MAT 1339-S14 Class 10 & 11

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1 MAT 1339-S14 Class 10 & 11 August 7 & 11, 2014 Contents 8 Lines and Planes Equations of Lines in Two-Space and Three-Space Equations of Planes Properties of Planes Intersection of Lines in Two-Space and Three-Space Intersection of Lines and Planes Intersection of Planes Lines and Planes 8.1 Equations of Lines in Two-Space and Three-Space Lines and planes are geometrical objects. Equations of lines and planes are their algebraic representations. The key to link the geometrical objects and their algebraic equations is that any point on the lines or planes satisfies the corresponding equations. Slope-Intercept Equation In two-sapce, a line can be defined by an equation in slope-intercept form as following y = mx + b where m is the slope of the line and b is the y-intercept. Scalar Equation The scalar equation (also called the standard equation) of a line in two-space is is a normal vector to the line. ax + by + c = 0, where n = [a, b] 1

2 Definition 8.1. A normal vector to a line is a vector that is perpendicular to the line. Example 8.2. Consider the line x + 2y + 1 = 0. How can we check that n = [1, 2] is perpendicular to the line? Solution: Choose any two points on the line, for example (0, 1 ) and ( 1, 0) are two points on 2 the line. Given two points, we get a vector [ m = [x 2 x 1, y 2 y 1 ] = 1, 1 ] 2 and this vector is parallel to the line. The dot product n m = [1, 2] [ 1, 1 ] = 0 implies that n is perpendicular to m. 2 We conclude that the n is perpendicular to the line x + 2y + 1 = 0. Vector Equation The vector equation of a line in two-space is where t R is a scalar, r = r 0 + t m or [x, y] = [x 0, y 0 ] + t[m 1, m 2 ] r = [x, y] is a vector corresponding to any unknown point on the line, r 0 = [x 0, y 0 ] is a vector corresponding to any known point on the line, m = [m 1, m 2 ] is a direction vector parallel to the line. Example 8.3. Write the vector equation of the line x + 2y + 1 = 0. Solution: The direction vector of this line is m = [x 2 x 1, y 2 y 1 ] = [ 1, 1 ] 2 and we know the point ( 1, 0) is on the line. So the vector equation of the line is r = [ 1, 0] + t[ 1, 1 ]. 2 Parametric Equation The parametric equation of a line in two-space is x = x 0 + tm 1 where t R is the parameter. y = y 0 + tm 2 2

3 Example 8.4. Consider the line a) Find two points on the line. b) Write the vector equation of the line. c) Write the scalar equation of the line. Solution: x = 3 + 2t y = 5 + 4t. a) Any values of t will produce a point on the line. Let t = 0, we get a point (3, 5). Let t = 1, we get another point (5, 1). b) The vector equation of the line is [x, y] = [3, 5] + t[2, 4]. c) To get the scalar equation of the line, we need to isolate t in the parametric equation. x = 3 + 2t = t = x 3 2 y = 5 + 4t = t = y x 3 = y + 5 4x 12 = 2y Thus the scalar equation of the line is Equations of Lines in Three-Space 2x y 11 = 0. The vector equation and the parametric equation of lines in two-space can be easily generalized to the vector equation and the parametric equation of lines in three-space. Vector Equation The vector equation of a line in three-space is where r = r 0 + t m or [x, y, z] = [x 0, y 0, z 0 ] + t[m 1, m 2, m 3 ] t R is a scalar, r = [x, y, z] is a vector corresponding to any unknown point on the line, r 0 = [x 0, y 0, z 0 ] is a vector corresponding to any known point on the line, m = [m 1, m 2, m 3 ] is a direction vector parallel to the line. 3

