Section 1

PAGE

1

Section 9.5: Equations of Lines and Planes

Practice HW from Stewart Textbook (not to hand in)

p. 673 # 3-15 odd, 21-37 odd, 41, 47

Lines in 3D Space

Consider the line L through the point

)

,

,

(

0

0

0

z

y

x

P

=

that is parallel to the vector

v = < a, b, c >

q

>

=<

z

y

x

r

,

,

)

,

,

(

0

0

0

z

y

x

P

=

>

=<

0

0

0

0

,

,

z

y

x

r

)

,

,

(

z

y

x

Q

=

)

,

,

(

z

y

x

Q

=

)

,

,

(

0

0

0

z

y

x

P

=

q

1

n

2

n

n

The line L consists of all points Q = (x, y, z) for which the vector

®

-

PQ

is parallel to v.

Now,

>

-

-

-

=<

®

-

0

0

0

,

,

z

z

y

y

x

x

PQ

Since

®

-

PQ

is parallel to v = < a, b, c > ,

®

-

PQ

= t v

where t is a scalar. Thus

®

-

>=

-

-

-

<

PQ

z

z

y

y

x

x

0

0

0

,

,

= t v = < t a, t b, t c >

Rewriting this equation gives

>

<

>=

<

-

>

<

c

b

a

t

z

y

x

z

y

x

,

,

,

,

,

,

0

0

0

Solving for the vector

>

<

z

y

x

,

,

gives

>

<

+

>

>=<

<

c

b

a

t

z

y

x

z

y

x

,

,

,

,

,

,

0

0

0

Setting r =

>

<

z

y

x

,

,

,

=

0

r

EMBED Equation.3

>

<

0

0

0

,

,

z

y

x

, and v = < a, b, c >, we get the following vector equation of a line.

Vector Equation of a Line in 3D Space

The vector equation of a line in 3D space is given by the equation

+

=

0

r

r

t v

where

0

r

=

>

<

0

0

0

,

,

z

y

x

is a vector whose components are made of the point

)

,

,

(

0

0

0

z

y

x

on the line L and v = < a, b, c > are components of a vector that is parallel to the line L.

If we take the vector equation

>

<

+

>

>=<

<

c

b

a

t

z

y

x

z

y

x

,

,

,

,

,

,

0

0

0

and rewrite the right hand side of this equation as one vector, we obtain

>

+

+

+

>=<

<

tc

z

tb

y

ta

x

z

y

x

0

0

0

,

,

,

,

Equating components of this vector gives the parametric equations of a line.

Parametric Equations of a Line in 3D Space

The parametric equations of a line L in 3D space are given by

tc

z

z

tb

y

y

ta

x

x

+

=

+

=

+

=

0

0

0

,

,

,

where

)

,

,

(

0

0

0

z

y

x

is a point passing through the line and v = < a, b, c > is a vector that the line is parallel to. The vector v = < a, b, c > is called the direction vector for the line L and its components a, b, and c are called the direction numbers.

Assuming

0

,

0

,

0

¹

¹

¹

c

b

a

, if we take each parametric equation and solve for the variable t, we obtain the equations

c

z

z

t

b

y

y

t

a

x

x

t

0

0

0

,

,

-

=

-

=

-

=

Equating each of these equations gives the symmetric equations of a line.

Symmetric Equations of a Line in 3D Space

The symmetric equations of a line L in 3D space are given by

c

z

z

b

y

y

a

x

x

0

0

0

-

=

-

=

-

where

)

,

,

(

0

0

0

z

y

x

is a point passing through the line and v = < a, b, c > is a vector that the line is parallel to. The vector v = < a, b, c > is called the direction vector for the line L and its components a, b, and c are called the direction numbers.

Note!! To write the equation of a line in 3D space, we need a point on the line and a parallel vector to the line.

Example 1: Find the vector, parametric, and symmetric equations for the line through the point (1, 0, -3) and parallel to the vector 2 i - 4 j + 5 k.

Example 2: Find the parametric and symmetric equations of the line through the points (1, 2, 0) and (-5, 4, 2)

Solution: To find the equation of a line in 3D space, we must have at least one point on the line and a parallel vector. We already have two points one line so we have at least one. To find a parallel vector, we can simplify just use the vector that passes between the two given points, which will also be on this line. That is, if we assign the point

P = (1, 2, 0) and Q = (-5, 4, 2), then the parallel vector v is given by

>

-

<

=

>

-

-

-

-

<

=

=

®

-

2

,

2

,

6

0

2

,

2

4

,

1

5

PQ

v

Recall that the parametric equations of a line are given by

tc

z

z

tb

y

y

ta

x

x

+

=

+

=

+

=

0

0

0

,

,

.

