MAT267_10.3_Jones

Kendall Painley Jones MAT 267 ONLINE B Spring 2024 Assignment Section 10.3 due 03/17/2024 at 11:59pm MST Problem 1. (1 point) Distance and Dot Products: Consider the vectors u = ⟨ 10 , 7 , 10 ⟩ and v = ⟨− 3 , 0 , − 10 ⟩ . Compute ∥ u ∥ = Compute ∥ v ∥ = Compute u · v = Solution: SOLUTION: ∥ u ∥ = p ( 10 ) 2 +( 7 ) 2 +( 10 ) 2 = √ 249 ∥ v ∥ = p ( − 3 ) 2 +( 0 ) 2 +( − 10 ) 2 = √ 109 u · v = 10 ( − 3 )+ 7 ( 0 )+ 10 ( − 10 ) = − 130 Correct Answers: • 15.7797338380595 • 10.4403065089106 • -130 Problem 2. (1 point) Find a · b if | a | = 10 , | b | = 9 , and the angle between a and b is π 7 radians. a · b = Solution: SOLUTION: a · b = | a || b | cos ( π 7 ) = 10 ( 9 ) cos ( π 7 ) Correct Answers: • 81.0871981112177 Problem 3. (1 point) If a = ⟨ 4 , − 5 , 1 ⟩ and b = ⟨− 5 , 1 , 1 ⟩ , then a · b = Is the angle between the vectors ”acute”, ”obtuse” or ”right”? Solution: SOLUTION: a · b = 4 ( − 5 ) − 5 ( 1 )+ 1 ( 1 ) = − 24 Since the dot product is negative, the angle between the vectors is obtuse. Correct Answers: • -24 • OBTUSE Problem 4. (1 point) Determine if the pairs of vectors below are ”parallel”, ”orthogo- nal”, or ”neither”. a = ⟨− 1 , − 3 , 5 ⟩ and b = ⟨ 4 , 12 , 8 ⟩ are a = ⟨− 1 , − 3 , 5 ⟩ and b = ⟨ 4 , 12 , − 20 ⟩ are a = ⟨− 1 , − 3 , 5 ⟩ and b = ⟨ 4 , 12 , − 20 ⟩ are Solution: SOLUTION: a · b = − 1 ( 4 ) − 3 ( 12 )+ 5 ( 8 ) = 0 , so the vectors are orthogonal. Because b = − 4 a , the vectors are parallel. Because b = − 4 a , the vectors are parallel. Correct Answers: • ORTHOGONAL • PARALLEL • PARALLEL 1

Problem 5. (1 point) Find a vector orthogonal to both ⟨− 1 , − 5 , 0 ⟩ and to ⟨ 0 , − 5 , − 1 ⟩ of the form ⟨ 1 , , ⟩ Solution: SOLUTION: Let a = ⟨ 1 , a 2 , a 3 ⟩ be a vector orthogonal to both ⟨− 1 , − 5 , 0 ⟩ and ⟨ 0 , − 5 , − 1 ⟩ . Then, the dot product of a with both vectors must be zero. Thus − 1 − 5 a 2 = 0 and − 5 a 2 − 1 a 3 = 0. Solving the two equations yields a 2 = − 1 5 and a 3 = 1. Thus a vector orthogonal to both ⟨− 1 , − 5 , 0 ⟩ and ⟨ 0 , − 5 , − 1 ⟩ is 1 , − 1 5 , 1 , . Correct Answers: • -0.2 • 1 Problem 6. (1 point) What is the angle in radians between the vectors a = ⟨− 10 , − 6 , 4 ⟩ and b = ⟨ 10 , − 4 , − 1 ⟩ ? Angle: (radians) Solution: SOLUTION: | a | = p ( − 10 ) 2 +( − 6 ) 2 +( 4 ) 2 = √ 152 , | b | = p ( 10 ) 2 +( − 4 ) 2 +( − 1 ) 2 = √ 117 , and a · b = − 10 ( 10 ) − 6 ( − 4 )+ 4 ( − 1 ) = − 80 . Then θ = arccos a · b | a || b | = arccos − 80 √ 152 √ 117 Correct Answers: • 2.21416622648682 Problem 7. (1 point) Find the scalar and vector projection of the vector b = ⟨− 1 , − 1 , − 5 ⟩ onto the vector a = ⟨− 1 , 0 , 5 ⟩ . Scalar projection (i.e., component): Vector projection ⟨ , , ⟩ Solution: SOLUTION: comp a b = a · b ∥ a ∥ = − 24 √ 26 proj