4 Parametric Equation The parametric equation of a line in two-space is x = x 0 + tm 1 where t R is the parameter. y = y 0 + tm 2 z = z 0 + tm 3 Example 8.5. A line passes through points (2, 1, 0) and (1, 0, 3). a) Write the vector equation of the line. b) Write the parametric equation of the line. c) Determine whether the point (3, 1, 2) lies on the line. Solution: a) Given two points (2, 1, 0) and (1, 0, 3), we get the direction vector Thus the vector equation of the line is b) The parametric equation of the line is [1, 0, 3] [2, 1, 0] = [ 1, 1, 3]. [x, y, z] = [2, 1, 0] + t[ 1, 1, 3]. x = 2 t y = 1 t z = 0 + 3t. c) If (3, 1, 2) lies on the line, then there exists some t R such that So [3, 1, 2] = [2, 1, 0] + t[ 1, 1, 3]. 3 = 2 t = t = 1 1 = 1 t = t = 0 2 = 0 + 3t = t = 2 3. The t-values are not equal, thus the point (3, 1, 2) does not lie on the line. Example 8.6. Find the vector equation of the line in two-space which is perpendicular to 4x 3y = 17 and passes through the point ( 2, 4). 4

5 Solution: We know that n = [4, 3] is a normal vector of 4x 3y = 17. So n is perpendicular to 4x 3y = 17 and n gives a direction vector of the line we want. The vector equation of the line is where t R is a scalar. [x, y] = [ 2, 4] + t[4, 3], Example 8.7. Find the parametric equation of the line in three-space which is parallel to the z-axis and passes through the point (1, 5, 10). Solution: The vector [0, 0, 1] is parallel to the z-axis, thus gives a direction vector of the line we want. The parametric equation of the line is Note x = 1 + 0t y = 5 + 0t z = t where t R is a parameter. a) A line in two-space or three-space is determined by two distinct points on the line, or a point on the line and a direction vector parallel to the line. b) There is no slope-intercept equation of a line in three-space because a line in threespace does not necessarily intersect with the axes. c) There is no scalar equation of a line in three-space because a line in three-sapce has infinite number of normal vectors which are not necessarily parallel to one another. 8.2 Equations of Planes Scalar Equation The scalar equation (also called the standard equation) of a plane in three-space is ax + by + cz + d = 0, where n = [a, b, c] is a normal vector to the plane. Definition 8.8. A normal vector to a plane is a vector that is perpendicular to the plane. Example 8.9. Consider the plane x + 2y + z + 1 = 0. How can we check that n = [1, 2, 1] 5

6 is perpendicular to the plane? Choose any two points on the plane, for example (0, 1, 0) and ( 1, 1, 2) are two 2 points on the plane. Given two points, we get a vector m = [x 2 x 1, y 2 y 1, z 2 z 1 ] = [ 1, 32 ], 2 and this vector is parallel to the plane. The dot product n m = [1, 2, 1] [ 1, 32 ], 2 = 0 implies that n is perpendicular to m. So n is perpendicular to the plane x+2y+z+1 = 0. Vector Equation The vector equation of a plane in three-space is r = r 0 + t a + s b or [x, y, z] = [x 0, y 0, z 0 ] + t[a 1, a 2, a 3 ] + s[b 1, b 2, b 3 ] where t, s R are scalars, r = [x, y, z] is a vector corresponding to any unknown point on the line, r 0 = [x 0, y 0, z 0 ] is a vector corresponding to any known point on the line, a = [a 1, a 2, a 3 ] and b = [b 1, b 2, b 3 ] are two non-collienar direction vectors parallel to the plane. Parametric Equation The parametric equation of a plane in three-space is x = x 0 + ta 1 + sb 1 where t, s R are the parameters. y = y 0 + ta 2 + sb 2 z = z 0 + ta 3 + sb 3 Example Consider the plane with direction vectors a = [1, 2, 3] and b = [2, 1, 4] passing through the point (2, 0, 1). a) Write the vector equation of the plane. b) Write the parametric equation of the plane. c) Find two other points on the plane. d) Find the y-intercept of the plane. 6

7 Solution: a) The vector equation of the plane is [x, y, z] = [2, 0, 1] + t[1, 2, 3] + s[2, 1, 4]. b) The parametric equation of the plane is x = 2 + t + 2s y = 0 + 2t s z = 1 3t + 4s. c) Any combination of values of the parameters t and s will produce a point on the plane. Let t = 0 and s = 1, we get a point (4, 1, 5). Let t = 1 and s = 0, we get another point (3, 2, 2). d) In order to find the y-intercept of the plane, we let x = 0 and z = 0 and solve for t and s. Let 0 = 2 + t + 2s 0 = 1 3t + 4s. we get t = 3 5 and s = Then y = 0 + 2t s = Properties of Planes For the previous example Example How to write the scaler equation of the plane! Note: A plane in three-space is determined by three non-collinear points on the plane or by a point on the plane and two non-collienar direction vectors parallel to the plane. 7

