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MATH 125 Final: Solutions for Additional Problems (up to Section 2.2)

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Department
Mathematics
Course
MATH 125
Professor
Scott
Semester
Fall

Description
MATH 125 (R1) Winter 2017 Additional Practice Problems for Midterm (up to x2.2) { Solutions Note: Some of the solutions here are just sketches (the main purpose is to enable you to check your own answers). In particular, row reduction of matrices is omitted completely. In your exam, however, you should show all your work, and give suitable justi▯cation for your answers. When row reducing a matrix, you should clearly indicate which row operations have been applied at each step of the process. 1. Let 2 3 2 3 ▯2 1 v = 4 1 5 and u = 4 0 5 : 2 ▯1 a) Compute: i) v ▯ u~ ii) vjj iii) ujj iv)v u ~ v) dv;u). b) Are v and u orthogonal? Justify your answer. c) Arev and u parallel? Justify your answer. 1 d) Is v a unit vector? If not, normv, i.e. compute ~v. jvjj 2 3 ▯11 4 5 Solution: a) i)v ▯ 3u = 4 . 11 ii) vjj p 3. iii)ujj = 2. iv)v ▯u = ▯p. v) dv;u) = 19. b) No, becausev u = ▯4 6= 0. c) No, because they are not scalar multiples of one another, i.e., we cannot ▯nd a scalar c such that 2 3 2 3 ▯2 1 c4 1 5 = 4 0 5 (c cannot be both ▯1=2 and 0) 2 ▯1 d) No, because vjj = 3 6= 1. The normalizatiov is ~ 2 3 2 3 1 1 ▯2 ▯2=3 v = 4 1 5 = 4 1=3 5 jvjj 3 2 2=3 ▯ ▯ ▯p ▯ 2. Let 1 3 ~a = p1 and b = 2 : 3 k For what value(s) of k: a) area and b orthogonal? b) area and b parallel? c) is b a unit vector? p Solution: a) We compute ~ a ▯ b = 3+ ( p1)k. The vectors ~a and b are othogonal 2 3 precisely when a ▯ b = 0. Solving for k, we get that the vectors are orthogonal if and only if k = ▯3=2. b) The vectors a and b are parallel if and only if there exists c 2 R such that ▯ ▯ ▯ p3▯ c 1 = 2 ; p k 3 or, in other words, if and only if there exists c 2 R such that p 3 1 c = 2 and p c = k: 3 Putting these together we see that the vectors are parallel if and only if p 1 3 1 k = p ▯ = : 3 2 2 c) Note that jjbjj = 1 if and only if jjbjj = 1. Now ! p 2 jjbjj = b ▯ b = 3 + k = 3 + k ; 2 4 so b is a unit vector if and only i+ k = 1. Solving for k, we see that b is a unit 4 vector if and only if k = 1=2 or k = ▯1=2. 3. Find parametric equations, a vector equation, a normal equation and a general equation 2 for the line ‘ in R which: a) contains the points A = (3;▯8) and B = (5;0). b) passes through the origin and is parallel to the line with general equation 2x▯5y = 4. ▯! ▯ 2▯ Solution: a) ‘ is parallel to AB = . To get a normal vector to ‘, take any 8 ▯ ▯ ▯! ▯8 non-zero vector orthogonal to AB. A natural choice here is then =ctor2~ (of course, this is just one example { there are in▯nitely many normal vectors to ‘). We now know i) a point of ‘ (A, say), ii) a direction vector for ‘ (AB) and iii) a normal vector to ‘n). We can therefore write down equations of ‘ of each form: ▯ 3 ▯ ▯ 2 ▯ Vector equation: x = + t (t 2 R). ▯8 8 Parametric equations: x = 3 + 2t (t 2 R). y = ▯8 + 8t ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯8 x ▯8 3 Normal equation: ▯ = ▯ . 2 y 2 ▯8 General equation: ▯8x + 2y = ▯40. Remark: These are not the only correct answers. Choosing a di▯erent point, direction vector or normal vector will lead to di▯erent equations of ‘. b) Parallel lines have the same normal vectors. A normal vector to ‘ is therefore given ▯ ▯ by n = 2 (we just read this from the equation for the second line). To get a ▯5 direction vector for ‘, just take any non-zero vector orthon. A natural (but ▯ ▯ not the only) choice here is d = . We now know i) a point of ‘ (the origin (0;0)), 2 ~ ii) a direction vector for ‘ (d) and iii) a normal vn). We can therefore write down equations of ‘ of each form: ▯ ▯ ▯ ▯ ▯ ▯ Vector equation: x = 0 + t 5 (t 2 R), or simpx = t 5 (t 2 R) 0 2 2 x = 5t Parametric equations: y = 2t (t 2 R). ▯ 2 ▯ ▯ x ▯ ▯ 2 ▯ ▯ 0 ▯ Normal equation: ▯ = ▯ . ▯5 y ▯5 0 General equation: 2x ▯ 5y = 0. 