6.07 - Vector Proofs

»You will need ˃Analytic Geometry ˃Properties of shapes ˃Properties of angles ˃Properties of vector operations » Distance Formula 𝑑 𝐴, 𝐵 = 𝑥𝐵 − 𝑥𝐴 2 + 𝑦𝐵 − 𝑦𝐴 2 » Pythagorean Theorem 𝑐 2 = 𝑎2 + 𝑏 2 » Midpoint Formula 𝑥𝐵 + 𝑥𝐴 𝑦𝐵 + 𝑦𝐴 𝑀𝐴𝐵 = , 2 2 » Division Point Formula 𝑎 𝑎 𝑃𝐴𝐵 = 𝑥𝐴 + 𝑥𝐵 − 𝑥𝐴 , 𝑦𝐴 + 𝑦𝐵 − 𝑦𝐴 𝑏 𝑏 Where 𝑏 is the full segment length of 𝐴𝐵, and 𝑎 is the portion away from 𝐴 to 𝐵 » Sum of Interior Angles » What makes a: ˃ Trapezoid a trapezoid ˃ Rectangle a rectangle ˃ Square a square ˃… ˃ Think proportions, parallels, # of sides … etc » » » » » » Vertically Opposite Alternate Interior Alternate Exterior Corresponding Complementary Supplementary 2 1 4 6 5 8 3 7 1. The sum of two vectors is a vector. 2. Commutativity 𝑢 + 𝑣റ = 𝑣റ + 𝑢 3. Associativity 𝑢 + 𝑣റ + 𝑤 = 𝑢 + (𝑣റ + 𝑤) 4. Identity Element 𝑢+0=𝑢 5. Additive Inverse (the opposite) 𝑢 + (−𝑢) = 0 Addition of Two Vectors 1. The product of a vector and a scalar is a vector. 2. Associativity 𝑘(𝑙𝑢) = 𝑘𝑙(𝑢) 3. Identity Element 1𝑢 = 𝑢 4. Distributivity over Vector Addition 𝑘(𝑢 + 𝑣) റ = 𝑘𝑢 + 𝑘𝑣റ 5. Distributivity over Scalar Addition (𝑘 + 𝑙)𝑢 = 𝑘𝑢 + 𝑙𝑢 Multiplication of a Vector by a Scalar 1. The scalar product of two vectors is a scalar. 2. Commutativity 𝑢 ⋅ 𝑣റ = 𝑣റ ⋅ 𝑢 3. Associativity 𝑘𝑢 ⋅ 𝑣റ = 𝑘 𝑢 ⋅ 𝑣റ - OR𝑘𝑢 ⋅ 𝑙 𝑣റ = (𝑘𝑙)(𝑢 ⋅ 𝑣) റ 4. Distributivity 𝑢 ⋅ 𝑣റ + 𝑤 = 𝑢 ⋅ 𝑣റ + (𝑢 ⋅ 𝑤) Dot Product of Two Vectors » To prove the validity of Parallelogram Method of addition, knowing that Chasles’ is true If 𝐴𝐵𝐶𝐷 is a parallelogram, prove that 𝐵𝐴 + 𝐵𝐶 = 𝐵𝐷 » Problem: Hypothesis: 𝐴𝐵𝐶𝐷 is a parallelogram Conclusion: 𝐵𝐴 + 𝐵𝐶 = 𝐵𝐷 » Proof: 1. 𝐵𝐴 + 𝐵𝐶 = 𝐵𝐴 + 𝐴𝐷 Since ABCD is a parallelogram, 𝐵𝐶 = 𝐴𝐷 2. 𝐵𝐴 + 𝐵𝐶 = 𝐵𝐷 By the Sum of two vectors » Using Addition properties to prove subtraction If 𝐴𝐵𝐶𝐷 is a parallelogram, show that 𝐴𝐷 − 𝐴𝐶 = 𝐷𝐴 − 𝐷𝐵 Problem: Hypothesis: 𝐴𝐵𝐶𝐷 is a parallelogram Conclusion: 𝐴𝐷 − 𝐴𝐶 = 𝐷𝐴 − 𝐷𝐵 Proof: 𝐴𝐷 − 𝐴𝐶 = 𝐴𝐷 + 𝐶𝐴 by definition of opposite vectors = 𝐴𝐵 + 𝐵𝐷 + 𝐶𝐷 + 𝐷𝐴 by sum of vectors = 𝐷𝐴 + 𝐵𝐷 + 𝐴𝐵 + 𝐶𝐷 by distributivity = 𝐷𝐴 − 𝐷𝐵 + 𝐴𝐵 + 𝐶𝐷 by commutativity & def’n of opposite vectors = 𝐷𝐴 − 𝐷𝐵 + 𝐴𝐵 − 𝐴𝐵 by hypothesis 𝐶𝐷 = 𝐴𝐵 = 𝐷𝐴 − 𝐷𝐵 » Linear Combinations and proving collinear state » Two vectors are defined by a linear combination of 𝑎Ԧ and 𝑏. Show that these two vectors are collinear. 