Lect analytical geometry in 3 d space
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Transcript of 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.
3-D COORDINATE SYSTEMS
• Look at any bottom corner of a room and call the corner the origin.
3-D COORDINATE SYSTEMS
• 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.
3-D COORDINATE SYSTEMS
• 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.
3-D COORDINATE SYSTEMS
• The z-axis runs up from the floor toward the ceiling along the intersection of the two walls.
3-D COORDINATE SYSTEMS
• 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.
3-D COORDINATE SYSTEMS
• 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.
3-D COORDINATE SYSTEMS
• 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.
3-D COORDINATE SYSTEMS
• The point P(a, b, c) determines a rectangular box.
3-D COORDINATE SYSTEMS
• 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.
3-D COORDINATE SYSTEMS
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
Spherical Coordinates
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
Spherical Coordinates
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
Spherical Coordinates
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 ,,