7.5lines and planes in space
TRANSCRIPT
-
8/3/2019 7.5Lines and Planes in Space
1/38
Chapter 7: Vectors and theGeometry of SpaceSection 7.5
Lines and Planes in Space
Written by Karen Overman
Instructor of Mathematics
Tidewater Community College, Virginia Beach Campus
Virginia Beach, VA
With Assistance from a VCCS LearningWare Grant
-
8/3/2019 7.5Lines and Planes in Space
2/38
In this lesson you will learn:
o Lines in Space
o Parametric equations for a line
o Symmetric equations for a line
o Relationships between lines in space
o Planes in Space
o Standard form and General form of a plane
o Sketching planes using traces
oThe line of intersection of two planes
o Distances in Space
o The distance between a point and a plane
o The distance between a point and a line
-
8/3/2019 7.5Lines and Planes in Space
3/38
Lines in Space
Previously you have studied lines in a two-dimensional coordinate system.
These lines were determined by a point and a direction or the slope.
In 3-dimensional space, a line will also be determined by a point and adirection, except in 3-dimensional space the direction will be given by aparallel vector, called the direction vector.
-
8/3/2019 7.5Lines and Planes in Space
4/38
Lines in Space
To determine the equation of the line passing through the point P
and parallel to the direction vector, , we will use our
knowledge that parallel vectors are scalar multiples. Thus, the vector
through P and any other point Q on the line is a scalar multiple
of the direction vector, .
000 ,, zyx
cbav ,,
zyx ,,
cbav ,,
In other words,
ctbtatzzyyxx
tcbat
,,,,
or
scalarnumberrealanyiswhere,,,PQ
000
-
8/3/2019 7.5Lines and Planes in Space
5/38
Equations of Lines in Space
Equate the respective components and there are three equations.
ctzzbtyyatxx
ctzzbtyyatxx
000
000
and,
or
and,
These equations are called the parametric equations of the line.
If the components of the direction vector are all nonzero, eachequation can be solved for the parameter t and then the three can beset equal.
c
zz
b
yy
a
xx000
These equations are called the symmetric equations of the line.
-
8/3/2019 7.5Lines and Planes in Space
6/38
Equations of Lines in Space
A line passing through the point P and parallel to the vector,
is represented by the parametric equations:
And if all three components of the direction vector are nonzero, the lineis also represented by the symmetric equations:
ctzzbtyyatxx 000 and,
cbav ,,
000 ,, zyx
c
zz
b
yy
a
xx000
-
8/3/2019 7.5Lines and Planes in Space
7/38
Example 1: Find the parametric and symmetric equations of the line passingthrough the point (2, 3, -4) and parallel to the vector, .
Solution: Simply use the parametric and symmetric equations for any line givena point on the line and the direction vector.
Parametric Equations:
tztytx 54and23,2
Symmetric Equations:
54
23
12
zyx
-
8/3/2019 7.5Lines and Planes in Space
8/38
Example 2: Find the parametric and symmetric equations of the line passingthrough the points (1, 2, -2) and (3, -2, 5).
Solution: First you must find the direction vector which is just finding thevector from one point on the line to the other. Then simply use theparametric and symmetric equations and either point.
7,4,225,22,13vectordirection v
tztytx 72and42,21:equationsparametric
7
2
4
2
2
1:equationssymmetric
zyx
Notes:1. For a quick check, when t= 0 the parametric equations give the point
(1, 2, -2) and when t= 1 the parametric equations give the point (3, -2, 5).2. The equations describing the line are not unique. You may have used the
other point or the vector going from the second point to the first point.
-
8/3/2019 7.5Lines and Planes in Space
9/38
Relationships Between Lines
In a 2-dimensional coordinate system, there were three possibilities when
considering two lines: intersecting lines, parallel lines and the two wereactually the same line.
In 3-dimensional space, there is one more possibility. Two lines may be skew,which means the do not intersect, but are not parallel. For an example seethe picture and description below.
If the red line is down in the xy-plane and the blue line is above thexy-plane, but parallel to the xy-plane the two lines never intersectand are not parallel.
-
8/3/2019 7.5Lines and Planes in Space
10/38
Example 3: Determine if the lines are parallel or identical.
tz
ty
tx
21
42
25:2Line
tz
ty
tx
4
22
3:1Line
Solution: First look at the direction vectors: 2,4,2and1,2,1 21 vv
Since , the lines are parallel.
