講者: 許永昌 老師 1. Contents Example Projection Axioms: Examples Normal Trajectory Shortest Distance Law of Cosines: C 2 =A 2 +B 2 +2ABcos . 2.

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講者: 許永昌 老師 1

Contents Example Projection Axioms: Examples Normal Trajectory Shortest Distance Law of Cosines: C 2 =A 2 +B 2 +2ABcos . 2

Example Assume that a block is putted on a frictionless incline, find out the acceleration of this block. 3

Projection 4 A B Acos  Bcos   還記得嗎? 交換律 結合律 分配律 Identity & inverses 還記得嗎? 交換律 結合律 分配律 Identity & inverses

Axioms of Scalar Product 5 If z=a+ib, the complex conjugate of z is defined as z * =a-ib where a and b are real numbers. If z=a+ib, the complex conjugate of z is defined as z * =a-ib where a and b are real numbers. P43 of Introduction to Mathematical Physics.

Scalar Product 6 *As long as the axioms are obeyed, you can get (A,B)=  i A i * B i.

Summary I 7

Dot Product A  B 8 AxAx AyAy A AA BB A B

Example P14~P17 Q: If a free motion particle whose position is r(t)=(x(t),y(t))=(-3t,4t), Please find that Its equation of trajectory. Its velocity at time t 0. The Concepts we need: A straight line (2D) and a plane (3D) can be describe by is the normal of a straight line (2D) or a plane (3D). r=r 0 +sT (2D) or r=r 0 +sl 1 +tl 2 (3D). T, l 1 and l 2 are the tangent vectors. Velocity: 9

Example P17~P18 Find the shortest distance of a rocket from an observer. i.e. |r(t)  r 0 | 2 =min. 10 Observer: r 0 r(t)r(t) r(t’)r(t’)

The tangent and normal of a curve If we know r 0 and the tangent vector v(r 0 ), we can get Tangent: r t =r 0 +v(r 0 )t. Normal: v(r 0 )  (r n  r 0 )=0. 11 tangent normal r0r0

Law of Cosines C 2 =A 2 +B 2 +2ABcos . Proof: C=A+B, C 2 =(A+B)  (A+B)=A 2 +A  B+B  A+B 2 =A 2 +B 2 +2ABcos . Proof2: C 2 =(A+Bcos  ) 2 +(Bsin  ) 2 =A 2 +B 2 +2ABcos . 12  A B C  A B C Bcos  Bsin 

Summary of 1.2 Scalar or Dot product is related to the geometric concept of projection of vectors. For 3D: A  B=B  A=ABcos  =A x B x +A y B y +A z B z. The applications shown here are: Use the normal to define a plane or a straight line algebraically: Finding the shortest distance Distance =|r 2 -r 1 |= Shortest distance: d|r 2 -r 1 |=0. 想一下為何數學上可以如此表示。 Scalar Product in general: (A,B)=(B,A)* and (B,a 1 A 1 +a 2 A 2 )=a 1 (B,A 1 )+a 2 (B,A 2 ). 13

Homework

1.2 Nouns 15