c4s1 - 1 4.1 Areas, Volumes, and Cross Products Chapter 4....

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Unformatted text preview: 1 4.1 Areas, Volumes, and Cross Products Chapter 4. Determinants 4.1 Areas, Volumes, and Cross Products Note. Area of a Parallelogram. Consider the parallelogram determined by two vectors a and b: Figure 4.1, Page 239. Its area is A = Area = (base) × (height) = a b sin θ 1 − cos2 θ. =ab Squaring both sides: A2 = a 2 b 2(1 − cos2 θ) = a 2 b 2 −a 2 b 2 cos2 θ 2 4.1 Areas, Volumes, and Cross Products a = 2 b 2 − (a · b)2. Converting to components a = [a1, a2] and b = [b1, b2] gives A2 = (a1b2 − a2b1)2 or A = |a1b2 − a2b1|. Definition. For a 2 × 2 matrix A = of A as a1 a2 b1 b2 det(A) = a1b2 − a2b1 = , define the determinant a1 a2 b1 b2 . Example. Page 249 number 26. Definition. For two vectors b = [b1, b2, b3] and c = [c1, c2, c3] define the cross product of b and c as b×c= b2 b3 c2 c3 ˆ− i b1 b3 c1 c3 ˆ+ j b1 b2 c1 c2 ˆ k. 4.1 Areas, Volumes, and Cross Products 3 Note. We can take dot products and find that b × c is perpendicular to both b and c. Note. If b, c ∈ R3 are not parallel, then there are two directions perpendicular to both of these vectors. We can determine the direction of b × c by using a “right hand rule.” If you curl the fingers of your right hand from vector b to vector c, then your thumb will point in the direction of b × c: Figure 4.3, Page 242. Example. Page 248 number 16. 4 4.1 Areas, Volumes, and Cross Products a1 a2 a3 b b b define the determiDefinition. For a 3 × 3 matrix A = 1 2 3 c1 c2 c3 nant as a1 a2 a3 det(A) = b1 b2 b3 = a1 c1 c2 c3 b2 b3 c2 c3 − a2 b1 b3 c1 c3 + a3 b1 b2 c1 c2 . Note. We can now see that cross products can be computed using determinants: ˆˆk ijˆ b × c = b1 b2 b3 . c1 c2 c3 5 4.1 Areas, Volumes, and Cross Products Theorem. The area of the parallelogram determined by b and c is b×c . Proof. We know from the first note of this section that the area squared is A2 = c b − (c · b)2. In terms of components we have A2 = (c2 + c2 + c2)(b2 + b2 + b2) − (c1b1 + c2b2 + c3b3)2. 1 2 3 1 2 3 Multiplying out and regrouping we find that A2 = b2 b3 c2 c3 2 + b1 b3 c1 c3 2 + b1 b2 c1 c2 Taking square roots we see that the claim is verified. 2 . QED Theorem. The volume of a box determined by vectors a, b, c ∈ R3 is V = |a1(b2c3 − b3c2) − a2(b1c3 − b3c1) + a3(b1c2 − b2c1 )| = |a · b × c|. Proof. Consider the box determined by a, b, c ∈ R3: Figure 4.5, Page 244. 6 4.1 Areas, Volumes, and Cross Products The volume of the box is the height times the area of the base. The area of the base is b × c by the previous theorem. Now the height is h = a | cos θ| = b×c a | cos θ| b×c = |( b × c ) · a | b×c . (Notice that if b × c is in the opposite direction as given in the illustration above, then θ would be greater than π/2 and cos θ would be negative. Therefore the absolute value is necessary.) Therefore V = (Area of base)(height) = b × c |(c × c) · a| b×c = |( b × c ) · a |. QED Note. The volume of a box determined by a, b, c ∈ R3 can be computed in a similar manner to cross products: a1 a2 a3 V = | det(A)| = b1 b2 b3 c1 c2 c3 Example. Page 249 number 37. . 7 4.1 Areas, Volumes, and Cross Products Theorem 4.1. Properties of Cross Product. Let a, b, c ∈ R3. (1) Anticommutivity: b × c = −c × b (2) Nonassociativity of ×: a × (b × c) = (a × b) × c (That is, equality does not in general hold.) (3) Distributive Properties: a × (b + c) = (a × b) + (a × c) (a + b) × c = (a × c) + (b × c) (4) Perpendicular Property: b · (b × c) = (b × c) · c = 0 (5) Area Property: b × c = Area of the parallelogram determined by b and c (6) Volume Property: a · (b × c) = (a × b) · c = ±Volume of the box determined by a, b, and c. (7) a × (b × c) = (a · c)b − (a · b)c Proof of (1). We have b×c = b2 b3 c2 c3 ˆ− i b1 b3 c1 c3 ˆ j+ b1 b2 c1 c2 ˆ k ˆ i j = (b2c3 − b3c2 )ˆ − (b1c3 − b3c1 )ˆ + (b1c2 − b2c1)k ˆ i j = − (b3c2 − b2c3 )ˆ − (b3c1 − b1c3)ˆ + (b2c1 − b1c2)k 8 4.1 Areas, Volumes, and Cross Products = − c2 c3 b2 b3 ˆ− i c1 c3 b1 b3 ˆ j+ c1 c2 b1 b2 ˆ k = −c × b QED Example. Page 249 number 56. ...
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This note was uploaded on 02/24/2012 for the course MATH 285 taught by Professor Igordolgachev during the Fall '04 term at University of Michigan-Dearborn.

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