Now we must determine if they are identical. So we need to determineif they pass through the same points. So we need to determine if thetwo sets of parametric equations produce the same points fordifferent values of t.
Let t=0 for Line 1, the point produced is (3, 2, 4). Set the x from Line
2 equal to the x-coordinate produced by Line 1 and solve for t.
12 2vv
122253 ttt
Now let t=1 for Line 2 and the point (3, 2, -1) is produced. Since the z-coordinates are not equal, the lines are not identical.
-
8/3/2019 7.5Lines and Planes in Space
11/38
Example 4: Determine if the lines intersect. If so, find the point ofintersection and the cosine of the angle of intersection.
tzty
tx
253
4:2Line
tz
ty
tx
42
23:1Line
Solution: Direction vectors:
Since , the lines are not parallel. Thus they either intersect orthey are skew lines.
1,5,11,2,2 21 vv
12vkv
Keep in mind that the lines may have a point of intersection or a commonpoint, but not necessarily for the same value of t. So equate each
coordinate, but replace the tin Line 2 with an s.
stz
sty
stx
24:
532:
423:System of 3 equations with 2unknowns Solve the first 2 andcheck with the 3rd equation.
-
8/3/2019 7.5Lines and Planes in Space
12/38
Solution to Example 4 Continued:
Solving the system, we get t= 1 and s= -1.
Line 1: t= 1 produces the point (5, -2, 3)Line 2: s= -1 produces the point (5, -2, 3)
The lines intersectat this point.
Recall from lesson 7.3 on the dot product,
numerator.theinvalueabsoluteuseweso
,90thanlessbeshouldlinesngintersectitwobetweenangleThe
andbetweenangletheishere,cos
vuwvu
vu
-
8/3/2019 7.5Lines and Planes in Space
13/38
Solution to Example 4 Continued:
706.039
11
279
1102cos
151122
1,5,11,2,2cos
222222
Thus,
-
8/3/2019 7.5Lines and Planes in Space
14/38
Planes in Space
In previous sections we have looked at planes in space. For example, we lookedat the xy-plane, the yz-plane and the xz-plane when we first introduced 3-dimensional space.
Now we are going to examine the equation for a plane. In the figure below P,, is a point in the highlighted plane and is the vector
normal to the highlighted plane. 000 ,, zyx cban ,,
n
P
Q
For any point Q,
in the plane, the vector from P
to Q ,
is also in the plane.
zyx ,,
000
,, zzyyxxPQ
-
8/3/2019 7.5Lines and Planes in Space
15/38
Planes in Space
Since the vector from P to Q is in the plane, are perpendicularand their dot product must equal zero.
n
P
Q
This last equation is theequation of the highlightedplane.
So the equation of any planecan be found from a point inthe plane and a vector normalto the plane.
nPQ and
0
0,,,,
0
000
000
zzcyybxxa
zzyyxxcba
PQn
-
8/3/2019 7.5Lines and Planes in Space
16/38
Standard Equation of a Plane
The standard equation of a plane containing the point and having
normal vector, is
Note: The equation can be simplified by using the distributive property and
collecting like terms.
This results in the general form:
000 ,, zyx
cban ,,
0000 zzcyybxxa
0 dczbyax
-
8/3/2019 7.5Lines and Planes in Space
17/38
Example 5: Given the normal vector, to the plane containing thepoint (2, 3, -1), write the equation of the plane in both standard form andgeneral form.
Solution: Standard Form 0000 zzcyybxxa
0123123 zyx
To obtain General Form, simplify.
01123
or
022363
zyx
zyx
-
8/3/2019 7.5Lines and Planes in Space
18/38
Example 6: Given the points (1, 2, -1), (4, 0,3) and (2, -1, 5) in a plane, find theequation of the plane in general form.
Solution: To write the equation of the plane we need a point (we have three) anda vector normal to the plane. So we need to find a vector normal to the plane.First find two vectors in the plane, then recall that their cross product will be avector normal to both those vectors and thus normal to the plane.
