1525 equations of lines in space
Download as PPT, PDF2 likes1,358 views
The document discusses lines and planes in 3D space. It defines lines as being determined by a point and direction vector, and gives parametric and symmetric equations to represent lines. Planes are defined by a point and normal vector, with standard and general forms for their equations. Methods are provided for finding the intersection of lines or planes, as well as the distance between a point and plane or line. Examples demonstrate finding equations of lines and planes, sketching planes, and determining relationships between lines or planes.
1525 equations of lines in space
- 1. Chapter 7: Vectors and the Geometry of Space Section 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
- 2. 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
- 3. 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 a direction, except in 3-dimensional space the direction will be given by a parallel vector, called the direction vector.
- 4. 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 =−−− =
- 5. 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, each equation can be solved for the parameter t and then the three can be set equal. c zz b yy a xx 000 − = − = − These equations are called the symmetric equations of the line.
- 6. 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 line is also represented by the symmetric equations: ctzzbtyyatxx +=+=+= 000 and, cbav ,,= ( )000 ,, zyx c zz b yy a xx 000 − = − = −
- 7. Example 1: Find the parametric and symmetric equations of the line passing through the point (2, 3, -4) and parallel to the vector, <-1, 2, 5> . Solution: 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 +−=+=−= Symmetric Equations: 5 4 2 3 1 2 + = − = − − zyx
- 8. Example 2: Find the parametric and symmetric equations of the line passing through the points (1, 2, -2) and (3, -2, 5). Solution: 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 +−=−=+= 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.
- 9. Relationships Between Lines In a 2-dimensional coordinate system, there were three possibilities when considering two lines: intersecting lines, parallel lines and the two were actually 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 see the picture and description below. If the red line is down in the xy- plane and the blue line is above the xy-plane, but parallel to the xy- plane the two lines never intersect and are not parallel.
- 10. 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 determine if they pass through the same points. So we need to determine 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 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.
- 11. Example 4: 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 Solution: 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.
- 12. 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 intersect at this point. Recall from lesson 7.3 on the dot product, numerator.theinvalueabsoluteuseweso ,90thanlessbeshouldlinesngintersectitwobetweenθangleThe andbetweenangletheishere,cos vuw vu vu θθ ⋅ ⋅ =
- 13. Solution to Example 4 Continued: ( ) ( ) ( ) ( ) 706.0 39 11 279 1102 cos 151122 1,5,11,2,2 cos 222222 ≈ − = ⋅ +−− = −++−⋅−+−+ −−⋅−− = θ θ Thus,
- 14. Planes in Space In previous sections 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 ,,= n P Q For any point Q, in the plane, the vector from P to Q , is also in the plane. ( )zyx ,, 000 ,, zzyyxxPQ −−−=
- 15. Planes in Space Since the vector from P to Q is in the plane, are perpendicular and their dot product must equal zero. 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
- 16. 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
- 17. Example 5: 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. Solution: Standard Form ( ) ( ) ( ) 0000 =−+−+− zzcyybxxa ( ) ( ) ( ) 0123123 =+−−+− zyx To obtain General Form, simplify. 01123 or 022363 =−−+ =−−−+− zyx zyx
- 18. Example 6: Given the points (1, 2, -1), (4, 0,3) and (2, -1, 5) in a plane, find the equation of the plane in general form. Solution: To write the equation of the plane we need a point (we have three) and a 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 a vector 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 > = <3,-2,4> From (1, 2, -1) to (2, -1, 5): < 2-1, -1-2, 5+1 > = <1,-3,6> Their cross product: kjkji kji 7147140 631 423 −−=−−= − − ( ) ( ) ( ) 032 or 021714 01721410 =−+ =+−− =+−−−− zy zy zyxEquation of the plane:
- 19. Sketching Planes in Space If a plane intersects all three coordinate planes (xy-plane, yz-plane and the xz-plane), part of the plane can be sketched by finding the intercepts and connecting them to form the plane. For example, let’s sketch the part of the plane, x + 3y + 4z – 12 = 0 that appears in the first octant. The x-intercept (where the plane intersects the x-axis) occurs when both y and 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 pair with a straight line in each coordinate plane. Each of these lines is called a trace. The sketch is shown on the next slide.
- 20. 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 triangular part of the plane that appears in the first octant.
- 21. 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.
- 22. 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.
- 23. 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 of these and connect each pair with a straight line.
- 24. 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 of these and connect each pair with a straight line. Hopefully you can see the part of the plane we have sketched appears on the negative side of the y-axis.
- 25. 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.
- 26. Intersecting Planes Any two planes that are not parallel or identical will intersect in a line and to find the line, solve the equations simultaneously. For example in the figure above, the white plane and the yellow plane intersect along the blue line.
- 27. Example 8: Find the line of intersection for the planes x + 3y + 4z = 0 and x – 3y +2z = 0. ( ) zyzy zyxzyx zyxzyx 3 1 or026 023023 0430431 − ==−− =+−→=+− =−−−→=++⋅− Back substitute y into one of the first equations and solve for x. zx zzx zzx 3 04 04 3 1 3 −= =+− =+ − ⋅+ Finally if you let z = t, the parametric equations for the line are tztytx = − =−= and 3 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.
- 28. 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.
- 29. Distance Between a Point and a Plane 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 section 7.3, n n nPQ PQprojn ⋅ ⋅ = 2 So taking the magnitude of this vector, we get: n nPQ n n nPQ n n nPQ PQprojn ⋅ =⋅ ⋅ =⋅ ⋅ = 22
- 30. Distance Between a Point and a Plane The distance from a plane containing the point P to a point Q not in the plane is n nPQ PQprojD n ⋅ == where n is a normal to the plane.
- 31. Example 9: Find the distance between the point Q (3, 1, -5) to the plane 4x + 2y – z = 8. Solution: 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. ( ) 4.2 21 11 1416 3212 124 1,2,43,1,3 222 ≈= ++ −+ = −++ −⋅ = ⋅ == D n nPQ PQprojD n
- 32. 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 ,, zyx 101010 ,, 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
- 33. 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 is 222 000 cba dczbyax PQprojD 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.
- 34. 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, θθ sinorsin PQD PQ D == θ θ
- 35. We know from Section 7.4 on cross products that .andbetweenangletheiswhere,sin vuvuvu θθ=× Thus, θ θ sin bysidesbothdividingor sin PQ u uPQ u uPQuPQ = × =× So if, then from above, .θsinPQD = u uPQ D × =
- 36. 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 uPQ D × = where u is the direction vector of the line and P is a point on the line.
- 37. Example 10: Find the distance between the point Q (1, 3, -2) and the line given 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 27 211 333 222 222 ≈== +−+ ++− = × = u uPQ D kji kji uPQ 333 211 541 ++−= − −=× Using the distance formula:
- 38. You have three sets of practice problems for this lesson in Blackboard under Chapter 7, Lesson 7.5 Parts A, B and C.