Lecture 5. Equations of Lines and Planes. Dan Nichols MATH 233, Spring 2018 University of Massachusetts.

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1 Lecture 5 Equations of Lines and Planes Dan Nichols nichols@math.umass.edu MATH 233, Spring 2018 Universit of Massachusetts Februar 6, 2018 (2) Upcoming midterm eam First midterm: Wednesda Feb. 21, 7:00-9:00 PM Location will be announced soon Practice eam problems will be available starting Thursda Covers all material through the end of net week If ou have a conflict, deadline to request a makeup eam is tomorrow!

2 (3) Last time: the cross product u v is orthogonal to both u and v Orientation determined b right-hand rule Magnitude: u v sin θ (area of parallelogram) Compute cross product using a (pseudo) determinant: i j k i j k u v = u 1 u 2 u 3 u 1 u 2 u 3 v 1 v 2 v 3 v 1 v 2 v 3 = u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1. (4) The equation of a line A line equation for a specific line L tells ou eactl which points lie on L. It s a rule that sas point (, ) is in L if and onl if and satisf this equation. In 2D space, we ve seen line equations like these: Point-slope formula: 0 = m( 0 ) Slope-intercept formula: = m + b In 2D space, ou need two things to specif a line: 1. One point on the line, ( 0, 0 ) 2. The slope (direction) of the line, m = Or reall just two numbers: slope and -intercept But in 3D space it s not so simple we can t epress the direction of a line using a single slope number.

3 (5) The equation of a line Instead of giving a line equation in terms of coordinates (, ), let s use r, the position vector of some point in space. From now on we ll often specif points b giving position vectors rather than coordinates We want to sa the point at the tip of the vector r lies on L if and onl if (some equation involving r) is true. Our equation can use vector arithmetic (i.e. dot product) This makes sense in an number of dimensions. In R 2, r =,. In R 3, r =,, z. (,, z) r =,, z (6) The equation of a line Let v be a vector parallel to the line Let a and b be the position vectors to an two points on the line Then b a is parallel to v, so it is a scalar multiple of v a b a b So we can write b a = tv for some scalar t. v

4 (7) The equation of a line Let r 0 be the position vector of some fied point on the line. Then ever other point on the line can be written with position vector r = r 0 + tv for some scalar t. r 0 + 2v r 0 0.5v r 0 r v r 0 + v v The line is the set of all points r for which r = r 0 + tv for some scalar t. This is called the vector equation of the line. (8) The equation of a line r = r 0 + tv What we just did is a form of parameterization. Each value of the parameter t gives us a point on the line Plot r = r 0 + tv for all values of t and ou get the whole line You can also separate the vector equation into its components. In R 3, let v = a, b, c and let r 0 = 0, 0, z 0. Then the equation is,, z = 0, 0, z 0 + t a, b, c, or = 0 + ta = 0 + tb z = z 0 + tc These are the parametric equations of the line L through the point P ( 0, 0, z 0 ) parallel to the vector v = a, b, c.

5 (9) The equation of a line: eample Eample 1: Find a vector equation for the line in R 3 which passes through the points P = (0, 1, 3) and Q = ( 1, 2, 2). v = P Q = 1, 1, 1 r 0 = 0, 1, 3 r = r 0 + tv = 0, 1, 3 + t 1, 1, 1 For eample, t = 0 gives us the point 0, 1, 3, and t = 4 gives us the point 4, 5, 1. (10) Direction numbers For each line L, there are infinitel man different vector equations. You can replace r 0 with an other point on L You can replace v with an other vector parallel to v However, an vector parallel to v is a scalar multiple of v, so the ratios between the components must be the same as v. For eample: v 1 = 2, 5, 3 v 2 = 1, 2.5, 1.5 The components of an vector parallel to L are called direction numbers of L; the ratios between them are like the slope of a line in 2D space.