Two vectors: From (1, 2, -1) to (4, 0, 3): < 4-1, 0-2, 3+1 > =
From (1, 2, -1) to (2, -1, 5): < 2-1, -1-2, 5+1 > =
Their cross product:
kjkji
kji
7147140
631
423
032
or
021714
01721410
zy
zy
zyxEquation of the plane:
-
8/3/2019 7.5Lines and Planes in Space
19/38
Sketching Planes in Space
If a plane intersects all three coordinate planes (xy-plane, yz-plane and thexz-plane), part of the plane can be sketched by finding the intercepts andconnecting them to form the plane.
For example, lets sketch the part of the plane, x + 3y + 4z 12 = 0 thatappears in the first octant.
The x-intercept (where the plane intersects the x-axis) occurs when both yand z equal 0, so the x-intercept is (12, 0, 0). Similarly the y-intercept is(0, 4, 0) and the z-intercept is (0, 0, 3).
Plot the three points on the coordinate system and then connect each pairwith a straight line in each coordinate plane. Each of these lines is called a
trace.
The sketch is shown on the next slide.
-
8/3/2019 7.5Lines and Planes in Space
20/38
Sketch of the plane x + 3y + 4z 12 = 0 with intercepts, (12, 0, 0), (0, 4, 0)and (0, 0, 3).
y
x
z
Now you can see the triangularpart of the plane that appearsin the first octant.
-
8/3/2019 7.5Lines and Planes in Space
21/38
y
x
z
Another way to graph the plane x + 3y + 4z 12 = 0 is by using the traces. Thetraces are the lines of intersection the plane has with each of the coordinateplanes.
The xy-trace is found by letting z = 0, x + 3y = 12 is a line the the xy-plane.Graph this line.
-
8/3/2019 7.5Lines and Planes in Space
22/38
y
x
z
Similarly, the yz-trace is 3y + 4z = 12, and the xz-trace is x + 4z = 12. Grapheach of these in their respective coordinate planes.
-
8/3/2019 7.5Lines and Planes in Space
23/38
Example 7: Sketch a graph of the plane 2x 4y + 4z 12 = 0.
Solution: The intercepts are (6, 0, 0), (0, -3, 0) and (0, 0, 3). Plot each ofthese and connect each pair with a straight line.
-
8/3/2019 7.5Lines and Planes in Space
24/38
y
x
z
Example 7: Sketch a graph of the plane 2x 4y + 4z 12 = 0.
Solution: The intercepts are (6, 0, 0), (0, -3, 0) and (0, 0, 3). Plot each ofthese and connect each pair with a straight line.
Hopefully you can see the part of theplane we have sketched appears on thenegative side of the y-axis.
-
8/3/2019 7.5Lines and Planes in Space
25/38
Not all planes have x, y and z intercepts. Any plane whose equation is missingone variable is parallel to the axis of the missing variable. For example,2x + 3y 6 = 0 is parallel to the z-axis. The xy trace is 2x + 3y = 6, the yz
trace is y = 2 and the xz trace is x = 3.
Part of the plane is outlined in red.
More on Sketching Planes
Any plane whose equation is missing two variables is parallel to the
coordinate plane of the missing variables. For example, 2x 6 = 0 or x = 3 isparallel to the yz-plane.
The plane is outlined inblue and is at the x value
of 3.
-
8/3/2019 7.5Lines and Planes in Space
26/38
Intersecting Planes
Any two planes that are not parallel or identical will intersect in a line and tofind the line, solve the equations simultaneously.
For example in the figure above, the white plane and the yellow plane intersectalong the blue line.
-
8/3/2019 7.5Lines and Planes in Space
27/38
Example 8: Find the line of intersection for the planes x + 3y + 4z = 0and x 3y +2z = 0.
zyzy
zyxzyx
zyxzyx
31or026
023023
0430431
Back substitute y into one of the first equations and solve for x.
zx
zzx
zzx
3
04
043
13
Finally if you let z = t, the parametric equations for the line are
tztytx
and3
1
,3
Solution: To find the common intersection, solve the equations simultaneously.Multiply the first equation by 1 and add the two to eliminate x.
-
8/3/2019 7.5Lines and Planes in Space
28/38
Distance Between a Point and a Plane
P
Q
n, normal
Projection of PQonto the normalto the plane
Thus the distance from Q to the plane is the length or the magnitude of the
projection of the vector PQ onto the normal.