8 Example 8.11 (Example 2 page 457). Find the scalar equation of the plane containing the points A( 3, 1, 2), B(4, 6, 2), and C(5, 4, 1). Key concepts: The scalar equation of a plane is three space is Ax + By + Cz + D = 0, where n = [A, B, C] is a normal vector to the plane. Any vector parallel to the normal vector of a plane is also normal to the plane. The coordinates of any point on the plane satisfy the scalar equation. A normal vector (for orientation) and a point (for position) can be used to define a plane. 8

9 Class 11 August 11, Intersection of Lines in Two-Space and Three-Space Intersection of Lines in Two Lines in Two-Space Geometrically there are three possibilities for the intersection of two lines in two-space. Intersect at a point Parallel Coincident Algebraically there are three possibilities for the solutions of the following system of equations. Unique solution No solution Infinitely many solutions a 1 x + b 1 y + c 1 = 0 a 2 x + b 2 y + c 2 = 0 Example (1) The system of equations x y + 1 = 0 2x + y + 4 = 0 9

10 has unique solution x = 5 3 and y = 2 3. We can also write the solution as a point (x, y) = ( 5, 2 ) or as a vector 3 3 [ [x, y] = 5 ] 3, 2. 3 (2) The system of equations x y + 1 = 0 x y + 1 = 0 has infinitely many solutions since all values of x = t, let y = t + 1, then is a solution for the system. (3) The system of equations [t, t + 1] x y + 1 = 0 x y 2 = 0 has no solution. There is no value of x and y satisfying both equations. Intersection of Two Lines in Three-Space Geometrically there are four possibilities for the intersection of two lines in three-space. Intersect at a point Coincident Parallel Skew (not parallel and not intersect) 10

11 Algebraically there are three possibilities for the solutions of the following system of equations. [x, y, z] = [x 0, y 0, z 0 ] + s[m 1, m 2, m 3 ] Unique solution (intersect at a point) Infinitely many solutions (coincident) No solution (parallel or skew) Example The system of equations [x, y, z] = [a 0, b 0, c 0 ] + t[n 1, n 2, n 3 ] [x, y, z] = [7, 2, 6] + s[2, 1, 3] 1 [x, y, z] = [3, 9, 13] + t[1, 5, 5] 2 has a unique solution. From 1 we get [x, y, z] = [7 + 2s, 2 + s, 6 3s]. From 2 we get [x, y, z] = [3 + t, 9 + 5t, t]. Thus we get another system of equations in terms of s and t as following s = 3 + t 2 + s = 9 + 5t 6 3s = t Solve this system: We get s = 3 and t = 2. Thus the unique solution of the original system is [x, y, z] = [7, 2, 6] + ( 3)[2, 1, 3] = [1, 1, 3] and the two lines intersect at the point (1, 1, 3). 11

12 8.5 Intersection of Lines and Planes Geometrically there are three possibilities for the intersection of a line and a plane in three-space. Intersect at a point The line lies on the plane Parallel Algebraically there are three possibilities for the solutions of the following system of equations. [x, y, z] = [x 0, y 0, z 0 ] + t[m 1, m 2, m 3 ] Unique solution (intersect at a point) ax + by + cz + d = 0 Infinitely many solutions (the line lies on the plane) No solution (parallel) Example The system of equations [x, y, z] = [5, 5, 2] + t[2, 5, 3] 9x + 13y 2z =

13 has a unique solution. From 1 we can write the parametric equation of the line x = 5 + 2t y = 5 5t z = 2 + 3t. Substitute the parametric equation into 2 we get 9(5 + 2t) + 13( 5 5t) 2(2 + 3t) = 29. Expand and solve for t we get t = 1. Thus [x, y, z] = [5, 5, 2] + ( 1)[2, 5, 3] = [3, 0, 1] is the unique solution. The line and the plane intersect at the point (3, 0, 1). Example 8.15 (The Distance From a Point to a Plane, Example 3, Page 477). Consider the plane with scalar equation 4x + 2y + z 16 = 0. (a) Determine if the point P = (10, 3, 8) is on the plane. (b) Write the vector and parametric equations of the line l, which is perpendicular to the plane and passes through the point Q = (10, 3, 8). (c) Find the intersection of the line l and the plane. 13

