Quiz Practice Problems Practice problems are similar, both in difficulty and in scope, to the type of problems you will see on the quiz. Problems marked with a are for your entertainment and are not essential. SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference section 11.3 of the recommended textbook (or the equivalent section in your alternative textbook/online resource) and your lecture notes. EXPECTED SKILLS: Know how to compute the dot product of two vectors. Be able to use the dot product to find the angle between two vectors and, in particular, be able to determine if two vectors are orthogonal. Know how to compute the direction cosines of a vector. Be able to decompose vectors into orthogonal components and know how to compute the orthogonal projection of one vector onto another. Be able to use the dot product to find the work W done by a constant force applied at a constant angle with the direction of motion to an object. 1
PRACTICE PROBLEMS: 1. For each of the following, compute u v based on the given information. (a) u = 3, 1 ; v =, 5 (b) u = 4, 5, 1 ; v = 3, 6, 1 (c) u = i 3j + 4k; v = 9i j (d) u = 3; v = 4; the angle between u and v is π 4.. Explain why the operation (u v) w does not make sense. 3. Determine whether the angle between v = 1,, 3 and w = 6, 4, 1 is acute, obtuse, or right. Explain. 4. Using the dot product only, give an example of a vector which is orthogonal to both v = 1, 1, 1 and w =, 0, 4. [Hint: Set up a system of equations.] 5. Consider the triangle with vertices A, B, and C. z 1 x 3 1 A 1 3 4 C B 5 y
Use vectors to compute the angle between the diagonal which extends from vertex A to vertex B and the line segment which extends from vertex A to vertex C. (Verify your answer with Quiz 1 Practice #3c.) 6. Consider the triangle, shown below, with vertices A(1,, 6), B(3, 0, 1), and C(, 1, 0). A z 6 5 4 3 1 1 1 1 1 3 1 x B C y Compute all three angles of the triangle. 7. Let v = 1, and b = 3, 4. (a) Find the vector component of v along b and the vector component of v orthogonal to b. (b) Sketch v, b, and the vector components that you found in part (a). 8. Express v = i+j+3k as the sum of a vector parallel to b = i+4j k and a vector orthogonal to b. 9. Suppose that u 0 and v 0. Under what condition will u + v = u v? Explain. 3
10. The following questions deal with finding the distance from a point to a line. (a) Given three points A, B, and P in -space or 3-space as shown in the picture below, describe two different ways that you could use the dot product to calculate the distance, d, between the point P and the line which contains A and B. (b) Use one of your methods from part (a) to compute the distance from the point P (5, 3, 0) to the line containing A(1, 0, 1) and B(, 3, 1). 11. (Pythagorean Theorem) Prove that for any two orthogonal vectors v and w, v + w = v + w. [Hint: Use properties of the dot product to expand v + w.] 1. Let A and B be endpoints of a diameter of a circle with radius r, and suppose that C is any other point on the circle, as shown below. C B O A 4
Prove that ABC is a right triangle. [Hint: Express each of CA and CB as the combination of CO and some other vector.] 13. (Cauchy-Schwarz Inequality:) Prove that for all vectors v and w, v w v w. 14. Let v = 1,, 3. (a) Compute the direction cosines of v. (b) Verify that cos α + cos β + cos γ = 1, where α, β, and γ are the angles between v and i, j, and k, respectively. 15. (Triangle Inequality:) Prove that for any two vectors v and w, v + w v + w. [Hint: Use problem #11 s hint and the Cauchy-Schwarz inequality.] 16. A constant force with vector representation F = 10i + 18j 6k moves an object along a straight line from the point (, 3, 0) to the point (4, 9, 15). Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons. 17. A woman exerts a horizontal force of 5 lb on a crate as she pushes it up a ramp that is 10 ft long and inclined at an angle of 0 above the horizontal. Find the work done on the box. An airplane flies at a constant airspeed c directly above a closed polygonal path in a plane, completing one circuit. Show that, compared to no wind, the presence of a wind of constant speed k < c and constant direction will increase the time required. 5
