Vectors #7 - Questions and Answers

Question 1 Dot product of two perpendicular vectors 𝑉 1 V 1 ​ and 𝑉 2 V 2 ​ : Answer 1 When two vectors 𝑉 1 V 1 ​ and 𝑉 2 V 2 ​ are perpendicular, the angle 𝜃 θ between them is 9 0 ∘ 90 ∘ . The dot product formula is: 𝑉 1 ⋅ 𝑉 2 = ∣ 𝑉 1 ∣ ∣ 𝑉 2 ∣ cos ⁡ ( 9 0 ∘ ) . V 1 ​ ⋅V 2 ​ =∣V 1 ​ ∣∣V 2 ​ ∣cos(90 ∘ ). Since cos ⁡ ( 9 0 ∘ ) = 0 cos(90 ∘ )=0, 𝑉 1 ⋅ 𝑉 2 = 0. V 1 ​ ⋅V 2 ​ =0. Answer 2 Reaffirming the dot product: 𝑉 1 ⋅ 𝑉 2 = 0 if the vectors are perpendicular . V 1 ​ ⋅V 2 ​ =0if the vectors are perpendicular. Question 2 Part A: Find the magnitude of the vector product 𝐴 × 𝐵 A×B. Part B: Find the direction of 𝐴 × 𝐵 A×B. Answer 1 Part A: Given 𝐴 = 2.80   cm A=2.80cm at 6 0 ∘ 60 ∘ and 𝐵 = 1.90   cm B=1.90cm at 6 0 ∘ 60 ∘ , the cross product is: 𝐴 × 𝐵 = ∣ 𝐴 ∣ ∣ 𝐵 ∣ sin ⁡ 𝜃   𝑘 ^ . A×B=∣A∣∣B∣sinθ k ^ . Here 𝜃 = 6 0 ∘ θ=60 ∘ , so: 𝐴 × 𝐵 = ( 2.80 ) ( 1.90 ) sin ⁡ ( 6 0 ∘ ) . A×B=(2.80)(1.90)sin(60 ∘ ). sin ⁡ ( 6 0 ∘ ) = 3 2 . sin(60 ∘ )= 2 3 ​ ​ . 𝐴 × 𝐵 = 2.80 × 1.90 × 3 2 . A×B=2.80×1.90× 2 3 ​ ​ . Simplifying: 𝐴 × 𝐵 = 4.61   cm 2   𝑘 ^ . A×B=4.61cm 2 k ^ . Part B: The direction of 𝐴 × 𝐵 A×B is determined by the right-hand rule. Curling fingers from 𝐴 A to 𝐵 B, the thumb points into the plane, which corresponds to the negative 𝑧 ^ z ^ -direction. Answer 2 Part C: Magnitude of 𝐴 × 𝐵 A×B: ∣ 𝐴 × 𝐵 ∣ = ∣ 𝐵 × 𝐴 ∣ = 4.61   cm 2 . ∣A×B∣=∣B×A∣=4.61cm 2 . Part D: Direction of 𝐴 × 𝐵 A×B is negative 𝑘 ^ k ^ -direction, and 𝐵 × 𝐴 B×A is in the positive 𝑘 ^ k ^ -direction. Question 3 Find the vector with representation given by the directed line segment 𝐴 𝐵 AB. Given 𝐴 ( − 2 , 1 ) A(−2,1) and 𝐵 ( 1 , 2 ) B(1,2). Answer 1 The vector representation is given by: 𝑎 = 𝐵 − 𝐴 . a=B−A. Subtract corresponding coordinates: 𝑎 = ⟨ 1 − ( − 2 ) , 2 − 1 ⟩ . a=⟨1−(−2),2−1⟩. Simplify: 𝑎 = ⟨ 3 , 1 ⟩ . a=⟨3,1⟩. Answer 2 Subtracting coordinates: 𝑎 = [ 1 2 ] − [ − 2 1 ] . a=[ 1 2 ​ ]−[ −2 1 ​ ]. Simplify: 𝑎 = [ 3 1 ] . a=[ 3 1 ​ ]. Answer 3 Vector representation: 𝑎 = ⟨ 3 , 1 ⟩ . a=⟨3,1⟩. Equivalent representation starting at the origin: The vector starts from ( 0 , 0 ) (0,0) and ends at ( 3 , 1 ) (3,1). Question 4 Subspace Verification for 𝐻 ⊆ 𝑀 2 × 4 H⊆M 2×4 ​ : Answer To determine if 𝐻 H is a subspace of 𝑀 2 × 4 M 2×4 ​ , verify the three subspace properties: Zero Vector: 𝐹 𝐴 = 0 FA=0 implies 𝐻 H contains the zero matrix. Closed Under Addition: If 𝐴 1 , 𝐴 2 ∈ 𝐻 A 1 ​ ,A 2 ​ ∈H, then 𝐹 ( 𝐴 1 + 𝐴 2 ) = 𝐹 𝐴 1 + 𝐹 𝐴 2 = 0 F(A 1 ​ +A 2 ​ )=FA 1 ​ +FA 2 ​ =0, so 𝐻 H is closed under addition. Closed Under Scalar Multiplication: If 𝐴 ∈ 𝐻 A∈H and 𝑐 ∈ 𝑅 c∈R, then 𝐹 ( 𝑐 𝐴 ) = 𝑐 ( 𝐹 𝐴 ) = 0 F(cA)=c(FA)=0. Conclusion: 𝐻 H satisfies all properties of a subspace of 𝑀 2 × 4 M 2×4 ​ . Question 5 Find the cross product 𝐴 × 𝐵 A×B: Given 𝐴 = 5 𝑖 + 𝑗 A=5i+j, 𝐵 = 2 𝑖 − 6 𝑗 + 2 𝑘 B=2i−6j+2k. Answer Using the cross product formula: 𝐴 × 𝐵 = ∣ 𝑖 𝑗 𝑘 5 1 0 2 − 6 2 ∣ . A×B= ​ i 5 2 ​ j 1 −6 ​ k 0 2 ​ ​ . Expand the determinant: 𝐴 × 𝐵 = 𝑖 ∣ 1 0 − 6 2 ∣ − 𝑗 ∣ 5 0 2 2 ∣ + 𝑘 ∣ 5 1 2 − 6 ∣ . A×B=i ​ 1 −6 ​ 0 2 ​ ​ −j ​ 5 2 ​ 0 2 ​ ​ +k ​ 5 2 ​ 1 −6 ​ ​ . Solve each minor: For 𝑖 i: ∣ 1 0 − 6 2 ∣ = ( 1 ) ( 2 ) − ( 0 ) ( − 6 ) = 2. ​ 1 −6 ​ 0 2 ​ ​ =(1)(2)−(0)(−6)=2. For 𝑗 j: ∣ 5 0 2 2 ∣ = ( 5 ) ( 2 ) − ( 0 ) ( 2 ) = 10. ​ 5 2 ​ 0 2 ​ ​ =(5)(2)−(0)(2)=10. For 𝑘 k: ∣ 5 1 2 − 6 ∣ = ( 5 ) ( − 6 ) − ( 1 ) ( 2 ) = − 30 − 2 = − 32. ​ 5 2 ​ 1 −6 ​ ​ =(5)(−6)−(1)(2)=−30−2=−32. Combine results: 𝐴 × 𝐵 = 2 𝑖 − 10 𝑗 − 32 𝑘 . A×B=2i−10j−32k. Components: 𝑥 x-component: 2 2 𝑦 y-component: − 10 −10 𝑧 z-component: − 32 −32.