We can use either point P or Q as our point on the line

)

,

,

(

0

0

0

z

y

x

. We choose the point P and assign

)

0

,

2

,

1

(

)

,

,

(

0

0

0

=

z

y

x

. The terms a, b, and c are the components of our parallel vector given by v = < -6, 2, 2 > found above. Hence a = -6, b = 2, and c = 2. Thus, the parametric equation of our line is given by

(2)

0

,

(2)

2

),

6

(

1

t

z

t

y

t

x

+

=

+

=

-

+

=

or

t

t, z

y

t

x

2

2

2

,

6

1

=

+

=

-

=

To find the symmetric equations, we solve each parametric equation for t. This gives

2

,

2

2

,

6

1

z

t

y

t

x

t

=

-

=

-

-

=

Setting these equations equal gives the symmetric equations.

2

2

2

6

1

z

y

x

=

-

=

-

-

The graph on the following page illustrates the line we have found

█

It is important to note that the equations of lines in 3D space are not unique. In Example 2, for instance, had we used the point Q = (-5, 4, 2) to represent the equation of the line with the parallel vector v = < -6, 2, 2 >, the parametric equations becomes

t

t, z

y

t

x

2

2

2

4

,

6

5

+

=

+

=

-

-

=

Example 3: Find the parametric and symmetric equations of the line passing through the point (-3, 5, 4) and parallel to the line x = 1 + 3t, y = -1 – 2t, z = 3 + t .

Solution:

█

Planes in 3D Space

Consider the plane containing the point

)

,

,

(

0

0

0

z

y

x

P

=

and normal vector n = < a, b, c >

perpendicular to the plane.

The plane consists of all points Q = (x, y, z) for which the vector

®

-

PQ

is orthogonal to the normal vector n = < a, b, c >. Since

®

-

PQ

and n are orthogonal, the following equations hold:

0

=

×

®

-

PQ

n

0

,

,

.

.

.

0

0

0

>=

-

-

-

<

×

>

<

z

z

y

y

x

x

c

b

a

0

)

(

)

(

)

(

0

0

0

=

-

+

-

+

-

z

z

c

y

y

b

x

x

a

This gives the standard equation of a plane. If we expand this equation we obtain the following equation:

0

Constant

0

0

0

=

-

-

-

+

+

4

4

4

3

4

4

4

2

1

d

cz

by

ax

cz

by

ax

Setting

0

0

0

cz

by

ax

d

-

-

-

=

gives the general form of the equation of a plane in 3D space

0

=

+

+

+

d

cz

by

ax

.

We summarize these results as follows.

Standard and General Equations of a Plane in the 3D space

The standard equation of a plane in 3D space has the form

0

)

(

)

(

)

(

0

0

0

=

-

+

-

+

-

z

z

c

y

y

b

x

x

a

where

)

,

,

(

0

0

0

z

y

x

is a point on the plane and n = < a, b, c > is a vector normal (orthogonal to the plane). If this equation is expanded, we obtain the general equation of a plane of the form

0

=

+

+

+

d

cz

by

ax

Note!! To write the equation of a plane in 3D space, we need a point on the plane and a vector normal (orthogonal) to the plane.

Example 4: Find the equation of the plane through the point (-4, 3, 1) that is perpendicular to the vector a = -4 i + 7 j – 2 k.

Solution:

█

Example 5: Find the equation of the plane passing through the points (1, 2, -3), (2, 3, 1), and (0, -2, -1).

Solution:

█

Intersecting Planes

Suppose we are given two intersecting planes with angle

q

between them.

Let

1

n

and

2

n

be normal vectors to these planes. Then

|

|

|

|

cos

2

1

2

1

n

n

n

n

×

=

q

Thus, two planes are

1. Perpendicular if

0

2

1

=

×

n

n

, which implies

2

p

q

=

.

2. Parallel if

1

2

cn

n

=

, where c is a scalar.

Notes

1. Given the general equation of a plane

0

=

+

+

+

d

cz

by

ax

, the normal vector is

n = < a, b, c >.

2.The intersection of two planes is a line.

Example 6: Determine whether the planes

3

4

3

=

-

+

z

y

x

and

4

12

3

9

=

+

-

-

z

y

x

are orthogonal, parallel, or neither. Find the angle of intersection and the set of parametric equations for the line of intersection of the plane.

Solution:

█

Example 7: Determine whether the planes

4

6

3

=

+

-

z

y

x

and

4

5

=

-

+

z

y

x

are orthogonal, parallel, or neither. Find the angle of intersection and the set of parametric equations for the line of intersection of the plane.

