7.5Lines and Planes in Space

8/3/2019 7.5Lines and Planes in Space

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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


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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


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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.


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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


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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.


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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


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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


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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.


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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.


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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.


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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.


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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


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Solution to Example 4 Continued:

706.039

11

279

1102cos

151122

1,5,11,2,2cos

222222

Thus,


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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


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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


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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


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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


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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:


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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.


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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.


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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.


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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.


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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.


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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.


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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.


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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.


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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.


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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.


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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


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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.


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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


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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


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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.


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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


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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


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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.


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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:


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You have three sets of practice problems for this lesson inBlackboard under Chapter 7, Lesson 7.5 Parts A, B and C.

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7.5Lines and Planes in Space

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