7.5 lines and_planes_in_space

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7.5 lines and_planes_in_space

  1. 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. 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 spaceo Planes in Space o Standard form and General form of a plane o Sketching planes using traces oThe line of intersection of two planeso Distances in Space o The distance between a point and a plane o The distance between a point and a line
  3. 3. Lines in SpacePreviously 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 adirection, except in 3-dimensional space the direction will be given by aparallel vector, called the direction vector.
  4. 4. Lines in SpaceTo determine the equation of the line passing through the point P ( x 0 , y 0 , z 0 )and parallel to the direction vector, v = a , b , c , we will use ourknowledge that parallel vectors are scalar multiples. Thus, the vectorthrough P and any other point Q ( x , y , z ) on the line is a scalar multipleof the direction vector, v = a , b , c .In other words, PQ = t a , b , c , where t is any real number ( scalar ) or x − x 0 , y − y 0 , z − z 0 = at , bt , ct
  5. 5. Equations of Lines in SpaceEquate the respective components and there are three equations. x − x 0 = at , y − y 0 = bt and z − z 0 = ct or x = x 0 + at , y = y 0 + bt and z = z 0 + ctThese equations are called the parametric equations of the line.If the components of the direction vector are all nonzero, eachequation can be solved for the parameter t and then the three can beset equal. x − x0 y − y0 z − z0 = = a b c These equations are called the symmetric equations of the line.
  6. 6. Equations of Lines in SpaceA line passing through the point P ( x 0 , y 0 , z 0 ) and parallel to the vector,v = a , b , c is represented by the parametric equations: x = x 0 + at , y = y 0 + bt and z = z 0 + ctAnd if all three components of the direction vector are nonzero, the lineis also represented by the symmetric equations: x − x0 y − y0 z − z0 = = a b c
  7. 7. Example 1: Find the parametric and symmetric equations of the line passingthrough the point (2, 3, -4) and parallel to the vector, <-1, 2, 5> .Solution: Simply use the parametric and symmetric equations for any line givena point on the line and the direction vector.Parametric Equations: x = 2 − t , y = 3 + 2t and z = −4 + 5t Symmetric Equations: x −2 y −3 z + 4 = = −1 2 5
  8. 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. direction vector v = 3 − 1, − 2 − 2, 5 + 2 = 2, − 4, 7 parametric equations : x = 1 + 2t , y = 2 − 4t and z = −2 + 7t x −1 y −2 z +2 symmetric equations : = = 2 −4 7Notes: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. 9. Relationships Between LinesIn a 2-dimensional coordinate system, there were three possibilities whenconsidering two lines: intersecting lines, parallel lines and the two wereactually 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 seethe 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. 10. Example 3: Determine if the lines are parallel or identical. Line 1 : x = 3 +t Line 2 : x = 5 − 2t y = 2 − 2t y = −2 + 4t z = 4 +t z = 1 − 2tSolution: First look at the direction vectors: v1 = 1,−2,1 and v 2 = − 2,4,−2 Sincev 2 = −2v 1 , 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. 3 = 5 − 2t → − 2 = − 2t → t = 1 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. 11. Example 4: Determine if the lines intersect. If so, find the point ofintersection and the cosine of the angle of intersection. Line 1 : x = 3 + 2t Line 2 : x = 4 −t y = −2t y = 3 + 5t z = 4 −t z = 2 −t Solution: Direction vectors: v1 = 2,− 2,− 1 v 2 = − 1,5,−1 Since v 2 ≠ k ⋅ v1 , 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. x : 3 + 2t = 4 − s System of 3 equations with 2 y : − 2t = 3 + 5s unknowns – Solve the first 2 and check with the 3rd equation. z : 4 −t = 2 − s
  12. 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) The lines intersect Line 2: s = -1 produces the point (5, -2, 3) at this point. Recall from lesson 7.3 on the dot product, u ⋅vcos θ = , where θ is the angle between u and v u ⋅vThe angle θ between two intersecting lines should be less than 90  ,so we use absolute value in the numerator.