Solution: For the plane

4

6

3

=

+

-

z

y

x

, the normal vector is

>

-

<

=

6

,

3

,

1

1

n

and for the plane

4

5

=

-

+

z

y

x

, the normal vector is

>

-

<

=

1

,

1

,

5

2

n

. The two planes will be orthogonal only if their corresponding normal vectors are orthogonal, that is, if

0

2

1

=

×

n

n

. However, we see that

0

4

6

3

5

)

1

)(

6

(

)

1

)(

3

(

)

5

)(

1

(

1

,

1

,

5

6

,

3

,

1

2

1

¹

-

=

-

-

=

-

+

-

+

>=

-

<

×

>

-

=<

×

n

n

Hence, the planes are not orthogonal. If the planes are parallel, then their corresponding normal vectors must be parallel. For that to occur, there must exist a scalar k where

2

n

= k

1

n

Rearranging this equation as k

1

n

=

2

n

and substituting for

1

n

and

2

n

gives

>

-

<

=

>

-

<

1

,

1

,

5

6

,

3

,

1

k

or

>

-

<

=

>

-

<

1

,

1

,

5

6

,

3

,

k

k

k

.

Equating components gives the equations

-1

6

1,

3

-

,

5

=

=

=

k

k

k

which gives

6

1

-

,

3

1

,

5

=

-

=

=

k

k

k

.

Since the values of k are not the same for each component to make the vector

2

n

a scalar multiple of the vector

1

n

, the planes are not parallel. Thus, the planes must intersect in a straight line at a given angle. To find this angle, we use the equation

|

|

|

|

cos

2

1

2

1

n

n

n

n

×

=

q

For this formula, we have the following:

4

6

3

5

)

1

)(

6

(

)

1

)(

3

(

)

5

)(

1

(

1

,

1

,

5

6

,

3

,

1

2

1

-

=

-

-

=

-

+

-

+

>=

-

<

×

>

-

=<

×

n

n

46

36

9

1

)

6

(

)

3

(

)

1

(

|

|

2

2

2

1

=

+

+

=

+

-

+

=

n

27

1

1

25

)

1

(

)

1

(

)

5

(

|

|

2

2

2

2

=

+

+

=

-

+

+

=

n

(continued on next page)

Thus,

27

46

4

cos

-

=

q

Solving for

q

gives

0

1

96.5

radians

68

.

1

27

46

4

cos

»

»

÷

÷

ø

ö

ç

ç

è

æ

-

=

-

q

.

To find the equation of the line of intersection between the two planes, we need a point on the line and a parallel vector. To find a point on the line, we can consider the case where the line touches the x-y plane, that is, where z = 0. If we take the two equations of the plane

4

6

3

=

+

-

z

y

x

4

5

=

-

+

z

y

x

‘

and substitute z = 0, we obtain the system of equations

4

3

=

-

y

x

(1)

4

5

=

+

y

x

(2)

Taking the first equation and multiplying by -5 gives

20

15

5

-

=

+

-

y

x

4

5

=

+

y

x

Adding the two equations gives 16y = -16 or

1

16

16

-

=

-

=

y

. Substituting

1

-

=

y

back into equation (1) gives

4

)

1

(

3

=

-

-

x

or

4

3

=

+

x

. Solving for x gives x = 4-3 = 1. Thus, the point on the plane is (1, -1, 0). To find a parallel vector for the line, we use the fact that since the line is on both planes, it must be orthogonal to both normal vectors

1

n

and

2

n

. Since the cross product

2

1

n

n

´

gives a vector orthogonal to both

1

n

and

2

n

,

2

1

n

n

´

will be a parallel vector for the line. Thus, we say that

k

j

i

k

j

i

k

j

i

k

j

i

n

n

v

16

31

3

)

15

1

(

)

30

1

(

)

6

3

(

1

5

3

-

1

1

5

6

1

1

1

6

3

1

1

5

6

3

1

2

1

+

+

-

=

-

-

+

-

-

-

-

=

+

-

-

-

-

=

-

-

=

´

=

(continued on next page)

Hence, using the point (1, -1, 0) and the parallel vector

k

j

i

v

16

31

3

+

+

-

=

, we find the parametric equations of the line are

t

z

t

y

t

x

16

,

31

1

,

3

1

=

+

-

=

-

=

The following shows a graph of the two planes and the line we have found.

█

Example 8: Find the point where the line x = 1 + t, y = 2t, and z = -3t intersects the plane

2

4

2

4

-

=

-

+

-

z

y

x

.

Solution:

█

Distance Between Points and a Plane

Suppose we are given a point Q not in a plane and a point P on the plane and our goal is to find the shortest distance between the point Q and the plane.

