\sqrt{3^2 + \left({-6}\right)^2 + 2^2}} = \frac{\left|{3x_N - 3}\right|}{\sqrt{9 + 36 + 4}} = \frac{\left|{3x_N - 3}\right|}{7}.$

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

$\sqrt{\left({x_N - 9}\right)^2 + 8} = \frac{\left|{3x_N - 3}\right|}{7}.$

Let's square both sides of the equation:

${\left({x_N - 9}\right)^2 + 8} = \frac{\left({3x_N - 3}\right)^2}{49}.$

Simplify:

$x_N^2 - 18x_N + 81 + 8 = \frac{9x_N^2 - 18x_N + 9}{49},\;\Rightarrow 49x_N^2 - 882x_N + 4361 = 9x_N^2 - 18x_N + 9,\;\Rightarrow 40x_N^2 - 864x_N + 4352 = 0.$

Divide both sides by 8:

$5x_N^2 - 108x_N + 544 = 0.$

Calculate the discriminant of the quadratic equation:

$D = \left({-108}\right)^2 - 4\cdot 5 \cdot 544 = 11664 - 10880 = 784.$

The roots of the quadratic equation are

$x_N = \frac{108 \pm \sqrt{784}}{2\cdot 5} = \frac{108\pm 28}{10} = 8,\frac{68}{5}.$

We found that two points satisfy the given condition:

$N_1\left({8,0,0}\right),\;N_2\left({\frac{68}{5},0,0}\right).$