SOLUTIONS 1. (a) (3)() + ( 1)( 5) = 6 + 5 = 11 (b) (4)(3) + ( 5)(6) + (1)( 1) = 1 30 1 = 19 (c) (1)(9) + ( 3)( ) + (4)(0) = 9 + 6 = 15 (d) (3)(4) cos(π/4) = 1 / = 6. The quantity u v is a scalar. We cannot take the dot product of a scalar with a vector. 3. Note that v w = (1)( 6) + ()(4) + (3)( 1) = 6 + 8 3 = 1 < 0 so that the cosine of the angle between v and w is negative, which means that the angle is obtuse. 4. Suppose u = a, b, c is a vector which is orthogonal to both v = 1, 1, 1 and w =, 0, 4. Then u v = 0 = u w. Hence a + b + c = 0 and a + 4c = 0. The second equation dictates that a = c. Plugging this into the first equation yields c + b + c = 0 so that b = c. Hence u = a, b, c = c, c, c = c, 1, 1. In other words, u must be a scalar multiple of, 1, 1. 5. Note that the vertices of the triangle are A(3,, 0), B(1, 5, 1), and C(1, 5, 0). Hence AB = 1 3, 5, 1 0 =, 3, 1 and AC = 1 3, 5, 0 0 =, 3, 0, and so cos( BAC) = = = AB AC AB AC ( )( ) + (3)(3) + (1)(0) ( ) + 3 + 1 ( ) + 3 + 0 4 + 9 + 0 4 + 9 + 1 4 + 9 + 0 = 13 14. 6
This matches the answer from Quiz 1 Practice Problem #3c. 6. For A, we have AB = 3 1, 0 ( ), 1 6 =,, 7 and AC = 1, 1 ( ), 0 6 = 3, 3, 6. Hence m A = cos 1 AB AC AB AC ( ) = cos 1 ()( 3) + ()(3) + ( 7)( 6) + + ( 7) ( 3) + 3 + ( 6) ( ) = cos 1 6 + 6 + 4 4 + 4 + 49 9 + 9 + 36 ( ) 4 = cos 1 57 54 For B, we have BC = 3, 1 0, 0 ( 1) = 5, 1, 1 and BA = AB =,, 7. Hence m A = cos 1 BA BC BA BC ( ) = cos 1 ( 5)( ) + (1)( ) + (1)(7) ( 5) + 1 + 1 ( ) + ( ) + 7 ( ) = cos 1 10 + 7 5 + 1 + 1 4 + 4 + 49 ( ) 15 = cos 1. 57 7 ( ) ( ) 4 15 Of course m C = π cos 1 cos 1 since, from 57 54 57 7 geometry, the sum of the angle measures of a triangle is π. 7. (a) The vector component of v along b is proj b v = v b (1)( 3) + ()(4) b = 3, 4 = 5 3, 4 = 3 b ( 3) + 4 5 5, 4. 5 7
The vector component of v orthogonal to b is orth b v = v proj b v = 1, 3 5, 4 8 = 5 5, 6. 5 (b) 8. Take proj b v = v b (1)( ) + ()(4) + (3)( 1) b = ( i+4j k) = 3 b ( ) + 4 + ( 1) 1 ( i+4j k) = 7 i+4 7 j 1 7 k and orth b v = v orth b v = (i+j+3k) ( 7 i + 47 j 17 ) k = 9 7 i 10 7 j+ 7 k. We get our desired decomposition v = ( 7 i + 4 7 j 1 7 k) + ( 9 7 i 10 7 j + 7 k). 9. Observe that 8
u + v = u v u + v = u v (u + v) (u + v) = (u v) (u v) u u + u v + v v = u u u v + v v 4u v = 0 u v = 0. Hence u and v must be orthogonal. 10. (a) OPTION 1: Once can compute AP and proj AP. Then use the Pythagorean theorem to calculate d. AB OPTION : One can find proj ABAP. Then d = AP proj ABAP. (b) This is an exercise left to the reader. In both cases, you should get d = 91/10. 11. Since v and w are orthogonal, we have that v w = 0, and so v + w = (v + w) (v + w) = v v + v w + w w = v + w. As you can see, this is a generalization of the theorem from geometry of the same name. 1. Notice that CA = CO + OA and CB = CO + OB. Yet OA and OB are collinear and are oppositely directed radii of O. Thus OB = OA, and so CB = CO OA. Therefore 9
CA CB = ( CO + OA) ( CO OA) = CO CO CO OA + OA CO OA OA = CO OA = r r = 0. Hence CA CB, and so ABC is a right triangle. 13. If θ is the angle between v and w, then v w = v w cos θ = v w cos θ v w since cos θ 1. 14. (a) Since v = 1 + + 3 = 1 + 4 + 9 = 14, cos α = v 1 = v 1/ 14, cos β = v = / 14, and cos γ = v 3 = 3/ 14. v v (b) Indeed cos α+cos β +cos γ = (1/ 14) +(/ 14) +(3/ 14) = 1/14 + 4/14 + 9/14 = 14/14 = 1. 15. As v + w, v, and w are positive for all vectors v and w, it suffices to prove the inequality holds on their squares. Indeed v + w = v v + v w + w w as shown in the solution of problem 11. Since x x for every real number x, it follows by the Cauchy-Schwarz inequality that 10
v v+v w+w w v + v w + w v + v w + w = ( v + w ) (Recall that vector addition follows the parallelogram law. In particular, the vectors v, w, and v +w, can be imagined as sides of a triangle. This is why the theorem is called the triangle inequality; it is a generalization of the inequality from geometry of the same name. As you may begin to notice from the last few problems, many theorems from geometry can be restated in the language of vectors. Can you come up with more?) 16. The displacement vector of the object is d = 4, 9 3, 15 0 =, 6, 15, and so the work done is F d = [(10)() + (18)(6) + ( 6)(15)] N-m = (0 + 108 30) J = 98 J. 17. W = (5 lb)(10 ft) cos(0 ) = 50 cos(0 ) ft-lb 34.9 ft-lb. 11