6 (11) Smmetric equations One last wa to represent a line: first, look at the parametric equations = 0 + ta = 0 + tb z = z 0 + tc. Eliminate the parameter (solve for t) and set them equal: 0 a = 0 b = z z 0 c These are called the smmetric equations of the line. This is a sstem of implicit equations, not a parameterization (no t). (12) Smmetric equations If one of the components a, b, c is zero, ou have to eliminate the parameter differentl. Suppose v = 2, 0, 4 and r 0 = 1, 1, 3. r = 1, 1, 3 + t 2, 0, 4 = 1 + 2t = t = = 1 + 0t = = 1 z = 3 4t = t = z = z 3 4, = 1.

7 (13) The equation of a line Let L be the line through P = ( 0, 0, z 0 ) parallel to v = a, b, c. We can describe L with: A vector equation: r = r 0 + tv (where r 0 = 0, 0, z 0 ) Parametric equations: = 0 + at, = 0 + bt, z = z 0 + bt Smmetric equations: (assuming a, b, c nonzero) 0 a = 0 b = z z 0 c The direction numbers of L are a, b, c, or an scalar multiple of this triple. (14) The equation of a line: eample Eample 2: Let L be the line passing through P (0, 1, 4) parallel to v = 3, 3, 0. Describe L using: a vector equation r = 0, 1, 4 + t 3, 3, 0 parametric equations = 3t, = 1 3t, z = 4 smmetric equations z = 4, 3 = 1 3

8 (15) Skew lines In R 2, an two lines must either be parallel or intersect at a point. Lines are parallel if their direction vectors differ b a scalar multiple. For eample: L 1 : r = 0, 0, 1 + t 3, 7, 1 L 2 : r = 16, 12, 3 + t 6, 14, 2 In R 3, two lines can be skew, meaning the are not parallel but never intersect. z (16) Skew lines To find out whether two lines intersect, use parametric equations. L 1 : r = 1, 1, z 1 + t a 1, b 1, c 1 L 2 : r = 2, 2, z 2 + s a 2, b 2, c 2 = 1 + a 1 t = 2 + a 2 s = 1 + b 1 t = 2 + b 2 s z = z 1 + c 1 t = z 2 + c 2 s Use t as the parameter for one line and s as the paramater for the other one If ou can find values of t and s that make all the parametric equations agree, then the lines intersect Otherwise, the lines are skew or parallel.

9 (17) Skew lines: eample Eample 3: Prove that these lines are parallel: L 1 : 1, 6, 1 + t 3, 2, 3 L 2 : 7, 10, 0 + t 1, 23, 1 The direction vectors are v 1 = 3, 2, 3 and v 2 = 1, 2 3, 1. Since v 1 = 3v 2, the direction vectors are parallel and therefore the lines are parallel. (18) Skew lines: eample Eample 4: Prove that these lines are skew: M 1 : r = 2, 0, 0 + t 0, 1, 2 M 2 : r = 1, 1, 1 + t 2, 3, 2 If the lines intersect, there must be a pair of numbers s and t such that 2, 0, 0 + t 0, 1, 2 = 1, 1, 1 + s 2, 3, 2 Put all the constants on one side and this becomes 3, 1, 1 = s 2, 3, 2 t 0, 1, 2 and then 3, 1, 1 = 2s, t 3s, 2t + 2s.

10 (19) Skew lines: eample Eample 4: (cont.) We want to show that there is no choice of s and t for which this equation is true: 3, 1, 1 = 2s, t 3s, 2t + 2s Let s split this into three scalar equations, one for each component: 3 = 2s 1 = t 3s 1 = 2t + 2s. This is a sstem of three linear equations in two unknowns. The first equation requires s = 3/2. When we plug in this value of s, the other two equations become 1 = t 9/2 1 = 2t + 3. No value of t satisfies both of these equations. The first one requires t = 11/2, while the second one requires t = 2. Therefore these two lines never intersect. Eample 5: Let N 1 and N 2 be two lines defined b these equations: N 1 : r = 5, 7, 1 + t 1, 3, 0 N 2 : r = 0, 0, 9 + s 2, 5, 4, Find the intersection point of these two lines. We need to find values of s and t which make this equation true: 5, 7, 1 + t 1, 3, 0 = 0, 0, 9 + s 2, 5, 4 5, 7, 8 = s 2, 5, 4 t 1, 3, 0 5, 7, 8 = 2s t, 5s + 3t, 4s Separate it into three scalar equations: 5 = 2s t 7 = 5s + 3t 8 = 4s (20) Line intersection: eample The onl solution to all three is s = 2, t = 1. Either plug in t = 1 to the N 1 equation or plug in s = 2 to the N 2 equation; the intersection point is (4, 10, 1).

