Distance from a Point to a Plane
The distance d from a point P1(x1, y1, z1) 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.
- (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
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 x1, y1, z1:
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(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3). 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(x0, y0, z0) = 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: