lect analytical geometry in 3 d space

ME 2304: 3D Geometry & Vector Calculus

TWO-DIMENSIONAL (2-D) COORDINATE SYSTEMS

• To locate a point in a plane, two numbers are essential.

– We know that any point in the plane can be represented as an ordered pair (a, b) of real numbers—where a is the x-coordinate and b is the y-coordinate.

– For this reason, a plane is called two-dimensional.

THREE-DIMENSIONAL (3-D) COORDINATE SYSTEMS

• To locate a point in space, three numbers are required.

– We represent any point in space by an ordered triple (a, b, c) of real numbers.

• In order to represent points in space, we first choose:

– A fixed point O (the origin)

– Three directed lines through O that are perpendicular to each other

3-D COORDINATE SYSTEMS

• The three lines are called the coordinate axes.

• They are labeled:

– x-axis– y-axis– z-axis

• Usually, we think of:

– The x- and y-axes as being horizontal

– The z-axis as being vertical

COORDINATE AXES

• We draw the orientation of the axes as shown.

COORDINATE AXES

• The direction of the z-axis is

determined by the right-hand rule,

illustrated in following slide.

COORDINATE AXES

• Curl the fingers of your right hand around the z-axis in the direction of a 90° counterclockwise rotation from the positive x-axis to the positive y-axis.

– Then, your thumb points in the positive direction of the z-axis.

COORDINATE AXES

• The three coordinate axes determine the three coordinate planes.

– The xy-plane contains

the x- and y-axes.

– The yz-plane contains

the y- and z-axes.

– The xz-plane contains

the x- and z-axes.

COORDINATE PLANES

• Many people have some difficulty visualizing diagrams of 3-D figures.

• Thus, you may find it helpful to do the following.


• Look at any bottom corner of a room and call the corner the origin.


• The wall on your left is in the xz-plane.• The wall on your right is in the yz-plane.• The floor is in the xy-plane.


• The x-axis runs along the intersection of the floor and the left wall.

• The y-axis runs along that of the floor and the right wall.


• The z-axis runs up from the floor toward the ceiling along the intersection of the two walls.


• Now, if P is any point in space, let:

– a be the (directed) distance from the yz-plane to P.

– b be the distance from the xz-plane to P.

– c be the distance from the xy-plane to P.


• We represent the point P by the ordered triple of real numbers (a, b, c).

• We call a, b, and c the coordinates of P.

– a is the x-coordinate.– b is the y-coordinate.– c is the z-coordinate.


• Thus, to locate the point (a, b, c), we can start at the origin O and proceed as follows:

– First, move a units along the x-axis.

– Then, move b units parallel to the y-axis.

– Finally, move c units parallel to the z-axis.


• The point P(a, b, c) determines a rectangular box.


• If we drop a perpendicular from P to the xy-plane, we get a point Q with coordinates (a, b, 0).

– This is called the projection of P on the xy-plane.

PROJECTIONS

• Similarly, R(0, b, c) and S(a, 0, c) are the projections of P on the yz-plane and xz-plane, respectively.

PROJECTIONS

• As numerical illustrations, the points (–4, 3, –5) and (3, –2, –6) are plotted here.


Lines in SpaceTo determine the equation of the line passing through the point P

and parallel to the direction vector, , we will use 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

y

z

x

cbav ,,

000 ,, zyx

zyx ,,

P

Q

Equations of Lines in SpaceEquate the respective components and solving for x, y and z gives 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, each equation can be solved for the parameter tt and then the three can be set equal.

czz

byy

axx 000

These equations are called the symmetric equations of the line.

Find the parametric and symmetric equations of the line passing through the point (2, 3, -4) and parallel to the vector, (-1, 2, 5) .

Simply use the parametric and symmetric equations for any line given a point on the line and the direction vector.

Parametric Equations:

tztytx 54and23,2

Notice that when t=0, we are at the point (2, 3, -4). As t increases or decreases from 0, we move away from this point parallel to the direction indicated by (-1, 2, 5).