  13. 13. Solution to Example 4 Continued:Thus, 2,−2,−1 ⋅ − 1,5,−1 cos θ = 22 + ( − 2) + ( − 1) ⋅ ( − 1 ) 2 + 52 + ( − 1 ) 2 2 2 − 2 − 10 + 1 − 11 cos θ = = ≈ 0.706 9 ⋅ 27 9 3
  14. 14. Planes in SpaceIn previous sections we have looked at planes in space. For example, we lookedat 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,( x 0 , y 0 , z 0 ) , is a point in the highlighted plane and n = a , b ,c is the vectornormal to the highlighted plane. n For any point Q, ( x , y , z ) in the plane, the vector from P Q to Q , PQ = x − x 0 , y − y 0 , z − z 0 P is also in the plane.
  15. 15. Planes in SpaceSince the vector from P to Q is in the plane, PQ and n are perpendicularand their dot product must equal zero. n ⋅ PQ = 0 a , b ,c ⋅ x − x 0 , y − y 0 , z − z 0 = 0 a ( x − x0 ) + b ( y − y0 ) + c (z − z 0 ) = 0 n This last equation is the equation of the highlighted plane. Q So the equation of any plane can be found from a point in P the plane and a vector normal to the plane.
  16. 16. Standard Equation of a PlaneThe standard equation of a plane containing the point ( x 0 , y 0 , z 0 ) and havingnormal vector, n = a , b , c is a ( x − x0 ) + b ( y − y0 ) + c (z − z0 ) = 0Note: The equation can be simplified by using the distributive property andcollecting like terms.This results in the general form: ax + by + cz + d = 0
  17. 17. Example 5: Given the normal vector, <3, 1, -2> to the plane containing thepoint (2, 3, -1), write the equation of the plane in both standard form andgeneral form.Solution: Standard Form a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 3( x − 2) + 1( y − 3) − 2( z + 1) = 0 To obtain General Form, simplify. 3x − 6 + y − 3 − 2z − 2 = 0 or 3x + y − 2z − 11 = 0
  18. 18. Example 6: Given the points (1, 2, -1), (4, 0,3) and (2, -1, 5) in a plane, find theequation of the plane in general form.Solution: To write the equation of the plane we need a point (we have three) anda 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 avector 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: i j k 3 − 2 4 = 0i − 14 j − 7k = −14 j − 7k 1 −3 6 Equation of the plane: 0( x − 1) − 14( y − 2) − 7 ( z + 1) = 0 − 14y − 7z + 21 = 0 or 2y + z − 3 = 0
  19. 19. Sketching Planes in SpaceIf a plane intersects all three coordinate planes (xy-plane, yz-plane and thexz-plane), part of the plane can be sketched by finding the intercepts andconnecting them to form the plane.For example, let’s sketch the part of the plane, x + 3y + 4z – 12 = 0 thatappears in the first octant.The x-intercept (where the plane intersects the x-axis) occurs when both yand 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 pairwith a straight line in each coordinate plane. Each of these lines is called atrace.The sketch is shown on the next slide.
  20. 20. Sketch of the plane x + 3y + 4z – 12 = 0 with intercepts, (12, 0, 0), (0, 4, 0)and (0, 0, 3). z y Now you can see the triangular part of the plane that appears in the first octant. x
  21. 21. Another way to graph the plane x + 3y + 4z – 12 = 0 is by using the traces. Thetraces are the lines of intersection the plane has with each of the coordinateplanes.The xy-trace is found by letting z = 0, x + 3y = 12 is a line the the xy-plane.Graph this line. z y x
  22. 22. Similarly, the yz-trace is 3y + 4z = 12, and the xz-trace is x + 4z = 12. Grapheach of these in their respective coordinate planes. z y x
  23. 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 ofthese and connect each pair with a straight line.
  24. 24. 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 ofthese and connect each pair with a straight line. z y Hopefully you can see the part of the plane we have sketched appears on the x negative side of the y-axis.
  25. 25. More on Sketching Planes 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.Any plane whose equation is missing two variables is parallel to thecoordinate plane of the missing variables. For example, 2x – 6 = 0 or x = 3 isparallel to the yz-plane. The plane is outlined in blue and is at the x value of 3.