By projecting the vector

®

-

PQ

onto the normal vector n (calculating the scalar projection

®

-

PQ

comp

n

), we can find the distance D.

|

|

|

|

|

|

D

plane

the

and

Between

Distance

n

n

PQ

PQ

comp

Q

n

×

=

=

=

®

-

®

-

Example 9: Find the distance between the point (1, 2, 3) and line

4

2

=

+

-

z

y

x

.

Solution: Since we are given the point Q = (1, 2, 3), we need to find a point on the plane

4

2

=

+

-

z

y

x

in order to find the vector

®

-

PQ

. We can simply do this by setting y = 0 and z = 0 in the plane equation and solving for x. Thus we have

4

2

=

+

-

z

y

x

4

0

0

2

=

+

-

x

2

2

4

4

2

=

=

=

x

x

Thus P = (2, 0, 0) and the vector

®

-

PQ

is

>

-

>=<

-

-

-

=<

®

-

3

,

2

,

1

0

3

,

0

2

,

2

1

PQ

.

Hence, using the fact that the normal vector for the plane is

>

-

=<

1

,

1

,

2

n

, we have

6

1

6

|

1

|

1

1

4

|

3

2

2

|

)

1

(

)

1

(

)

2

(

1

,

1

,

2

3

,

2

,

1

|

|

|

|

|

plane

the

and

Q

Between

Distance

2

2

2

=

-

=

+

+

+

-

-

=

+

-

+

>

-

<

×

>

-

<

=

×

=

®

-

n

n

PQ

Thus, the distance is

6

1

.

█

� EMBED Equation.3 ���

� EMBED Equation.3 ���

n = < a, b, c >

� EMBED Equation.3 ���

� EMBED Equation.3 ���

� EMBED Equation.3 ���

x

y

z

L

v = < a, b, c >

z

� EMBED Equation.3 ���

� EMBED Equation.3 ���

y

x

� EMBED Equation.3 ���

� EMBED Equation.3 ���

� EMBED Equation.3 ���

Plane 2

Plane 1

� EMBED Equation.3 ���

Q

P

_1224423582.unknown

_1224448275.unknown

_1224504655.unknown

_1224508771.unknown

_1224511071.unknown

_1224515631.unknown

_1224516284.unknown

_1224516552.unknown

_1224516811.unknown

_1224516810.unknown

_1224516357.unknown

_1224516005.unknown

_1224516234.unknown

_1224515696.unknown

_1224512053.unknown

_1224511815.unknown

_1224509102.unknown

_1224509462.unknown

_1224509950.unknown

_1224509144.unknown

_1224508823.unknown

_1224509053.unknown

_1224508797.unknown

_1224505396.unknown

_1224508603.unknown

_1224508649.unknown

_1224508577.unknown

_1224504744.unknown

_1224505389.unknown

_1224504715.unknown

_1224502798.unknown

_1224503023.unknown

_1224504449.unknown

_1224504544.unknown

_1224503281.unknown

_1224502912.unknown

_1224502960.unknown

_1224502822.unknown

_1224502761.unknown

_1224502769.unknown

_1224502793.unknown

_1224502699.unknown

_1224502718.unknown

_1224448316.unknown

_1224448296.unknown

_1224444460.unknown

_1224445747.unknown

_1224446104.unknown

_1224448169.unknown

_1224446071.unknown

_1224444810.unknown

_1224445712.unknown

_1224444481.unknown

_1224444040.unknown

_1224444459.unknown

_1224444053.unknown

_1224444214.unknown

_1224427755.unknown

_1224443573.unknown

_1224443850.unknown

_1224443977.unknown

_1224443214.unknown

_1224423598.unknown

_1224416039.unknown

_1224418767.unknown

_1224423029.unknown

_1224423355.unknown

_1224423391.unknown

_1224423072.unknown

_1224421254.unknown

_1224422754.unknown

_1224418960.unknown

_1224416276.unknown

_1224418655.unknown

_1224418754.unknown

_1224416648.unknown

_1224416744.unknown

_1224416383.unknown

_1224416161.unknown

_1222973379.unknown

_1222975966.unknown

_1222976183.unknown

_1222976427.unknown

_1224415874.unknown

_1222976194.unknown

_1222976265.unknown

_1222975769.unknown

_1222975757.unknown

_1222972007.unknown

_1222973078.unknown

_1222973337.unknown

_1222972585.unknown

_1222972440.unknown

_1222971562.unknown

_1222971571.unknown

_1222971480.unknown

_1222970754.unknown

_1222971002.unknown

_1222970210.unknown