11 (21) Line intersection: eample N 1 : r = 5, 7, 1 + t 1, 3, 0 N 2 : r = 0, 0, 9 + s 2, 5, 4, Eample 5: (cont.) Now write an equation of the line passing through the intersection point, perpendicular to both N 1 and N 2. We need a point on the line and a direction vector (a vector parallel to the line) We alread know that the intersection point (4, 10, 1) lies on this line If L is perpendicular to both N 1 and N 2, then a direction vector of L must be orthogonal to both direction vectors 1, 3, 0 and 2, 5, 4 We can find a direction vector b taking the cross product of these, which is 1, 3, 0 2, 5, 4 = 12, 4, 11 So the line we want is r = 4, 10, 1 + t 12, 4, 11. (22) The equation of a plane A plane is a flat surface. z A AB B w v We sa that a vector v is in the plane Π if v = AB for some points A and B in Π. In this picture, w is not in the plane.

12 (23) The equation of a plane We can t describe a plane just b giving a point on the plane and a vector inside it. We d need a point and two vectors in the plane. But in R 3, we can describe a plane b giving a normal vector. Definition A vector that is orthogonal to ever vector contained in a plane is called a normal vector of that plane. z n (24) The equation of a plane For a given plane Π: Let n be a normal vector for Π Let r 0 be the position vector of a fied point on Π Let r be the position vector of an arbitrar point on Π The vector r r 0 lies in Π, so it s orthogonal to n r r r 0 r 0 n z Therefore n (r r 0 ) = 0. So the point with position vector r lies on the plane Π if and onl if n (r r 0 ) = 0, or equivalentl n r = n r 0.

13 (25) The equation of a plane n (r r 0 ) = 0 is called the vector equation of a plane. Write n = a, b, c and r 0 = 0, 0, z 0 and epand the dot product to get a scalar equation of the plane: a( 0 ) + b( 0 ) + c(z z 0 ) = 0 Collect terms (combine all constants) to get a linear equation of the plane: ( where d = a 0 b 0 cz 0 ) a + b + cz + d = 0 These are all implicit equations which ou can use to test whether or not a given point (,, z) is on the plane. A linear equation is the nicest and most compact wa to describe a plane. But usuall we have to start b finding a vector equation. (26) The equation of a plane: eample Eample 6: Write vector, scalar, and linear equations for the plane containing point P = (2, 3, 1) with normal vector n = 6, 2, 5. Vector equation: 6, 2, 5 (r 2, 3, 1 ) = 0 Scalar equation: 6( 2) + 2( 3) 5(z + 1) = 0 Linear equation: z 23 = 0

14 (27) The equation of a plane: eample Eample 7: Write an equation for the plane containing the points P 1 = (0, 1, 3), P 2 = (1, 2, 2), and P 3 = (4, 0, 1). This plane contains the vectors P 1 P 2 = 1, 1, 1 and P 1 P 3 = 4, 1, 4 The normal vector must be orthogonal to both, so we take n = P 1 P 2 P 1 P 3 = 5, 0, 5 We might as well use the simpler parallel vector 1, 0, 1. Equation of the plane: 1, 0, 1 ( ),, z 0, 1, 3 = 0 1( 0) + 0( 1) + 1(z 3) = 0 + z 3 = 0. (28) The equation of a plane: eample Eample 7: (cont.) z P 1 P 1 P 3 P 2 P 1 P 2 P 3

15 (29) Homework Paper homework #5 is due net Tuesda because we still need to cover some additional material. You can do the first question now though. Homework 12.3, 12.4 due Wednesda night, 11:59 PM Remember, tomorrow is the deadline to request a makeup for the first midterm!