4. Let ‘ be the line in R with general equation 3x ▯ 2y = 3. Find the point where ‘ 1 2 1 intersects the line 2 in R given by: a) ‘2: x ▯ 3y = ▯6. ▯ ▯ ▯ ▯ 5 2 b) ‘2: x = + t . ▯1 4 Solution: a) The intersection point of the two lines is the unique solution of the linear system 3x ▯ 2y = 3 x ▯ 3y = ▯6 Solving this system, we get x = 3, y = 3. Hence (3;3) is the point of intersection. b) Let A be the unique intersection point. Since A lies on ‘ 2 we have that A = (5 + 2t;▯1 + 4t) for some t 2 R. On the other hand, since A lies on ‘ ,1we have 3(5 + 2t) ▯ 2(▯1 + 4t) = 3: Solving for t, we get t = 7. Hence A = (5 + 2(7);▯1 + 4(7)) = (19;27) is the point of intersection of the two lines in this case. 2 5. Consider the line ‘1: x + 2y = 2 in R . a) Find a normal vector ~ n for 1 . b) Does the point (2;1) lie on the line ‘1? Justify your answer. c) Let ‘ 2e another line in R with normal vector ▯ ▯ 2 n2= 1 Do ‘ 1nd ‘ i2tersect? If so, why? If not, why not? ▯ ▯ 1 Solution: a) ~n = is a normal vector to ‘1. 2 b) No, because x = 2, y = 1 is not a solution of the equation of ‘ : 2 + 2(1) = 5 6= 2. 1 c) Yes: Non-parallel lines must intersect, and these lines are not parallel: Indeed, the respective normal vectors ~ n and n~2are not parallel (not scalar multiples of one another), so the lines themselves are not parallel (parallel lines have parallel normal vectors). Alternatively, note th2t ‘ has general equation 2x+y = c, where c is some (unknown) scalar. One can then check that, irrespective of the value of c, the system x + 2y = 2 2x + y = c is consistent (just bring the augmented matrix of the system to row echelon form). Since the solutions of the system correspond to the intersection points of the line, it follows that the lines do indeed intersect. 6. Let the following two vectors be given: 2 3 2 3 1 ▯1 v = 415 and w~ = 4 1 5 1 0 a) Are v and ~ orthogonal? Now let P be the plane in R which passes through the origin and which is perpendic- ular tv.~ b) Find a normal equation of P. c) Is~ parallel to P? d) Find parametric equations and a vector equation of P. Solution: a) Yes, sinv ▯~ = 0. b) Sincev is a normal vector to P, and since P passes through the origin (0;0;0), a normal equation of P is given by 2 32 3 2 3 2 3 1 x 1 0 4154 y 5 = 4 1 ▯ 0 5: 1 z 1 0 c) Yes, sinc~ is orthogonal to the normal vecv (part a)). d) By part b), a general equation of P is given by x + y + z = 0: We have two free variables. Setting y = s, z = t, we ▯nd that x = ▯s ▯ t y = s (s;t 2 R) z = t is a set of parametric equations of P. A vector equation of P is then given by 2 3 2 3 ▯1 ▯1 x = s4 15 + t4 0 5 (s;t 2 R): 0 1 7. Consider the plane P in R with vector equation 2 1 3 2▯1 3 223 4 5 4 5 4 5 x = ▯2 + s 1 + t 1 (s;t 2 R) 3 1 1 a) Find a normal equation of P. b) Give an example of a point that is not on the plane P. c) Two planes are parallel if they have the same normal vectors. Find a normal equation of the plane which passes through the origin and is parallel to P. Solution: a) This is the problem solved in Q3 of Written Assignment 3. Our aim is to ▯nd a normal vector to P, i.e., a non-zero vector perpendicular to P. Let 2 3 2 3 ▯1 2 u = 4 1 5 and v = 415 1 1 2 3 x (i.eu and v are the given direction vectors for P). x =, let ~be a general z 3 vector in R . Thex is perpendicular to P if and onx is orthogonal to the two given direction vecu and v, i.e., if and ox ▯u = 0 andx v = 0. Now, x ▯u = ▯x + y + z and x ▯v = 2x + y + z; and sox is perpendicular to P if and only if it is a solution of the linear system ▯x + y + z = 0 2x + y + z = 0 After applying the row reduction algorithm, we ▯nd that the reduced row echelon form of the augmented matrix of this system is ▯ ▯ 1 0 0 0 0 1 1 0 The system is thus consistent, and has one free variable, namely the variable z. As- signing the parametric value t to z, and re-writing the above matrix as a system of equations, we see that the general solution of the system is x = 0 y = ▯t ; z = t or, in vector form, 2 3 0 x = t ▯1 5: 1 2 3 0 In particular, we seen =at ~1 5 is a normal vector to P. Since the plane passes 1 through the point (1;▯2;3) (given in the question), it follows that 2 3 2 3 2 3 2 3 0 x 0 1 4 ▯1 5 ▯ y 5= 4 ▯1 5▯4 ▯2 5 1 z 1 3 is a normal equation of P. b) From part a), we just need to choose a point (a;b;c) with the property that 2 0 3 2 a 3 2 0 3 2 1 3 4 ▯1 5▯4 b5 6=4 ▯1 5 ▯ ▯2 5 ; 1 c 1 3 i.e., such that ▯b+c 6= 5. For instance, we can take
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