Symmetric Equations:

czz

byy

axx 000

Thus , we will have:

54

23

12

zyx

Find the parametric and symmetric equations of the line passing through the points (1, 2, -2) and (3, -2, 5).

First you must find the direction vector which is just finding the vector from one point on the line to the other. Then simply use the parametric and symmetric equations and either point.

7,4,225,22,13vectordirection v

tztytx 72and42,21:equationsparametric

72

42

21

: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.

Find the parametric equation of the line passing through the point P(5, -2, 4) and parallel to the vector, a(1/2, 2, -2/3) .

Parametric Equations: To avoid fraction, using vector b=6a=(3,12,-4)

t-t, z-t, yx 4412235 Where does the line intersect the xy-plane.

The line will intersect xy-plane at a point, say R(x,y,z) if z=4-4t=0,

Which implies t=1. Putting t=1 in parametric equations yields x andy co-ordinates of R

, )(- y and )(x 1011228135

Hence, R is the point with co-ordinates (8,10,0)

Sketch the position vector a and the line,(say l) for previous example

y

z

x

3/2,2,2/1 a

0,10,8R

4 ,2 ,5 P

Relationships Between LinesIn a 2-dimensional coordinate system, there were three possibilities when considering two lines: intersecting lines, parallel lines and identical lines i.e. the two were actually the same line, as passing through the same points.

In 3-dimensional space, there is one more possibility, i.e. two lines may be skewskew, which means the lines do not intersect, but are not parallel as well. For an example,

If the red line is down in the xy-plane i.e. z=0, and the blue line is above the xy-plane, but parallel to the xy-plane the two lines never intersect and are not parallel.

Determine if the lines are parallel or identical.

tz

ty

tx

4

22

3:1Line

tz

ty

tx

21

42

25:2Line

First look at the direction vectors:

2,4,2and1,2,1 21 vv

Since , the lines are parallel. 12 2vv

Now we must determine if they are identical.So we need to determine if they pass through the same points.This can be verified if the two sets of parametric equations produce the same points for different 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 i.e. x=3 and solve for t.

122253 tttNow let t=1 for Line 2 and the point (3, 2, -1) is produced.

By inspecting point (3, 2, 4) on line 1 and point (3, 2, -1) for line 2, it can be seen that the z-coordinates are not equal, implying the lines are not identical.

Determine if the lines intersect. If so, find the point of intersection.

tt z-t y-: xLine 2145321

vv zv y: xLine 342312

Direction vectors:

Since , the lines are not parallel. Thus they either intersect or they are skew lines.

3,2,12,4,3 21 vv

12 vkv

The line will intersect if there are values of t (for Line 1) and v (for line 2) that give us the same point.

Hence, we will consider the following system of three equations as follows:

33 132 vtv or tx:

224 2345 vtv or ty:

532 3421 vtv or tz:

Solving the first two equations simultaneously yields t=2 and v= -3

Substituting into the third equation, 2t+3v= -5, we will have: 2(2)+3(-3)= -5. => - 5 = - 5. Since, t =2 and v= -3 satisfy all three equations, which

implies that lines intersect. IMPORTANT: If the solution of first two equations i.e. t=2 & v=-3, had not

satisfied the third equation, then lines would have no point of intersection.

Determine if the lines intersect. If so, find the point of intersection and the cosine of the angle of intersection.

tz

ty

tx

2

53

4:2Line

tz

ty

tx

4

2

23:1Line

Direction vectors:

Since , the lines are not parallel. Thus they either intersect or they are skew lines.

1,5,11,2,2 21 vv

12 vkv

Keep in mind that the lines may have a point of intersection or a common point, but not necessarily for the same value of t. So equate each coordinate, but replace the t in Line 2 with an s.

stz

sty

stx

24:

532:

423: System of 3 equations with 2 unknowns – Solve the first 2 and check with the 3rd equation.

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 intersect at this point.

Recall from previous lecture, the dot product,

ba

ba

1cos

Thus,

706.039

11

279

1102cos

151122

1,5,11,2,2cos

222222

Planes in SpaceIn previous lectures we have looked at planes in space. For example, we looked at 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 ,,

For any point Q, in the plane, the vector from P to Q , i.e.

is also in the plane.

zyx ,,

n

P

Q000 ,, zzyyxxPQ

Planes in Space (contd)Since the vector from P to Q is in the plane, are perpendicular and their dot product must be equal zero, i.e.

n

P

Q

This last equation is the equation of the highlighted plane.So the equation of any plane can be found from a point in the plane and a vector normal to the plane.

nPQ and

0

0,,,,

0

000

000

zzcyybxxa

zzyyxxcba

PQn

Standard & general Equation of a PlaneThe standard equation of a plane containing the point and having normal vector, is

Note: The equation can be simplified by collecting like terms.

This results in the general form:

000 ,, zyx

cban ,,

0000 zzcyybxxa

0 dczbyax

Given the normal vector, <3, 1, -2> to the plane containing the point (2, 3, -1), write the equation of the plane in both standard form and general form.

Standard Form Equation 0000 zzcyybxxa

0123123 zyx

To obtain General Form, simplify.

01123

or

022363

zyx

zyx

Example

Given the points a (1, 2, -1), b (4, 0,3) and c (2, -1, 5) in a plane, find the equation of the plane in general form.

To write the equation of the plane we need a point (we have three) and a vector normal to the plane.

Example: 2

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 a vector normal to both those vectors and thus normal to the plane.

Two vectors: From a (1, 2, -1) to b (4, 0, 3): ab = < 4-1, 0-2, 3+1 > = <3,-2,4> From a (1, 2, -1) to c(2, -1, 5): ac= < 2-1, -1-2, 5+1 > = <1,-3,6>

Their cross product:

kjkji

kji

7147140

631

423

032

or

021714

01721410

zy

zy

zyx

Equation of the plane using a (1, 2, -1):

Exercise

Prove that planes a: 2x - 3y - z- 5 = 0 and b: -6x + 9y + 3z +2 =0 are parallel. The planes aa and bb have normal vector (2, -3, -1) and (-6, 9, 3) respectively.

This implies b= -3a. i.e. the vectors a and b representing corresponding normal vectors to the planes aa and bb are parallel, and hence planes are parallel as well.

ExerciseFind an equation of the plane through P (5, -2, 4) that is parallel to 3x + y – 6z +8= 0 and

The plane (3x +y -6z +8 =0) has a normal vector (3, 1, -6) Hence, an equation of a parallel plane can be given as:

3x + y – 6z + d =0 ; for some real number d

if P (5, -2, 4) is on this plane, then it must satisfy the equation of the plane i.e.

3x + y – 6z + d =0 => 3(5)+1(-2)-6(4)+d=0 => d= 11

Thus equation of parallel plane is 3x + y – 6z + 11 =0

Sketching Planes in Space it is really easy to find three points on plane i.e. the points

of intersection of the plane with the coordinate axes.

For example, let’s sketch the plane, x + 3y + 4z – 12 = 0

The x-intercept (where the plane intersects the x-axis) occurs when both yy and zz equal to 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 pair with a straight line in each coordinate plane. Each of these lines is called a trace.

Sketch of the plane x + 3y + 4z – 12 = 0 with intercepts, (12, 0, 0), (0, 4, 0) and (0, 0, 3).

y

x

z

Trace in xy-plane

y

x

z

Another way to graph the plane x + 3y + 4z – 12 = 0 is by using the traces.traces. The traces are the lines of intersection the plane has with each of the coordinate planes.

The xy-trace is found by letting z = 0, x + 3y = 12 is a line the the xy-plane. Graph this line.

y

x

z

Similarly, the yz-trace is 3y + 4z = 12, and the xz-trace is x + 4z = 12. Graph each of these in their respective coordinate planes.

Sketch a graph of the plane 2x – 4y + 4z – 12 = 0.

