# notes - 1 Review: Vector Analysis REFERENCES: Arya Chapter...

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1 Review: Vector Analysis REFERENCES: Arya Chapter 5. 1.1 Vectors A scalar has only a magnitude. A vector has both a magnitude and a direction. In this course, vectors will be represented in two ways (bold, or overarrow): R = -→ R, (1) or in the special case of a unit vector, with a hat (e.g. ˆ i ). The magnitude of a vector is represented R = | R | = | R | (2) Vectors can be written in components. For example, in two dimensions: R =( R x ,R y ) (3) and R x = R cos( θ ) R y = R sin( θ ) (4) where R = ± R 2 x + R 2 y tan θ = R y R x (5) We also denote vectors with unit vectors : For example in 3D cartesian coordinates: ˆ i = (1 , 0 , 0) , ˆ j = (0 , 1 , 0) , ˆ k = (0 , 0 , 1) (6) where ˆ i is the unit vector associated with the x-axis direction, ˆ j the y-axis direction, and ˆ k the z-axis direction. Each has unit magnitute | ˆ i | = | ˆ j | = | ˆ k | = 1 (7) A general vector in 3D can then be represented as: R R x y z )= R x ˆ i + R y ˆ j + R z ˆ k (8) Adding and subtracting vectors can be done component-wise. From this, certain proper- ties follow. Consider vectors A , B and C and scalar ϕ : 1. A i + B i = B i + A i (commutative law) 2. A i +( B i + C i ) = ( A i + B i )+ C i (associative law) 3. ϕ A = B is a vector 1

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1.2 Scalar product The scalar product is defned to be a scalar quantity obtained by multiplying the magnitudes oF two vectors: S = A · B = AB cos θ = ± i A i B i (9) where θ is the angle between the vectors. The magnitude oF a vector can be calculated | R | = R · R = ² ± i R 2 i (10) The scalar product has certain useFul properties: 1. A · B = B · A (commutative). 2. A · B = 0 iF A or B = 0, or θ = 90 degrees (the vectors are orthogonal ). 3. A · ( B + C )= A · B + A · C (distributive) 4. ( ϕ A ) · B = ϕ ( A · B ) (associative with scalar multiplication) example: Two position vectors are A = ˆ i +2 ˆ j - 2 ˆ k and B =4 ˆ i ˆ j - 3 ˆ k . ±ind the magnitude oF the vector From point A to B, the angle θ between A and B , and the component oF B in the direction oF A solution: The vector From point A to B is B - A . By components: B - A = (4 - 1) ˆ i + (2 - 2) ˆ j +( - 2 + 3) ˆ k =3 ˆ i - ˆ k (11) The magnitude oF this vector is | B - A | = 9 + 1 = 10 (12) ±rom the scalar product Eq. 9: cos θ = A · B AB = 4 + 4 + 6 9 29 =0 . 8666 (13) The component oF B in the direction oF A is B cos θ , so B cos θ = A · B A = 14 3 . 67 (14) 2
1.3 Vector product The vector (or cross) product is defned to be a vector quantity: C = A × B (15) i.e. C has a magnitude and direction. The magnitude oF C is C = AB sin θ (16) (For 0 < θ < π ), and it’s direction is determined by the right-hand rule, defned along the normal direction ˆ n . Thus, For example, one can defne the area oF a parallelogram with two sizes B and C as area A = B × C =( BC sin θ n (17) The components oF C are defned by the relation C i ± j,k ε ijk A j B k (18) where ε ijk is the Levi-Civita symbol, defned: ε 123 = ε 231 = ε 312 = 1 (19) ε 132 = ε 213 = ε 321 = - 1 ε ijk = 0 otherwise . Alternatively, the vector product can be represented by a determinate: C = ˆ i ˆ j ˆ k A x A y A z B x B y B z (20) in other words C x = A y B z - A z B y (21) C y = A z B x - A x B z C z = A x B y - A y B x The vector product has certain useFul properties: 1. A × B = - B × A (anticommutative).

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## This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.

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notes - 1 Review: Vector Analysis REFERENCES: Arya Chapter...

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