Like this presentation? Why not share!

# Lect analytical geometry in 3 d space

## by Haris Jawed, GOOGLE at MYTH on Sep 25, 2011

• 2,528 views

By the Students Of SZABIST...

By the Students Of SZABIST...

### Views

Total Views
2,528
Views on SlideShare
2,528
Embed Views
0

Likes
1
Downloads
79
Comments
0

No embeds

### Upload Details

Uploaded via SlideShare as Microsoft PowerPoint

### Usage Rights

© All Rights Reserved

### Report content

Edit your comment

## Lect analytical geometry in 3 d spacePresentation Transcript

• 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 Space In other words, To 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, . y z x P Q
• Equations of Lines in Space Equate the respective components and solving for x, y and z gives three equations. 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 t and then the three can be set equal . 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:
• 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: Thus , we will have:
• 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.
• Notes :
• 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) 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 and y co-ordinates of R 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
• Relationships Between Lines
• In 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 skew , 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 . First look at the direction vectors: Since , the lines are parallel .
• 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 .
Now 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. Direction vectors: Since , the lines are not parallel. Thus they either intersect or they are skew lines. 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:
• 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. Direction vectors: Since , the lines are not parallel. Thus they either intersect or they are skew lines. 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 . System of 3 equations with 2 unknowns – Solve the first 2 and check with the 3 rd 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, Thus,
• Planes in Space In 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. For any point Q, in the plane, the vector from P to Q , i.e. is also in the plane. n P Q
• 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.
• Standard & general 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 collecting like terms. This results in the general form :
• 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 To obtain General Form, simplify. 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: 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 a and b 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 a and b are parallel, and hence planes are parallel as well.
• Exercise Find 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 y and z 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. 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 Planes Any 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. Back substitute y into one of the first equations and solve for x. By letting z take on every real value t , i.e. z = t , the parametric equations for the line are 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 z take on every real value t , i.e. z = t , the parametric equations for the line are:
• 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. n 1 (2, -4, 4) and n 2 (6, 2, -3) will yield the angle between two planes.
• Recall, from previous lectures, dot product formula:
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: Now, recall from previous lecture, So taking the magnitude of this vector, we get:
• 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 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.
• 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, So now the dot product of PQ and n becomes: 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:
• 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: 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 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 So,
• We know from notes on cross products that Thus, So if, then from above, .
• 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 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: 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: Using the distance formula:
• Cylindrical Coordinates As 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 R 3 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.
• Likewise, if we have a point in Cartesian coordinates the cylindrical coordinates can be found by using the following conversions. Let’s take a quick look at some surfaces in cylindrical coordinates. Cylindrical Coordinates (contd.)
• Spherical Coordinates
• 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 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
• 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: Spherical Coordinates
• Example: 1
• So, the spherical coordinates of this point will are Example: 1
• Example: 2 (a) Cartesian to Cylindrical Coordinates
• Example: 2 (Contd.) (b) Cartesian to Spherical Coordinates So, the spherical coordinates of this point will are