Course Name : Physics I Course # PHY 107

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Course Name : Physics I Course # PHY 107 Lecture-2 : Representation of Vectors and the Product Rules Abu Mohammad Khan Department of Mathematics and Physics North South University http://abukhan.weebly.com Copyright: It is unlawful to distribute or use this lecture material without the written permission from the author

Lecture-2: Topics to be covered 1. Unit vectors and their properties 2. Addition/Subtraction rules 3. Product rules for vectors: a) Dot/Scalar product b) Cross/Vector product 4. Polar form of product rules and the geometrical interpretations 5. Examples

Unit Vectors: Definition: any vector whose length or magnitude is one is called a unit vector. In Cartesian Coordinate system, the unit vector along the x-axis is denoted by ^i, and similarly ^j, are the unit vectors along the y- and z- axes. ^k In three dimensions, any vector is expressed as A=( A x, A y, A z )= A x ^i + A y ^j+ A z ^k The unit vectors are mutually perpendicular. For any two vectors, the addition/subtraction is given by A± B=( A x, A y, A z )±(B x, B y, B z )=( A x ±B x )^i +( A y ±B y ) ^j+( A z ±B z ) ^k The polar form of the sum can be obtained by Pythagorean Theorem.

Product Rules for Vectors: There are two rules for product between two vectors: 1) Dot or Scalar Product: It is defined for the unit vectors as ^i ^i=^j ^j= ^k ^k=1 unit vectors, ^i ^j=^j ^k=^k ^i=0. Orthogonality of vectors 2) Cross or Vector Product: It is defined for the unit vectors as ^i ^i=^j ^j=^k ^k=0, (Parallel Vector properties) ^i ^j= ^k, ^j ^k=^i, ^k ^i=^j, (The Right-hand Rule) ^j ^i= ^k, ^k ^j= ^i, ^i ^k= ^j (Note the change in direction)

In Cartesian form, these two products between any two vectors are very well known and easy to remember: The Dot product between any two vector is now given by A B=( A x ^i + A y ^j+ A z ^k) (Bx ^i +B y ^j+b z ^k)= A x B x + A y B y + A z B z Scalar The Cross product between any two vectors is now given by A B=^i ( A y B = z A z B y ) ^j( A x B z A z B x )+ ^k ( A x B y A y B x ) ^i ^j ^k A x A y A z quantity B x B y B z Vector

Polar form of the Dot product (Geometrical Interpretation) A B= A ( B cosθ )=( A cosθ ) B = A B cosθ. So, it is really a scalar multiplication (multiplying a number by another number), also known as scaling. The result is a scalar.

Polar Form of the Cross Product (Geometrical Interpretation) A B= A B sinθ ^n =( AB sinθ ) ^n =(Area of the Parallelogran) ^n Area of the Parallelogram= AB sinθ Note that: A surface or plane is a vector quantity The area is the magnitude of the surface and the direction of the surface or plane is given by ^n which is known as the normal unit vector.

Volume or Scalar Triple product: Area of the bottom face= A B = AB sin 90 = AB Volume of the parallelopiped = C ( A B) = C x C y C z A x A y A z B x B y B z Scalar quantity Using the properties of Determinant, it is to show that: A ( B C)= C ( A B)= B ( C A).

Example: Vectors C and D have magnitudes of 3 and 4 units respectively. What is the angle between the directions of C and D if the magnitude of the vector product is zero? 12 units? C D By definition of thr cross product : C D=CD sinθ Applying the given values, we obtain 0=(3)(4)sinθ θ =0,180 These are parallel or anti-parallel vectors. For the 2nd case, applying the given values, we obtain 12=(3)(4)sinθ θ =π /2 rad or 90 These are orthogonal or perpendicular vectors.

Example: The two vectors a and b in Figure have equal magnitudes of 10.0 m and the angles are θ 1 =30 O and θ 2 =105. O. Find the (a) x and (b) x components of their vector sum r, (c) the magnitude of r, and (d) the angle r makes with the positive direction of the x axis. Solution: r x =a x +b x =acosθ 1 +b cos(θ 1 +θ 2 ) =(10cos 30 O +10 cos135 O )m=1.59 m. r y =a y +b y =asinθ 1 +b sin(θ 1 +θ 2 ) =(10sin 30 O +13sin 105 O )m=12.1m. Therefore, by Pythagorean Theorem, r= r x 2 +r y 2 = (1.59) 2 +(12.1) 2 m=12.2 m. θ r =tan 1 (r y /r x )=tan 1 (12.1/1.59)=82.5 O.

Example: For the vectors in the figure below, with a = 4, b = 3, and c = 5, what are (a) the magnitude and (b) the direction of a x b, (c) the magnitude and (d) the direction of a x c, and (e) the magnitude and (f) the direction of b x c? (The z axis is not shown, but it is perpendicular to the page and outward). Solution: From the diagram, th egiven vectors are: a=4 ^i=(4,0)=(4,0 O ), b=3 ^j=(0,3)=(3,90 O ), c= 4 ^i 3 ^j=(4,3)=(5,217 O ). Firstly, a b=(4 ^i) (3 ^j)=12(^i ^j)=12 ^k. Therefore, (a) magnitude of a b=12 and (b) Direction of a b= ^k. Similarly, b c=(3 ^j) ( 4 ^i 3 ^j) =( 12 ^j ^i)+( 9 ^j j) = 12( ^k)+( 9) 0=12 ^k. Therefore, (e) magnitude of b c=12 and (f) Direction of a b= ^k.