The intercepts are (6, 0, 0), (0, -3, 0) and (0, 0, 3). Plot each of these and connect each pair with a straight line.

y

x

z

The intercepts are (6, 0, 0), (0, -3, 0) and (0, 0, 3). Plot each of these and connect each pair with a straight line.

Not all planes have x, y and z intercepts. Any plane whose equation is missing one 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 is parallel to the yz-plane.

The plane is outlined in blue and is at the x value of 3.

Intersection of PlanesAny two planes that are not parallel or identical will intersect in a line and to find the line, we need to solve their equations simultaneously.

For example in the above figure , the white plane and the yellow plane intersect along the blue line.

Find the parametric equation of line of intersection for the planes x + 3y + 4z = 0 and x – 3y +2z = 0.

zyzy

zyxzyx

zyxzyx

31

or026

023023

0430431

Back substitute y into one of the first equations and solve for x.

zx

zzx

zzx

3

04

0431

3

By letting zz take on every real value tt, i.e. z = t, the parametric equations for the line are

tztytx

and31

,3

To find the common intersection, solve the equations simultaneously. Multiply the first equation by –1 and add the two to eliminate x.

Exercise

Find the parametric equation of line of intersection for the planes 2x - y + 4z = 4 and x + 3y - 2z = 1.

By letting zz take on every real value tt, i.e. z = t, the parametric equations for the line are:

t; zt and t, yx 7

8

7

2

7

10

7

13

ANGLE OF INTERSECTION BETWEEN TWO PLANES

Find the angle of intersection between the planes 2x - 4y + 4z = 7 and 6x + 2y - 3z = 2. The angle between corresponding normals to two

planes i.e. n1(2, -4, 4) and n2 (6, 2, -3) will yield the angle between two planes.

Recall, from previous lectures, dot product formula:

21

21coscos 11

nn

nn

ba

ba

Answer: 79 Degrees

Distance Between a Point and a Plane

P

Q

n, normal

Projection of PQ onto the normal to 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 are interested in finding the distance from the point Q to the plane that contains the point P.

We can find the distance between the point, Q, and the plane by projecting the vector from P to Q onto the normal to the plane and then finding its magnitude or length.

Distance Between a Point and a Plane (contd)If the distance from Q to the plane is the length or the magnitude of the projection of the vector PQ onto the normal, we can write that mathematically:

PQprojn

planethetoQfromDistance

Now, recall from previous lecture,

nn

nPQPQproj

n

2

So taking the magnitude of this vector, we get:

n

nPQn

n

nPQn

n

nPQPQproj

n

22

Distance Between a Point and a Plane: Summary

The distance from a plane containing the point P to a point Q not in the plane is

n

nPQPQprojD

n

where n is a normal to the plane.

Find the distance between the point Q (3, 1, -5) to the plane 4x + 2y – z = 8.We know the normal to the plane is <4, 2, -1> from the general form of a plane. We can find a point in the plane simply by letting x and y equal 0 and solving for z: P (0, 0, -8) is a point in the plane.

Thus the vector, PQ = <3-0, 1-0, -5-(-8)> = <3, 1, 3>

Now that we have the vector PQ and the normal, we simply use the formula for the distance between a point and a plane.

4.221

11

1416

3212

124

1,2,43,1,3222

D

n

nPQPQprojD

n

Let’s look at another way to write the distance from a point to a plane. If the equation of the plane is ax + by + cz + d = 0, then we know the normal to the plane is the vector, <a, b, c> .

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 the plane, so

and the dot product can be rewritten:111111 or0 czbyaxddczbyax

dczbyaxnPQ 000

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 Q can be written as:

222

000

cba

dczbyaxPQprojD

n

000 ,, zyx

Now that you have two formulas for the distance between a point and a plane, let’s consider the second case, the distance between a point and a line.