a b = a · b ∥ a ∥ 2 a = − 24 26 ⟨− 1 , 0 , 5 ⟩ = 12 13 , 0 , − 60 13 Correct Answers: • -4.70678724331642 • 0.923076923076923 • 0 • -4.61538461538462 Problem 8. (1 point) A rectangular box has length 2 inches, width 6 inches, and a height of 18 inches. Find the angle between the diagonal of the box and the diagonal of its base. The angle should be measured in radians. Angle = Solution: SOLUTION: For convenience, consider the box positioned so that its back left corner is at the origin, and its edges lie along the coordinate axis. The diagonal of the box begins at the origin and ends at the point ( 2 , 6 , 18 ) and has vector representation ⟨ 2 , 6 , 18 ⟩ . The diagonal of the base has vector representation ⟨ 2 , 6 , 0 ⟩ . The angle between these two vectors is given by θ = arccos ⟨ 2 , 6 , 18 ⟩·⟨ 2 , 6 , 0 ⟩ |⟨ 2 , 6 , 18 ⟩||⟨ 2 , 6 , 0 ⟩| = arccos √ 40 √ 364 ! Correct Answers: • 1.23290671562904 2

Problem 13. (1 point) v = − 3 i − 8 j + 10 k w = − 7 j + 9 k Compute the dot product v · w . Solution: Solution: v · w = ( − 3 i − 8 j + 10 k ) · ( − 7 j + 9 k ) = ( − 3 · 0 )+( − 8 ·− 7 )+( 10 · 9 ) = 146 Correct Answers: • 146 Problem 14. (1 point) The force on an object is ⃗ F = − 23 ˜ j . For the vector ⃗ v = − 2 ˜ i − 4 ˜ j , find: (a) The component of ⃗ F parallel to ⃗ v : (b) The component of ⃗ F perpendicular to ⃗ v : The work, W , done by force ⃗ F through displacement ⃗ v : Solution: SOLUTION We first find the unit vector in direction ⃗ v . Since || ⃗ v || = p ( − 2 ) 2 +( − 4 ) 2 = √ 20, the unit vector in direction of ⃗ v is ⃗ u = ⃗ v / √ 20. Then (a) ⃗ F || = ( ⃗ F · ⃗ u ) ⃗ u = ( 92 / √ 20 ) ⃗ u = − 184 20 ⃗ i + − 368 20 ⃗ j . (b) We have ⃗ F ⊥ = ⃗ F − ⃗ F || = ( − 23 ˜ j ) − ( − 184 20 ⃗ i + − 368 20 ⃗ j ) ≈ 9 . 2 ˜ i − 4 . 6 ˜ j . (c) Since work is the dot product of the force and displacement vectors, we have W = ⃗ F · ⃗ v = 92 . Correct Answers: • -9.2i-18.4j • 9.2i-4.6j • 92 Problem 15. (1 point) Consider the vectors ⃗ a = 2 ˜ i + ˜ j − ˜ k , ⃗ b = ⃗ i + 2 ⃗ j + 3 ⃗ k , ⃗ c = − ˜ i − 2 ˜ j + ˜ k ⃗ d = ˜ i + 2 ˜ j + ˜ k , ⃗ g = 2 ˜ i − ˜ j + ˜ k . Which pairs (if any) of these vectors are (a) Are perpendicular? (Enter none or a pair or list of pairs, e.g., if ⃗ a is perpendicular to ⃗ b and ⃗ c, enter (a,b),(a,c) .) (b) Are parallel? (Enter none or a pair or list of pairs, e.g., if ⃗ a is parallel to ⃗ b and ⃗ c, enter (a,b),(a,c) .) (c) Have an angles less than π / 2 between them? (Enter none or a pair or list of pairs, e.g., if ⃗ a is at an angle less than pi/2 from ⃗ b and ⃗ c, enter (a,b),(a,c) .) (d) Have an angle of more than π / 2 between them? (Enter none or a pair or list of pairs, e.g., if ⃗ a is at an angle greater than pi/2 from ⃗ b and ⃗ c, enter (a,b),(a,c) .) Solution: SOLUTION (a) Perpendicular vectors have a dot product of 0. Since this is not the case for any of the pairs of vectors here, there are no perpen- dicular pairs. (b) Parallel vectors are multiples of one another. Since this is not the case for any of the pairs of vectors here, there are no parallel pairs. (c) Since ⃗ v · ⃗ w = ∥ ⃗ v ∥∥ ⃗ w ∥ cos θ , the dot product of the vectors we want is positive. Since ⃗ a · ⃗ b = ( 2 ) · ( 1 )+( 1 ) · ( 2 )+( − 1 ) · ( 3 ) > 0 and similarly ⃗ a · ⃗ d > 0, ⃗ a · ⃗ g > 0, ⃗ b · ⃗ d > 0, ⃗ b · ⃗ g > 0, ⃗ c · ⃗ g > 0, ⃗ d · ⃗ g > 0, the pairs we want are ( ⃗ a , ⃗ b ) , ( ⃗ a , ⃗ d ) , ( ⃗ a ,⃗ g ) , ( ⃗ b , ⃗ d ) , ( ⃗ b ,⃗ g ) , ( ⃗ c ,⃗ g ) , ( ⃗ d ,⃗ g ) . (d) Vectors with an angle of more than π / 2 between them have a negative dot product. Since ⃗ a · ⃗ c = ( 2 ) · ( − 1 )+( 1 ) · ( − 2 )+( − 1 ) · ( 1 ) < 0 and similarly ⃗ b · ⃗ c < 0, ⃗ c · ⃗ d < 0, the pairs we want are ( ⃗ a ,⃗ c ) , ( ⃗ b ,⃗ c ) , ( ⃗ c , ⃗ d ) . Correct Answers: • none • none • (a, b), (a, d), (a, g), (b, d), (b, g), (c, g), (d, g) 4

• (a, c), (b, c), (c, d) Problem 16. (1 point) Perform the following operations on the vectors ⃗ u = ⟨− 2 , − 5 , 1 ⟩ , ⃗ v = ⟨ 4 , − 5 , 1 ⟩ , and ⃗ w = ⟨− 1 , 1 , 0 ⟩ . ⃗ u · ⃗ w = ( ⃗ u · ⃗ v ) ⃗ u = (( ⃗ w · ⃗ w ) ⃗ u ) · ⃗ u = ⃗ u · ⃗ v + ⃗ v · ⃗ w = Solution: SOLUTION ⃗ u · ⃗ w = < − 2 , − 5 , 1 > · < − 1 , 1 , 0 > = − 3 ( ⃗ u · ⃗ v ) ⃗ u = 18 < − 2 , − 5 , 1 > = < − 36 , − 90 , 18 > (( ⃗ w · ⃗ w ) ⃗ u ) · ⃗ u = ( 2 < − 2 , − 5 , 1 > ) · < − 2 , − 5 , 1 > = < − 4 , − 10 , 2 > · < − 2 , − 5 , 1 > = 60 ⃗ u · ⃗ v + ⃗ v · ⃗ w = 18 − 9 = 9 Correct Answers: • -3 • <-36,-90,18> • 60 • 9 Problem 17. (1 point) Several unit vectors ⃗ r ,⃗ s , ⃗ t ,⃗ u ,⃗ n , and ⃗ e in the xy-plane (not three-dimensional space) are shown in the figure. Using the geometric definition of the dot product, are the following dot products positive, negative, or zero? You may assume that angles that look the same are the same. ? 1. ⃗ s · ⃗ t ? 2. ⃗ e · ⃗ s ? 3. ⃗ e · ⃗ r ? 4. ⃗ n · ⃗ e ? 5. ⃗ t · ⃗ u ? 6. ⃗ n · ⃗ t ? 7. ⃗ r · ⃗ u ? 8. ⃗ r · ⃗ s (Click on graph to enlarge) Solution: SOLUTION 1. The angle between the vectors is acute. Thus the dot product is positive. 2. The vectors are perpendicular. Thus the dot product is zero. 3. The angle between the vectors is obtuse. Thus the dot product is negative. 4. The vectors are perpendicular. Thus the dot product is zero. 5. The angle between the vectors is acute. Thus the dot product is positive. 6. The angle between the vectors is obtuse. Thus the dot product is negative. 7. The angle between the vectors is obtuse. Thus the dot product is negative. 8. The angle between the vectors is acute. Thus the dot product is positive. 5