Vectors #6 - Questions and Answers

Question 1 Determine whether the given vectors 𝑢 u and 𝑣 v are parallel. If they are, express 𝑣 v as a scalar multiple of 𝑢 u. Answer To check if vectors are parallel, express 𝑣 v as a scalar multiple of 𝑢 u ( 𝑣 = 𝜆 𝑢 v=λu): a) Given 𝑢 = ⟨ 3 , − 2 , 4 ⟩ u=⟨3,−2,4⟩ and 𝑣 = ⟨ − 6 , − 4 , − 8 ⟩ v=⟨−6,−4,−8⟩: 𝑣 = − 2 𝑢 . v=−2u. Conclusion: Vectors 𝑢 u and 𝑣 v are parallel. b) Given 𝑢 = ⟨ − 9 , − 6 , 12 ⟩ u=⟨−9,−6,12⟩ and 𝑣 = ⟨ 12 , 8 , − 16 ⟩ v=⟨12,8,−16⟩: Divide 𝑢 u by − 3 −3: 1 3 𝑢 = ⟨ − 3 , − 2 , 4 ⟩ . 3 1 ​ u=⟨−3,−2,4⟩. Compare with 𝑣 v: 𝑣 = − 4 3 𝑢 . v=− 3 4 ​ u. Conclusion: Vectors 𝑢 u and 𝑣 v are parallel. c) Given 𝑢 = 𝑖 + 𝑗 + 𝑘 u=i+j+k and 𝑣 = 2 𝑖 + 2 𝑗 − 2 𝑘 v=2i+2j−2k: No scalar multiple exists. Conclusion: Vectors 𝑢 u and 𝑣 v are not parallel. Question 2 Find the tangential and normal components of the acceleration vector 𝑟 ( 𝑡 ) = ( 3 𝑡 − 𝑡 3 ) 𝑖 + 3 𝑡 2 𝑗 r(t)=(3t−t 3 )i+3t 2 j. Answer Given: 𝑟 ( 𝑡 ) = ( 3 𝑡 − 𝑡 3 ) 𝑖 + 3 𝑡 2 𝑗 . r(t)=(3t−t 3 )i+3t 2 j. Differentiate 𝑟 ( 𝑡 ) r(t) to get velocity 𝑟 ′ ( 𝑡 ) r ′ (t): 𝑟 ′ ( 𝑡 ) = ( 3 − 3 𝑡 2 ) 𝑖 + 6 𝑡 𝑗 . r ′ (t)=(3−3t 2 )i+6tj. Differentiate again to get acceleration 𝑟 ′ ′ ( 𝑡 ) r ′′ (t): 𝑟 ′ ′ ( 𝑡 ) = − 6 𝑡 𝑖 + 6 𝑗 . r ′′ (t)=−6ti+6j. Magnitude of 𝑟 ′ ( 𝑡 ) r ′ (t): ∣ 𝑟 ′ ( 𝑡 ) ∣ = ( 3 − 3 𝑡 2 ) 2 + ( 6 𝑡 ) 2 . ∣r ′ (t)∣= (3−3t 2 ) 2 +(6t) 2 ​ . Simplify: ∣ 𝑟 ′ ( 𝑡 ) ∣ = 9 − 18 𝑡 2 + 9 𝑡 4 + 36 𝑡 2 = 3 ( 𝑡 2 + 1 ) . ∣r ′ (t)∣= 9−18t 2 +9t 4 +36t 2 ​ =3(t 2 +1). Tangential component 𝑎 𝑇 a T ​ : 𝑎 𝑇 = 𝑟 ′ ( 𝑡 ) ⋅ 𝑟 ′ ′ ( 𝑡 ) ∣ 𝑟 ′ ( 𝑡 ) ∣ . a T ​ = ∣r ′ (t)∣ r ′ (t)⋅r ′′ (t) ​ . Dot product: 𝑟 ′ ( 𝑡 ) ⋅ 𝑟 ′ ′ ( 𝑡 ) = [ ( 3 − 3 𝑡 2 ) ( − 6 𝑡 ) ] + ( 6 𝑡 ) ( 6 ) = − 18 𝑡 + 18 𝑡 3 + 36 𝑡 . r ′ (t)⋅r ′′ (t)=[(3−3t 2 )(−6t)]+(6t)(6)=−18t+18t 3 +36t. Thus: 𝑎 𝑇 = 18 𝑡 3 + 18 𝑡 3 ( 𝑡 2 + 1 ) = 6 𝑡 . a T ​ = 3(t 2 +1) 18t 3 +18t ​ =6t. Normal component 𝑎 𝑁 a N ​ : 𝑎 𝑁 = ∣ 𝑟 ′ ( 𝑡 ) × 𝑟 ′ ′ ( 𝑡 ) ∣ ∣ 𝑟 ′ ( 𝑡 ) ∣ . a N ​ = ∣r ′ (t)∣ ∣r ′ (t)×r ′′ (t)∣ ​ . Cross product: 𝑟 ′ ( 𝑡 ) × 𝑟 ′ ′ ( 𝑡 ) = ∣ 𝑖 𝑗 𝑘 3 − 3 𝑡 2 6 𝑡 0 − 6 𝑡 6 0 ∣ . r ′ (t)×r ′′ (t)= ​ i 3−3t 2 −6t ​ j 6t 6 ​ k 0 0 ​ ​ . Solve determinant: 𝑟 ′ ( 𝑡 ) × 𝑟 ′ ′ ( 𝑡 ) = 𝑘 [ ( 3 − 3 𝑡 2 ) ( 6 ) − ( 6 𝑡 ) ( − 6 𝑡 ) ] = 𝑘 [ 18 − 18 𝑡 2 + 36 𝑡 2 ] . r ′ (t)×r ′′ (t)=k[(3−3t 2 )(6)−(6t)(−6t)]=k[18−18t 2 +36t 2 ]. Simplify: 𝑟 ′ ( 𝑡 ) × 𝑟 ′ ′ ( 𝑡 ) = 𝑘 [ 18 + 18 𝑡 2 ] . r ′ (t)×r ′′ (t)=k[18+18t 2 ]. Magnitude: ∣ 𝑟 ′ ( 𝑡 ) × 𝑟 ′ ′ ( 𝑡 ) ∣ = 18 ( 1 + 𝑡 2 ) . ∣r ′ (t)×r ′′ (t)∣=18(1+t 2 ). Thus: 𝑎 𝑁 = 18 ( 1 + 𝑡 2 ) 3 ( 1 + 𝑡 2 ) = 6. a N ​ = 3(1+t 2 ) 18(1+t 2 ) ​ =6. Final Result: 𝑎 𝑇 = 6 𝑡 , 𝑎 𝑁 = 6. a T ​ =6t,a N ​ =6. Question 3 How do you find the unit vector in the direction of the given vector 𝑣 = ⟨ 5 , − 12 ⟩ v=⟨5,−12⟩? Answer Magnitude of 𝑣 v: ∣ 𝑣 ∣ = 5 2 + ( − 12 ) 2 = 25 + 144 = 169 = 13. ∣v∣= 5 2 +(−12) 2 ​ = 25+144 ​ = 169 ​ =13. Unit vector: 𝑢 = 𝑣 ∣ 𝑣 ∣ = 1 13 ⟨ 5 , − 12 ⟩ = ⟨ 5 13 , − 12 13 ⟩ . u= ∣v∣ v ​ = 13 1 ​ ⟨5,−12⟩=⟨ 13 5 ​ ,− 13 12 ​ ⟩. Question 4 Vector Dot Product Learning Goal: Determine which operations can be computed. Given vectors: 𝐴 = ( 2 , 1 , − 4 ) , 𝐵 = ( − 3 , 0 , 1 ) , 𝐶 = ( − 1 , − 1 , 2 ) . A=(2,1,−4),B=(−3,0,1),C=(−1,−1,2). Which of the following can be computed? 𝐴 ⋅ 𝐵 ⋅ 𝐶 A⋅B⋅C 𝐴 ⋅ ( 𝐵 ⋅ 𝐶 ) A⋅(B⋅C) 𝐴 ⋅ ( 𝐵 + 𝐶 ) A⋅(B+C) 3 𝐴 3A. Answer 𝐴 ⋅ 𝐵 ⋅ 𝐶 A⋅B⋅C: Not defined; the dot product of two vectors gives a scalar, and a scalar cannot perform another dot product. 𝐴 ⋅ ( 𝐵 ⋅ 𝐶 ) A⋅(B⋅C): Not defined; 𝐵 ⋅ 𝐶 B⋅C is a scalar, and the dot product requires two vectors. 𝐴 ⋅ ( 𝐵 + 𝐶 ) A⋅(B+C): Defined. Compute 𝐵 + 𝐶 B+C: 𝐵 + 𝐶 = ( − 3 , 0 , 1 ) + ( − 1 , − 1 , 2 ) = ( − 4 , − 1 , 3 ) . B+C=(−3,0,1)+(−1,−1,2)=(−4,−1,3). Dot product: 𝐴 ⋅ ( 𝐵 + 𝐶 ) = ( 2 , 1 , − 4 ) ⋅ ( − 4 , − 1 , 3 ) . A⋅(B+C)=(2,1,−4)⋅(−4,−1,3). Simplify: 𝐴 ⋅ ( 𝐵 + 𝐶 ) = ( 2 ) ( − 4 ) + ( 1 ) ( − 1 ) + ( − 4 ) ( 3 ) = − 8 − 1 − 12 = − 21. A⋅(B+C)=(2)(−4)+(1)(−1)+(−4)(3)=−8−1−12=−21. 3 𝐴 3A: Defined as scalar multiplication. Conclusion: 𝐴 ⋅ ( 𝐵 + 𝐶 ) A⋅(B+C) and 3 𝐴 3A can be computed. Question 5 Dot product of two parallel vectors 𝑉 1 V 1 ​ and 𝑉 2 V 2 ​ : Answer If 𝑉 1 V 1 ​ and 𝑉 2 V 2 ​ are parallel, the angle 𝜃 = 0 ∘ θ=0 ∘ , so: 𝑉 1 ⋅ 𝑉 2 = ∣ 𝑉 1 ∣ ∣ 𝑉 2 ∣ cos ⁡ ( 0 ∘ ) = ∣ 𝑉 1 ∣ ∣ 𝑉 2 ∣ . V 1 ​ ⋅V 2 ​ =∣V 1 ​ ∣∣V 2 ​ ∣cos(0 ∘ )=∣V 1 ​ ∣∣V 2 ​ ∣.