Distance Between a Point and a LineIn 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

So, sinorsin PQDPQ

D

We know from notes on cross products that

.andbetweenangletheiswhere,sin vuvuvu

Thus,

sin

bysidesbothdividingor

sin

PQu

uPQ

u

uPQuPQ

So if, then from above, .sinPQD u

uPQD

Summary: Distance Between a Point and a Line

The distance, D, between a line and a point Q not on the line is given by

u

uPQD

where u is the direction vector of the line and P is a point on the line.

Find the distance between the point Q (1, 3, -2) and the line given by the parametric equations:

tztytx 23and1,2

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:

12.229

6

27

211

333222

222

u

uPQD

kji

kji

uPQ 333

211

541

Using the distance formula:

Cylindrical CoordinatesAs with two dimensional space the standard coordinate system is called the Cartesian coordinate system.

In the last two sections of this sections we’ll be looking at some alternate coordinates systems for three dimensional space.

This one is fairly simple as it is nothing more than an extension of polar coordinates into three dimensions.

All that we do is add a z on as the third coordinate. The r and θ are the same as with polar coordinates.

Cylindrical Coordinates (contd.)Here is a sketch of a point in R3

The conversions for x and y are the same conversions that we used back in when we were looking at polar coordinates.

So, if we have a point in cylindrical coordinates the Cartesian coordinates can be found by using the following conversions.

The third equation is just an acknowledgement that the z-coordinate of a point in Cartesian and polar coordinates is the same. zz

θry

θrx

sin

cos

Likewise, if we have a point in Cartesian coordinates the cylindrical coordinates can be found by using the following conversions.

zz

x

y

yxryxr

1

2222

tan

or

Let’s take a quick look at some surfaces in cylindrical coordinates.

Cylindrical Coordinates (contd.)

Spherical Coordinates

0.

require will weandpoint the to

origin thefrom distance theis This

r.by

denoted above line theand

axis- xpositive ebetween th

angle theisIt s.coordinate

ndricalpolar/cyliin saw we

that angle same theis This

0 require will We

point. theorigin to thefrom line theand

axis-z positive ebetween th angle theis This

We should first derive some conversion formulas. Let’s first start with a point in spherical coordinates and ask what the cylindrical coordinates of the point are. So, we know

Of course we really only need to find r and z since, theta is the same in both coordinate systems


zr ,, find what toand ,,

that,seecan weso, is and zbetween angle

that thesee egeometry w little a With sketch.our in aboveshown

ngleright tria at the lookingby our work of all do toable be willWe

sin

cos

r

z and these are exactly the formulas that we were looking for

So, given a point in spherical coordinates the cylindrical coordinates of the point will be:

It can noted that

Or


cos

sin

z

r

Next, let’s find the Cartesian coordinates of the same point. To do this we’ll start with the cylindrical conversion formulas from the previous section.

Now all that we need to do is use the formulas from above for r and z to get:

zz

θry

θrx

sin

cos

cos

sinsin

cossin

z

θry

θrx


get ;r

know wesince that note 222 weyx

Also

2222 zyx

Example: 1

s.coordinate spherical

tolcylindrica from 2,4,6 int potheConvert

:follows as and find letsNext

. with thatanything do toneedt don’ weso and systems coordinate

both in same theis that ingacknowledgby start llWe’

22826 22 zr

cos usecan we, find zTo

3222cosz cos 1-1-

So, the spherical coordinates of this point will are

Example: 1

3,

4,22 ,,

Example: 2

s.coordinate spherical (b) and lCylindrica (a)

toCartesian from 2,-2,1 int potheConvert

(a) Cartesian to Cylindrical Coordinates

)1,4

,22(z,r, are scoordinate lCylindrica Hence

1z i.e. sameremain willcoordinate z Whereas,

4tan

22

1

22

x

y

yxr

Example: 2 (Contd.)(b) Cartesian to Spherical Coordinates

:follows as and find letsNext

. with thatanything do toneedt don’ weso and systems coordinate

both in same theis that ingacknowledgby start llWe’

3 222 zyx

cos usecan we, find zTo

5.7031cosz cos 1-1-

So, the spherical coordinates of this point will are

5.70,

4,3 ,,

lect analytical geometry in 3 d space

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