𝑣Ԧ = −4𝑎Ԧ + 3𝑏 and 𝑢 = 2𝑎Ԧ − 1.5𝑏 » 𝑣Ԧ = −4𝑎Ԧ + 3𝑏 by definition » 𝑣Ԧ = −2(2𝑎Ԧ − 1.5𝑏 ) by distributivity » 𝑣Ԧ = −2(𝑢 ) by definition » Therefore 𝑢 and 𝑣Ԧ are collinear, since 𝑣Ԧ = 𝑘(𝑢 ), where 𝑘 = −2 » Linear Combinations and Orthonormal Basis » Orthonormal: orthogonal + magnitudes of 1 » Two vectors are defined by a linear combination of orthonormal vectors 𝑖Ԧ and 𝑗Ԧ. Show that these two vectors are orthogonal. 𝑢 = 3Ԧ𝑖 + 2Ԧ𝑗 and 𝑣Ԧ = −4Ԧ𝑖 + 6Ԧ𝑗 1. 𝑢 ⋅ 𝑣Ԧ = 3Ԧ𝑖 + 2Ԧ𝑗 ⋅ −4Ԧ𝑖 + 6Ԧ𝑗 by substitution 2. 𝑢 ⋅ 𝑣Ԧ = −12 𝑖Ԧ ⋅ 𝑖Ԧ + 18 𝑖Ԧ ⋅ 𝑗Ԧ − 8 𝑗Ԧ ⋅ 𝑖Ԧ + 12 𝑗Ԧ ⋅ 𝑗Ԧ by distributivity 3. 𝑢 ⋅ 𝑣Ԧ = −12 𝑖Ԧ 2 + 18(0) − 8(0) + 12 𝑗Ԧ since 𝑖Ԧ and 𝑗Ԧ are orthogonal 4. 𝑢 ⋅ 𝑣Ԧ = −12 + 12 since 𝑖Ԧ and 𝑗Ԧ both have a magnitude of 1 5. 𝑢 ⋅ 𝑣Ԧ = 0 » Therefore 𝑢 and 𝑣Ԧ are orthogonal, since they have a zero scalar-product. 2 » Proofs and Triangles » In a given triangle ABC, point M is the midpoint of side BC. Show that: 1 1 𝐴𝑀 = 𝐴𝐵 + 𝐴𝐶 2 2 𝐵 𝑀 𝐴 » Problem: Hypothesis: ABC is a triangle M is the midpoint of 𝐵𝐶 1 1 Conclusion: 𝐴𝑀 = 𝐴𝐵 + 𝐴𝐶 2 2 𝐶 » Problem: Hypothesis: ABC is a triangle M is the midpoint of 𝐵𝐶 1 1 Conclusion: 𝐴𝑀 = 𝐴𝐵 + 𝐴𝐶 2 2 𝐵 𝐴 𝑀 𝐶 » Proof: 1. 𝐴𝑀 = 𝐴𝐵 + 𝐵𝑀 by Chasles method 𝐴𝑀 = 𝐴𝐶 + 𝐶𝑀 2. 2𝐴𝑀 = 𝐴𝐵 + 𝐵𝑀 + 𝐴𝐶 + 𝐶𝑀 by adding the vectors defined in 1. 3. 2𝐴𝑀 = 𝐴𝐵 + 𝐵𝑀 + 𝐴𝐶 − 𝐵𝑀 by hypothesis 𝐶𝑀 = −𝐵𝑀 4. 2𝐴𝑀 = 𝐴𝐵 + 𝐴𝐶 + 0 by adding opposite vectors 5. 𝐴𝑀 = 1 2 1 𝐴𝐵 + 𝐴𝐶 1 6. 𝐴𝑀 = 𝐴𝐵 + 𝐴𝐶 2 2 by dividing by 2 by distributivity To prove that quadrilateral ABCD is a parallelogram, you can prove that 𝐴𝐵 = 𝐷𝐶 𝑎𝑛𝑑 𝐴𝐷 = 𝐵𝐶 𝐴 𝐵 𝐷 𝐶 To prove that line segments 𝐴𝐵 and 𝐶𝐷 are parallel you can prove that 𝐴𝐵 and 𝐶𝐷 are collinear. (show equal slopes or 𝐴𝐵 = 𝑘(𝐶𝐷) 𝐴𝐵 𝐶𝐷 To prove that line segments 𝐴𝐵 and 𝐶𝐷 are perpendicular, you can prove that 𝐴𝐵 and 𝐶𝐷 are orthogonal 𝐴𝐵 ⋅ 𝐶𝐷 =0 To prove that M is the midpoint of segment 𝐴𝐵, you can prove that 𝐴𝑀 = 𝑀𝐵 (Equipollent) or 𝐴𝑀 = −𝐵𝑀 (Opposite)