  26. 26. Intersecting PlanesAny two planes that are not parallel or identical will intersect in a line and tofind 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. 27. Example 8: Find the line of intersection for the planes x + 3y + 4z = 0and x – 3y +2z = 0.Solution: To find the common intersection, solve the equations simultaneously. Multiply the first equation by –1 and add the two to eliminate x. − 1 ⋅ ( x + 3y + 4z = 0 ) → − x − 3y − 4z = 0 x − 3y + 2z = 0 → x − 3y + 2z = 0 −1 − 6y − 2z = 0 or y = z 3 Back substitute y into one of the first equations and solve for x. −1  x + 3 ⋅  z  + 4z = 0  3  x − z + 4z = 0 x = −3z Finally if you let z = t, the parametric equations for the line are −1 x = −3 , y = t t and z =t 3
  28. 28. Distance Between a Point and a Plane 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. n, normal QProjection of PQonto the normalto the plane P Thus the distance from Q to the plane is the length or the magnitude of the projection of the vector PQ onto the normal.
  29. 29. Distance Between a Point and a PlaneIf the distance from Q to the plane is the length or the magnitude of theprojection of the vector PQ onto the normal, we can write thatmathematically: Distance from Q to the plane = projn PQ  Now, recall from section 7.3,  PQ ⋅ n  projn PQ =  2  ⋅n  n    So taking the magnitude of this vector, we get:   PQ ⋅ n PQ ⋅ n  PQ ⋅ n  projn PQ =  2  ⋅n = 2 ⋅n =  n  n n  
  30. 30. Distance Between a Point and a PlaneThe distance from a plane containing the point P to a point Q not in theplane is PQ ⋅ n D = projn PQ = n where n is a normal to the plane.
  31. 31. Example 9: Find the distance between the point Q (3, 1, -5) to the plane4x + 2y – z = 8.Solution: We know the normal to the plane is <4, 2, -1> from the generalform of a plane. We can find a point in the plane simply by letting x and yequal 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 formulafor the distance between a point and a plane. PQ ⋅ n 3,1,3 ⋅ 4,2,−1 D = projn PQ = = n 42 + 22 + ( −1) 2 12 + 2 − 3 11 D = = ≈ 2.4 16 + 4 + 1 21
  32. 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 = ( x1 , y1 , z 1 ) and Q be the point not in the plane, Q = ( x 0 , y 0 , z 0 ) . Then the vector, PQ = x 0 − x1 , y 0 − y1 , z 0 − z 1 So now the dot product of PQ and n becomes: PQ ⋅ n = a , b , c ⋅ x 0 − x1 , y 0 − y1 , z 0 − z 1 = a ( x 0 − x1 ) + b ( y 0 − y1 ) + c ( z 0 − z 1 ) = ax 0 + by 0 + cz 0 − ax1 − by1 − cz 1Note that since P is a point on the plane it will satisfy the equation of theplane, so ax1 + by1 + cz 1 + d = 0 or d = − ax1 − by 1 − cz 1 and the dotproduct can be rewritten: PQ ⋅ n = ax 0 + by 0 + cz 0 + d
  33. 33. Thus the formula for the distance can be written another way:The Distance Between a Point and a PlaneThe distance between a plane, ax + by + cz + d = 0 and a point Q ( x 0 , y 0 , z 0 )is ax 0 + by 0 +cz 0 +d D = projn PQ = a 2 + b 2 +c 2Now 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. 34. 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, uis a direction vector for the line and θ is the angle between u and PQ. Q D = Distance from Q to the line θ P u D Obviously, sin θ = or D = PQ sin θ PQ
  35. 35. We know from Section 7.4 on cross products that u ×v = u v sin θ, where θ is the angle between u and v .Thus, PQ ×u = PQ u sin θ or dividing both sides by u PQ ×u = PQ sin θ u PQ × uSo if, D = PQ sin θ then from above, D = . u
  36. 36. Distance Between a Point and a LineThe distance, D, between a line and a point Q not on the line is given by PQ × u D = uwhere u is the direction vector of the line and P is a point on the line.
  37. 37. Example 10: Find the distance between the point Q (1, 3, -2) and the linegiven by the parametric equations: x = 2 + t , y = −1 − t and z = 3 + 2tSolution: 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: i j k PQ × u = 1 − 4 5 = −3i + 3 j + 3k 1 −1 2Using the distance formula: PQ × u ( − 3) 2 + 32 + 32 27 9 D = = = = ≈ 2.12 u 12 + ( − 1) + 22 2 6 2
  38. 38. You have three sets of practice problems for this lesson inBlackboard under Chapter 7, Lesson 7.5 Parts A, B and C.

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