14 (d) Determine the distance between the point Q = (10, 3, 8) and the plane. Exercise: Do the same for the point B = (2, 2, 4) 8.6 Intersection of Planes Intersection of Two Planes in Three-Space Geometrically there are three possibilities for the intersection of two planes in three-space. Intersect in a line Coincident Parallel 14

15 Algebraically there are two possibilities for the solutions of the following system of equations. a 1 x + b 1 y + c 1 z + d 1 = 0 a 2 x + b 2 y + c 2 z + d 2 = 0 Infinitely many solutions (intersect in a line or coincident) No solution (parallel) Example 8.16 (Example 1, Page 483). Describe how the planes in each pair intersect.. a) π 1 : 2x y + z 1 = 0 1 π 2 : x + y + z 6 = 0 2 The normal vector for the plane 1 is n 1 = [2, 1, 1] and the normal vector for the plane 2 is n 2 = [1, 1, 1]. Since n 2 is not a scalar multiple of n 1, we know that n 2 is not parallel to n 1. Thus the two planes must intersect. To solve for the intersection points. We expect the result to be a line of intersection. We want to write the parametric equation and the vector equation of the expected intersection line. Now, 1 2, we get x 2y + 5 = 0 3. We introduce a parameter by letting y = t. Then by 3 we get x = 5 + 2t. By 2 we have z = 6 x y = 6 ( 5 + 2t) t = 11 3t. So the parametric equation of the line is The vector equation of the line is x = 5 + 2t y = 0 + t z = 11 3t. [x, y, z] = [ 5, 0, 11] + t[2, 1, 3]. Question: Is the Two planes π 1 and π 2 are perpendicular? why? 15

16 b) π 3 : 2x 6y + 4z 7 = 0 1 π 4 : 3x 9y + 6z 2 = 0 2 c) π 5 : x + y 2z + 2 = 0 1 π 6 : 2x + 2y 4z + 4 =

17 Intersection of Three Planes in Three-Space Geometrically there are eight possibilities for the intersection of two planes in three-space. 1 Interest at a point 2 Intersect in a line Note about the normals: 3 Three planes are coincident. 4 Two planes are coincident and the third plane is not parallel. Note about the normals: 5 Three planes are parallel. 6 Two planes are coincident and the third plane is parallel to the first two planes. 17

18 Note about the normals: 7 Two planes are parallel and the third plane is not parallel. 8 Pairs of planes intersect in lines that are parallel. Note about the normals: Theorem Three normal vectors n 1, n 2 and n 3 are coplanar if n 1 n 2 n 3 = 0 Why? what cases does this cover? Algebraically there are three possibilities for the solutions of the following system of equations. a 1 x + b 1 y + c 1 z + d 1 = 0 a 2 x + b 2 y + c 2 z + d 2 = 0 a 3 x + b 3 y + c 3 z + d 3 = 0 18

19 Unique solution (1) Note about the normals: Infinitely many solutions (2-4) Note about the normals: No solution (5-8) Note about the normals: Example 8.18 (Example 2 page 486). For each set of planes, describe the number of solutions and how the planes intersect. a) π 1 : x 5y + 2z 10 = 0 1 π 2 : x + 7y 2z + 6 = 0 2 π 3 : 8x + 5y + z 20 = 0 3 Solution: Discuss normals: Let 2 1, we get 12y 4z + 16 = 0 4. Let ( 8) 1 + 3, we get 45y 15z + 60 = 0 5. Let , we get 0 = 0 0 which is true for all values of y and z. 19

20 ** = ** ( 2 1 ) (( 8) ) = 0 which implies 3 = 15 We introduce a parameter by letting z = t. Then y = t by 4 or 5. By 1, 3 3 we get x = 5y 2z + 10 = 5( t) 2t + 10 = 1 t. So the parametric equation of the line is The vector equation of the line is b) [x, y, z] = x = t y = t z = 0 + t. [ ] 10 3, 4 3, 0 + t [ 13, 13 ], 1. π 4 : 2x + y + 6z 7 = 0 1 π 5 : 3x + 4y + 3z + 8 = 0 2 π 6 : x 2y 4z 9 =