Lines and Planes in 3D Space
Equation of a line in 3D Space
As we know, we can uniquely determine a line in 3D space by specifying a point on the line and the direction of the line. Suppose a line passing through the point and is in the same direction as the vector (this non-zero vector is called the direction vector of ). Let be any point on line and . We have
i.e. the position vector of .
As is parallel to , for some real number . Therefore, we have
The parametric equation of line is as follows:
where is any real number.
The line is illustrated in the applet below.
Remarks
- The parametric equation of a line is not unique i.e. different parametric equations can represent the same line. For example, you can choose another point on the line or another direction vector that is a scalar multiple of .
- If are all non-zero, we can eliminate the parameter and write the equation of line in the following symmetric form: .
Exercise: Find the parametric equation of the line that passes through and .
Equation of a plane in 3D space
For any plane in 3D space, we can identify its "direction" by its normal vector - a non-zero vector that is perpendicular to any vector contained in the plane. We can uniquely determine a plane if we are given a point and a normal vector . Let be any point on the plane . Then we have
(because is contained in ), which implies that
Therefore, the following is the equation of the plane :
This is called the point-normal form of the equation of the plane.
Expanding the left hand side of the above equation, we get
where . This is called the general form of the equation of the plane.
The plane is illustrated in the applet below.
Two distinct planes are parallel if their normal vectors are parallel. Moreover, they do not intersect each other. If two distinct planes are non-parallel, they intersect at a line, called the line of intersection.
Two distinct planes are orthogonal if their normal vectors are orthogonal vectors. More generally, the angle between two planes is the angle between the normal vectors. If we specify that such angle, called , is smaller or equal to , we have the following formula:
, where are the normal vectors of the two given planes.
Exercise: Find the plane that passes through and is parallel to the plane .
Exercise: Find the plane that passes through , , and . (Hint: Use the cross product to find a normal vector of the plane.)
Exercise: Find the line of intersection of the planes and . (Hint: The direction vector of the line of intersection is orthogonal to the normal vectors of the two planes.)
You can use the applet below to visualize the planes and lines in the exercises.
Distance Problem
Given a plane and a point not lying on the plane. How can we find the (shortest) distance from the point to the plane?
Suppose point is the foot of the perpendicular from point to the plane. By definition, the distance between and is the distance from to the plane. Let be a point on the plane and be the normal vector of the plane. Then we have
and the equation of the plane is , where .
Hence, the distance formula from point to the plane is as follows:
The distance from a point to a plane is illustrated in the applet below.
This formula can also be used to compute the distance between two parallel planes and the distance between two non-parallel lines.
Exercise: Given the planes and . (a) Show that these two planes are parallel. (b) Find the distance between these two planes (Hint: Find a point on one of the planes and use the distance formula.)
Exercise: Let be the line passing through with the direction vector and be the line passing through with the direction vector . Find the distance between and . (Hint: Find the plane containing with the normal vector orthogonal to both and and then use the distance formula.)