Grade 12 → Algebra → Vectors and 3D Geometry ↓
Distance between lines and planes
In three-dimensional geometry, understanding the concept of distance between lines and planes is essential for solving various mathematical and real-world problems. This topic involves calculating the shortest path between elements such as a line and a plane, which is a fundamental concept in vector and plane geometry.
Distance between a point and a plane
The first basic concept we need to understand is the distance between a point and a plane. Consider a plane defined by the equation:
Ax + By + Cz + D = 0
and a point P(x₁, y₁, z₁)
. The formula to calculate the distance d
from the point P
to the plane is given by:
d = |Ax₁ + By₁ + Cz₁ + D| / sqrt(A² + B² + C²)
Given a plane 2x + 3y - 4z + 5 = 0
and a point P(1, -2, 2)
, let us find the distance between them. By applying the formula:
d = |2(1) + 3(-2) - 4(2) + 5| / sqrt(2² + 3² + (-4)²)
d = |2 - 6 - 8 + 5| / sqrt(4 + 9 + 16)
d = |-7| / sqrt(29)
d = 7 / sqrt(29)
Distance between a line and a plane
To find the distance between a line and a plane, it is important to establish their position in space. Let's say we have a line represented parametrically:
x = x₀ + at y = y₀ + bt z = z₀ + ct
Here, (x₀, y₀, z₀)
is a point on the line and (a, b, c)
denotes the direction ratio of the line. Consider the plane with the equation:
Ax + By + Cz + D = 0
The line is parallel to the plane if the dot product of the direction vector (a, b, c)
of the line and the normal vector (A, B, C)
of the plane is zero:
aA + bB + cC = 0
If the lines are not parallel to the plane, they may intersect, and the possible distance is zero.
Suppose we have a line with a parametric equation:
x = 2 + 3t y = -1 + 4t z = 5t
And a plane:
3x - y + 2z - 6 = 0
First, we find the direction vector of the line (3, 4, 5)
and the normal vector of the plane (3, -1, 2)
. The dot product is:
3(3) + 4(-1) + 5(2) = 9 - 4 + 10 = 15
Because the dot product is not zero, the line and the plane intersect, that is, the distance is zero.
Distance between parallel line and plane
If a line is parallel to a plane, then the minimum distance between them is measured along the perpendicular drawn from the line to the plane. Given that the direction vector of the line is (a, b, c)
and the plane is represented by Ax + By + Cz + D = 0
, the distance d
is calculated as follows:
d = |Ax₀ + By₀ + Cz₀ + D| / sqrt(A² + B² + C²)
It is important to note that the point (x₀, y₀, z₀)
can be any point on the line.
Suppose we have a line:
x = -1 + ty = 2 + 2t z = 3
And a plane:
4x - 2y + 5z + 1 = 0
Find out if the line is parallel to the plane:
The direction vector of the line is (1, 2, 0)
and the normal vector of the plane is (4, -2, 5)
.
Dot product: 4(1) - 2(2) + 5(0) = 4 - 4 + 0 = 0
Since the dot product is equal to zero, the line is parallel to the plane.
Using (x₀, y₀, z₀) = (-1, 2, 3)
, the distance from the plane is calculated as:
d = |4(-1) - 2(2) + 5(3) + 1| / sqrt(4² + (-2)² + 5²)
d = |-4 - 4 + 15 + 1| / sqrt(16 + 4 + 25)
d = 8 / sqrt(45)
Simplified, d = 8 / 3sqrt(5)
Visualization example of distance calculation
In the visualization above, the black line represents the line in 3D space and the blue rectangle represents the plane. The red dashed line represents the shortest distance, which is perpendicular to both, showing how orthogonal projections determine the minimum distance.
Conclusion
It is necessary to leverage vector algebra and geometric principles to understand the concepts of distance between lines and planes. When calculating distances, determining whether lines are parallel, intersecting, or skew is crucial for applying the proper formulas. By mastering these concepts, one can efficiently solve complex problems in 3D space, which can be used in a variety of fields, including physics, engineering, and computer graphics.