Let P be a point in the plane and let Q be a point not in the plane. We areinterested in finding the distance from the point Q to the plane thatcontains the point P.
We can find the distance between the point, Q, and the plane by projectingthe vector from P to Q onto the normal to the plane and then finding itsmagnitude or length.
-
8/3/2019 7.5Lines and Planes in Space
29/38
Distance Between a Point and a Plane
If the distance from Q to the plane is the length or the magnitude of theprojection of the vector PQ onto the normal, we can write that
mathematically:
PQprojn
planethetoQfromDistance
Now, recall from section 7.3,
nn
nPQPQprojn
2
So taking the magnitude of this vector, we get:
n
nPQn
n
nPQn
n
nPQPQproj
n
22
-
8/3/2019 7.5Lines and Planes in Space
30/38
Distance Between a Point and a Plane
The distance from a plane containing the point P to a point Q not in theplane is
n
nPQPQprojD n
where n is a normal to the plane.
-
8/3/2019 7.5Lines and Planes in Space
31/38
Example 9: Find the distance between the point Q (3, 1, -5) to the plane4x + 2y z = 8.
Solution: We know the normal to the plane is from the generalform of a plane. We can find a point in the plane simply by letting x and yequal 0 and solving for z: P (0, 0, -8) is a point in the plane.
Thus the vector, PQ = =
Now that we have the vector PQ and the normal, we simply use the formulafor the distance between a point and a plane.
4.221
11
1416
3212
124
1,2,43,1,3222
D
n
nPQPQprojD
n
-
8/3/2019 7.5Lines and Planes in Space
32/38
Lets look at another way to write the distance from a point to a plane. Ifthe equation of the plane is ax + by + cz + d = 0, then we know the normalto the plane is the vector, .
Let P be a point in the plane, P = and Q be the point not in the
plane, Q = . Then the vector,
111 ,, zyx
000
,, zyx101010 ,, zzyyxxPQ
So now the dot product of PQ and n becomes:
111000
101010
101010,,,,
czbyaxczbyax
zzcyybxxa
zzyyxxcbanPQ
Note that since P is a point on the plane it will satisfy the equation of theplane, so and the dotproduct can be rewritten:
111111 or0 czbyaxddczbyax
dczbyaxnPQ 000
-
8/3/2019 7.5Lines and Planes in Space
33/38
Thus the formula for the distance can be written another way:
The Distance Between a Point and a Plane
The distance between a plane, ax + by + cz + d = 0 and a point Qis
222
000
cba
dczbyaxPQprojD n
000 ,, zyx
Now that you have two formulas for the distance between a point and a plane,lets consider the second case, the distance between a point and a line.
-
8/3/2019 7.5Lines and Planes in Space
34/38
Distance Between a Point and a Line
In the picture below, Q is a point not on the line , P is a point on the line, u
is a direction vector for the line and is the angle between u and PQ.
P
Q
u
D = Distance from Q to the line
Obviously, sinorsin PQDPQD
-
8/3/2019 7.5Lines and Planes in Space
35/38
We know from Section 7.4 on cross products that
.andbetweenangletheiswhere,sin vuvuvu
Thus,
sin
bysidesbothdividingor
sin
PQu
uPQ
u
uPQuPQ
So if, then from above, .sinPQD u
uPQD
-
8/3/2019 7.5Lines and Planes in Space
36/38
Distance Between a Point and a Line
The distance, D, between a line and a point Q not on the line is given by
uuPQD
where u is the direction vector of the line and P is a point on the line.
-
8/3/2019 7.5Lines and Planes in Space
37/38
Example 10: Find the distance between the point Q (1, 3, -2) and the linegiven by the parametric equations:
tztytx 23and1,2
Solution: From the parametric equations we know the direction vector, u is< 1, -1, 2 > and if we let t = 0, a point P on the line is P (2, -1, 3).
Thus PQ = < 2-1, -1-3, 3-(-2) > = < 1, -4, 5 >
Find the cross product:
12.2
2
9
6
27
211
333222
222
u
uPQD
kji
kji
uPQ 333
211
541
Using the distance formula:
-
8/3/2019 7.5Lines and Planes in Space
38/38
You have three sets of practice problems for this lesson inBlackboard under Chapter 7, Lesson 7.5 Parts A, B and C.