The Cross Product. MATH 311, Calculus III. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan The Cross Product

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Transcription:

The Cross Product MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Fall 2011

Introduction Recall: the dot product of two vectors is a scalar. There is another binary operation on vectors called the cross product whose result is yet another vector.

Introduction Recall: the dot product of two vectors is a scalar. There is another binary operation on vectors called the cross product whose result is yet another vector. Definition The determinant of a 2 2 matrix of real numbers is defined by a 1 a 2 b 1 b 2 = a 1b 2 a 2 b 1.

Examples Find the following 2 2 determinants. 1 1 2 3 1 2 2 1 5 4 3 3 0 4 2

Examples Find the following 2 2 determinants. 1 1 2 3 1 = (1)(1) (2)(3) = 5 2 2 1 5 4 = (2)(4) ( 1)( 5) = 3 3 3 0 4 2 = (3)(2) (0)(4) = 6

3 3 Determinants Definition The determinant of a 3 3 matrix of real numbers is defined as a combination of three 2 2 determinants, as follows: a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c 2

3 3 Determinants Definition The determinant of a 3 3 matrix of real numbers is defined as a combination of three 2 2 determinants, as follows: a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c 2 Remark: this method of calculating the 3 3 determinant is called expansion along the first row.

Examples Find the following 3 3 determinants. 1 2 1 1 3 0 1 5 4 2 3 2 2 2 1 1 5 2 1 6

Examples Find the following 3 3 determinants. 1 2 1 1 3 0 1 5 4 2 = 38 3 2 2 2 1 1 5 2 1 6 = 41

Cross Product Definition For two vectors a = a 1, a 2, a 3 and b = b 1, b 2, b 3, we define the cross product or vector product of a and b to be i j k a b = a 1 a 2 a 3 b 1 b 2 b 3 = a 2 a 3 b 2 b 3 i a 1 a 3 b 1 b 3 j+ a 1 a 2 b 1 b 2 k

Cross Product Definition For two vectors a = a 1, a 2, a 3 and b = b 1, b 2, b 3, we define the cross product or vector product of a and b to be i j k a b = a 1 a 2 a 3 b 1 b 2 b 3 = a 2 a 3 b 2 b 3 i a 1 a 3 b 1 b 3 j+ a 1 a 2 b 1 b 2 k Note: the cross product is another vector in V 3.

Examples If a = 1, 3, 4 and b = 2, 7, 5 then evaluate the following. 1 a b 2 b a 3 a 0

Examples If a = 1, 3, 4 and b = 2, 7, 5 then evaluate the following. 1 a b = 43, 13, 1 2 b a = 43, 13, 1 3 a 0 = 0, 0, 0

Properties of the Cross Product (1 of 2) Theorem For any vector a in V 3, a a = 0 and a 0 = 0.

Properties of the Cross Product (1 of 2) Theorem For any vector a in V 3, a a = 0 and a 0 = 0. Proof. a a = a 0 = i j k a 1 a 2 a 3 a 1 a 2 a 3 i j k a 1 a 2 a 3 0 0 0 =? =?

Properties of the Cross Product (2 of 2) Theorem For any vectors a and b in V 3, a b is orthogonal to both a and b.

Properties of the Cross Product (2 of 2) Theorem For any vectors a and b in V 3, a b is orthogonal to both a and b. Proof. (a b) a = (a b) b = i j k a 1 a 2 a 3 b 1 b 2 b 3 i j k a 1 a 2 a 3 b 1 b 2 b 3 a =? b =?

Cross Product Algebra Theorem For any vectors a, b, and c in V 3 and any scalar d, the following hold: 1 a b = (b a) (anticommutativity) 2 (da) b = d(a b) = a (db) 3 a (b + c) = a b + a c (distributive law) 4 (a + b) c = a c + b c (distributive law) 5 a (b c) = (a b) c (scalar triple product) 6 a (b c) = (a c)b (a b)c (vector triple product)

Geometric Interpretation (1 of 3) The vector a b is perpendicular to the plane in R 3 containing both a and b.

Geometric Interpretation (1 of 3) The vector a b is perpendicular to the plane in R 3 containing both a and b. Use the right-hand rule to determine the direction in which a b points.

Geometric Interpretation (2 of 3) Theorem For nonzero vectors a and b in V 3, if θ is the angle between a and b where 0 θ π, then a b = a b sin θ.

Geometric Interpretation (2 of 3) Theorem For nonzero vectors a and b in V 3, if θ is the angle between a and b where 0 θ π, then Proof. a b = a b sin θ. a b 2 = (a 2 b 3 a 3 b 2 ) 2 + (a 1 b 3 a 3 b 1 ) 2 + (a 1 b 2 a 2 b 1 ) 2 = a 2 2 b2 3 2a 2a 3 b 2 b 3 + a 2 3 b2 2 + a2 1 b2 3 2a 1a 3 b 1 b 3 + a 2 3 b2 1 + a2 1 b2 2 2a 1a 2 b 1 b 2 + a 2 2 b2 1 = (a 2 1 + a2 2 + a2 3 )(b2 1 + b2 2 + b2 3 ) (a 1b 1 + a 2 b 2 + a 3 b 3 ) 2

Geometric Interpretation (3 of 3) Corollary Two nonzero vectors a and b in V 3 are parallel if and only if a b = 0.

Geometric Interpretation (3 of 3) Corollary Two nonzero vectors a and b in V 3 are parallel if and only if a b = 0. Proof. a b = a b sin θ

Area of a Parallelogram a a sin Θ Θ b Area = (base)(altitude) = b a sin θ = a b

Example Find the area of the parallelogram with two adjacent sides formed by the vectors a = 3, 2, 2 and b = 5, 4, 6.

Example Find the area of the parallelogram with two adjacent sides formed by the vectors a = 3, 2, 2 and b = 5, 4, 6. Area = a b = 4, 8, 2 = 2 21

Distance from a Point to a Line Q PQ sin Θ P Θ R d = PQ sin θ = PQ PR PR

Example Find the distance from the point with coordinates (2, 1, 1) to the line through the points with coordinates ( 2, 1, 3) and (3, 1, 2).

Example Find the distance from the point with coordinates (2, 1, 1) to the line through the points with coordinates ( 2, 1, 3) and (3, 1, 2). Let P = ( 2, 1, 3), R = (3, 1, 2), and Q = (2, 1, 1), then d = PQ PR PR 4, 0, 2 5, 2, 5 = 5, 2, 5 4, 10, 8 = 5, 2, 5 10 = 3

Volume of a Parallelepiped a b b c a Volume = (Area of base)(altitude) = a b compa b c c (a b) = a b = c (a b) a b

Example Find the volume of the parallelepiped with three adjacent sides formed by the vectors a = 3, 2, 2, b = 5, 4, 6, and c = 2, 3, 4.

Example Find the volume of the parallelepiped with three adjacent sides formed by the vectors a = 3, 2, 2, b = 5, 4, 6, and c = 2, 3, 4. V = c (a b) = 2, 3, 4 ( 3, 2, 2 5, 4, 6 ) = 2, 3, 4 4, 8, 2 = 24

Torque Consider a wrench being used to tighten a bolt. A force vector F is applied at a point specified by a position vector r. The torque vector is defined to be τ = r F.

Magnus Force (1 of 2) The spinning motion of a ball creates a force called the Magnus force. ω: angular velocity of the spin (radians per second) spin vector s: vector parallel to the axis of spin with magnitude ω. v: velocity vector of the ball where c > 0 is constant. F m = c(s v)

Magnus Force (2 of 2)

Homework Read Section 10.4. Pages 825 828: 1 67 odd.