Introduction to the Intersection of Lines and Planes - Angles
Angles and Distance
Let’s see how the angles between intersecting planes are defined, and how these angles behave. Let’s also see how we can define the angles between an intersecting line and plane, and how these angles behave.
Angles Between Intersecting Planes
Suppose two planes X and Y intersect in a common line L . Consider line M in plane X and line N in plane Y , such that M 
L and N L , as shown in Fig. 7-7. The angle between the intersecting planes X and Y is called a dihedral angle , and can be represented in two ways. The first angle, whose measure is denoted by u , is the smaller angle between lines M and N . The second angle, whose measure is denoted by v , is the larger angle between lines M and N .
If only one angle is mentioned, the “angle between two intersecting planes” is usually considered to be the smaller angle u . Therefore, the angle of intersection is larger than zero but less than or equal to a right angle. That is, 0° < u ≤ 90° (0 < u ≤ π /2).
Adjacent Angles Between Intersecting Planes
Suppose two planes intersect, and their angles of intersection are u and v as defined above. Then if u and v are specified in degrees, u + v = 180°. If u and v are specified in radians, then u + v = π .
Fig. 7-7 . The dihedral angle between the intersecting planes X and Y can be represented by u , the acute angle between lines M and N , or by v , the obtuse angle between lines M and N .
Perpendicular Planes
Suppose two planes X and Y intersect in a common line L . Consider line M in plane X and line N in plane Y , such that M L and N L , as shown in Fig. 7-7. Then X and Y are said to be perpendicular planes if the angles between lines M and N are right angles, that is, u = v = 90° ( π /2 rad). Actually, it suffices to say that either u = 90° ( π /2 rad) or v = 90° ( π /2 rad).
Normal Line To A Plane
Let plane X be determined by lines L and M , which intersect at point S . Then line N that passes through plane X at point S is normal (also called perpendicular or orthogonal) to plane X if and only if N L and N M . This is shown in Fig. 7-8. Line N is the only line normal to plane X at point S . Furthermore, line N is perpendicular to any line, line segment, or ray that lies in plane X and runs through point S .
Fig. 7-8 . Line N through plane X at point S is normal to X if and only if N L and N M , where L and M are lines in plane X that intersect at point S .
Angles Between An Intersecting Line And Plane
Let X be a plane. Suppose a line O , which is not normal to plane X , intersects plane X at some point S as shown in Fig. 7-9. In order to define an angle at which line O intersects plane X , we must construct some objects. Let N be a line normal to plane X , passing through point S . Let Y be the plane determined by the intersecting lines N and O . Let L be the line formed by the intersection of planes X and Y . The angle between line O and plane X can be represented in two ways. The first angle, whose measure is denoted by u , is the smaller angle between lines L and O as determined in plane Y . The second angle, whose measure is denoted by v , is the larger angle between lines L and O as determined in plane Y .
If only one angle is mentioned, the “angle between a line and a plane that intersect” is considered to be the smaller angle u . Therefore, the angle of intersection is larger than zero but less than or equal to a right angle. That is, 0° < u ≤ 90° (0 < u ≤ π /2).
Fig. 7-9 . Angles u and v between a plane X and a line O that passes through X at point S .
Adjacent Angles Between An Intersecting Line And Plane
Suppose a line and a plane intersect, and their angles of intersection are u and v as defined above. Then if u and v are specified in degrees, u + v = 180°. If u and v are specified in radians, then u + v = π .
Dropping A Normal To A Plane
Suppose that R is a point near, but not in, a plane X . Then there is exactly one line N through point R , intersecting plane X at some point S , such that line N is normal to plane X , as shown in Fig. 7-10. Any lines in plane X that pass through point S , such as L and M shown in the figure, must necessarily be perpendicular to line N .
Fig. 7-10 . Line N through point R is normal to plane X at point S . The distance between R and X is equal to the length of line segment RS .
Distance Between A Point And Plane
Suppose that R is a point near, but not in, a plane X . Let N be the line through R that is normal to plane X . Suppose line N intersects plane X at point S . Then the distance between point R and plane X is equal to the length of line segment RS (Fig. 7-10).
Plane Perpendicular To Line
Imagine a line N in space. Imagine a specific point S on line N . There is exactly one plane X containing point S , such that line N is normal to plane X at point S (Fig. 7-10). Any lines in plane X that pass through point S , such as L and M shown in the figure, must necessarily be perpendicular to line N .
