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Parallel Lines and Planes Help (page 2)

By — McGraw-Hill Professional
Updated on Oct 3, 2011

Corresponding Angles For Intersecting Planes

Suppose that X is a plane that passes through two parallel planes Y and Z , intersecting Y and Z in lines L and M . Let points P, Q, R, S, T, U, V , and W be as shown in Fig. 7-17, such that:

  • Point V lies on lines L, P, Q , and RS
  • Point W lies on lines M, P, Q , and TU
  • Points P and Q lie in plane X
  • Points R and S lie in plane Y
  • Points T and U lie in plane Z
  • Lines PQ and RS are perpendicular to line L
  • Lines PQ and TU are perpendicular to line M

 

An Expanded Set of Rules More Facts Corresponding Angles For Intersecting Planes

Fig. 7-17 . Corresponding angles between intersecting planes.

Then the following pairs of angles are corresponding angles, and each pair has equal measures:

∠QWU = ∠QVS

∠PWT = ∠PVR

∠UWP = ∠SVP

∠TWQ = ∠RVQ

Parallel Principles for Lines and Planes

Parallel Principle For Planes

Suppose X is a plane and R is a point not on X . Then there exists one and only one plane Y through R such that plane Y is parallel to plane X . This is the 3D counterpart of the parallel principle for Euclidean plane geometry.

The denial of this principle, which is in fact a postulate, can take two forms:

  • There exist no planes Y through point R such that plane Y is parallel to plane X
  • There exist infinitely many planes Y through point R such that plane Y is parallel to plane X

Both of these hypotheses give rise to forms of non-Euclidean geometry in which 3D space is “curved” or “warped.” Albert Einstein was one of the first scientists to envision a universe in which space is non-Euclidean.

Parallel Principle For Lines And Planes

Suppose X is a plane and R is a point not on X . Then there exist an infinite number of lines through R that are parallel to plane X . All of these lines lie in the plane Y through R such that plane Y is parallel to plane X .

The denial of the parallel principle for planes, defined in the previous paragraph, can result in there being no lines through R that are parallel to plane X . In certain specialized instances, it can even result in there being exactly one line through R that is parallel to plane X . If you have trouble imagining this, don’t be concerned. You must think in 4D, and that is a mental trick that few humans can do until they become armed with the power of non-Euclidean mathematics.

Parallel Lines and Planes Practice Problems

PROBLEM 1

Suppose you are standing inside a large warehouse. The floor is flat and level, and the ceiling is flat and is at a uniform height of 5.455 meters above the floor. You have a flashlight with a narrow beam, and hold it so its bulb is 1.025 meters above the floor. You shine the beam upward toward the ceiling. The center of the beam strikes the ceiling 9.577 meters from the point on the ceiling directly above the bulb. How long is the shaft of light representing the center of the beam? Round your answer off to two decimal places.

SOLUTION 1

Let’s call the flashlight bulb point A , the point at which the center of the light beam strikes the ceiling point B , and the point directly over the flashlight bulb point C as shown in Fig. 7-18. Call the lengths of the sides opposite these points a, b , and c . Then ΔABC is a right triangle because line segment AC (whose length is b ) is normal to the ceiling at point C , and therefore is perpendicular to line segment BC . The right angle is ∠ACB . From this, we know that the lengths of the sides are related according to the Pythagorean formula:

a 2 + b 2 = c 2

We want to know the length of side c; therefore:

c = ( a 2 + b 2 ) 1/2

 

An Expanded Set of Rules More Facts Parallel Principle For Lines And Planes

Fig. 7-18 . Illustration for Problem 7-5.

Calling meters “units” so we don’t have to write the word “meters” over and over, the length of side a is given as 9.577. The length of side b is equal to the height of the ceiling above the floor, minus the height of the bulb above the floor:

b = 5.455 − 1.025 = 4.430

Therefore:

c = (9.577 2 + 4.430 2 ) 1/2

= (91.719 + 19.625) 1/2

= 111.344 1/2

= 10.55

The center of the beam is 10.55 meters long.

Practice problems for these concepts can be found at:  Points, Lines, and Planes Practice Test.

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