Line Parallel To Plane
A line L is parallel to a plane X if and only if the following two conditions hold true:
Line L does not lie in plane X- Line L does not intersect plane X
- nder these conditions, there is exactly one line M in plane X , such that lines L and M are parallel. Any line N in plane X , other than line M , is a skew line relative to L (Fig. 7-11).
Fig. 7-11 . A line L parallel to a plane X . There is exactly one line M in plane X such that M is parallel to L; all other lines N in plane X are skew lines relative to L .
Distance Between A Parallel Line And Plane
Suppose line L is parallel to plane X . Let R be some (any) point on line L . Then the distance between line L and plane X is equal to the distance between point R and plane X , which has already been defined.
Addition And Subtraction Of Angles Between Intersecting Planes
Angles between intersecting planes add and subtract just like angles between intersecting lines (or line segments). Here is how we can prove it, based on knowledge we already have.
Suppose three planes X, Y , and Z intersect in a single, common line L . Let S be a point on line L . Let P, Q , and R be points on planes X, Y , and Z respectively, such that line segments SP, SQ , and SR are all perpendicular to line L . Let ∠XY be the angle between planes X and Y, ∠YZ be the angle between planes Y and Z , and ∠XZ be the angle between planes X and Z . This is diagrammed in Fig. 7-12. From the preceding definition of the angle between two planes, we know that:
∠XY = ∠PSQ
∠YZ = ∠QSR
∠XZ = ∠PSR
Fig. 7-12 . Addition and subtraction of angles between planes.
Line segments SP, SQ , and SR all lie in a single plane because they all intersect at point S and they are all perpendicular to line L . Therefore, we know from the rules for addition of angles in a plane, that the following hold true for the measures of the angles between the line segments:
∠PSQ + ∠QSR = ∠PSR
∠PSR − ∠QSR = ∠PSQ
∠PSR − ∠PSQ = ∠QSR
Substituting the angles between the planes for the angles between the line segments, we see that the following hold true for the measures of the angles between the planes:
∠XY + ∠YZ = ∠XZ
∠XZ − ∠YZ = ∠XY
∠XZ − ∠XY = ∠YZ
Intersection of Lines and Planes Problems and Solutions
PROBLEM 1
Suppose a communications cable is strung above a fresh-water lake. Imagine that the cable does not sag at all, and is attached at the tops of a set of utility poles. Suppose the engineering literature recommends that the cable be suspended 10 meters above “effective ground,” and that “effective ground” is, on average, 2 meters below the average level of the surface of a body of fresh water. How tall should the poles be? Assume they are all perfectly vertical.
SOLUTION 1
Because the poles are perfectly vertical, they are perpendicular to the surface of the lake. This means that the poles should each be tall enough so their tops are 10 meters above “effective ground,” so they should each extend 10 − 2 meters, or 8 meters, above the water surface. The actual height of each pole depends on the depth of the lake at the point where it is placed. It is assumed that the lake is small enough, and/or weather conditions reasonable enough, so the lake does not acquire waves so high that they inundate the cable!
PROBLEM 2
Imagine that you are flying a kite over a perfectly flat field. The kite is of a design that flies at a “high angle.” Suppose the kite line does not sag, and the kite flies only 10° away from the vertical. (Some kites can actually fly straight overhead!) Imagine that it is a sunny day, and the sun is shining down from exactly the zenith. What is the angle between the kite string (also called the kite line) and its shadow on the flat field?
SOLUTION 2
Suppose you stand at point S on the surface of the field, which we call plane X . The kite line and its shadow lie along lines SR and ST , as shown in Fig. 7-13. The sun shines down along a line QS that is normal to plane X . Lines SQ, SR , and ST all lie in a common plane Y , which is perpendicular to plane X . We know that the measure of ∠QSR is 10°, because we are given this information. We also know that the measure of ∠QST is 90°, because line QS is normal to plane X , and line ST lies in plane X . Because lines SR, ST , and SQ all lie in the same plane Y , we know that the measures of the angles among them add as follows:
Fig. 7-13 . Illustration for Problem 2
∠QSR + ∠RST = ∠QST
and therefore
∠RST = ∠QST − ∠QSR
The measure of ∠RST , which represents the angle between the kite line and its shadow, is equal to 90° − 10°, or 80°.
Practice problems for these concepts can be found at: Points, Lines, and Planes Practice Test.
Celebrate Memorial Day! Worksheets and Activities About American History 
May Workbooks are Here!
Get Outside! 10 Playful Activities
Add your own comment