VECTORS Mickey Frost Frances Coronel ! Pre-Calculus Mrs. Zborowski 

Definition directed line segment where the length represents magnitude and the direction represents direction used to represent quantities with both magnitude and direction (i.e. force, velocity) 

Visualize Initial Point:(1,2) Terminal Point:(4,4) Definition-Pictures Initial Point:(1,2) Terminal Point:(4,4) 1 2 3 4 1 2 3 4 5 6 7 8 9 7 5 6 8 9 10 

Distance Formula 

Ex.1: Vector Representation by Directed Line Segments Let u be represented by the directed line segment from P=(0,0) to Q=(3,2) Let v be represented by the directed line segment from R=(1,2) to S=(4,4) Show that u=v Definition-Pictures Initial Point:(1,2) Terminal Point:(4,4) 1 2 3 4 1 2 3 4 5 6 7 8 9 7 5 6 8 9 10 

Solution to Ex.1 Distance= The distance formula shows PQ and RS have same magnitude. ✓ IIPQII =√((3-0)²+(2-0)²)=√13 ✓ IIRSII =√((4-1)²+(4-2)²)=√13 Both line segments have same direction because they are both directed towards upper right on lines having slope of 2/3. SO IIPQII and IIRSII have same magnitude and direction so u=v 

Component Form of Vector Component Form of vector with initial point P(p1,p2) and terminal point Q(q1,q2) is given by: PQ= ==v Magnitude (length) is given by: IIvII= √((q₁-p₁)²+(q₂-p₂)²)= √v₁²+V₂² If IIvII is 1, v --->unit vector IIvII=o if and only if v is the zero vector 0 

Ex.2: Component Form of Vectors Find component form and magnitude of vector v that has initial point (4,-7) and terminal point (-1,5) PQ= <-1-4, 5-(-7)>=<-5,12>=v PQ= ==v So P=(4,-7)=(p₁,p₂) and Q=(-1,5)=(q₁,q₂) Finding component form SO v=<-5,12> and magnitude of v is: IIvII=√v₁²+V₂² IIvII=√(-5)²+(12)² IIvII=√169 IIvII=13 Finding magnitude IIvII=√(25+144) 

Two Basic Vector Operations 1. Scalar Multiplication 2. Vector Addition 

Definitions of Vector Operations Let u= and v= be vectors and let k be a scalar (a real number). Then the sum of u and v is the vector u+v = and the scalar multiple of k times u is the vector ku=k= SUM Scalar Multiple 

The negative of v= is -v=(-1)v =<-v₁, -v₂> and the difference of u and v is u-v= u +(-v) = Definitions of Vector Operations Continued Negative Add (-v) Differenc 

Ex. 3: Vector Operations Let v=<-2, 5> and w=<3,4> and find each of the following vectors. a. 2v b. w-v c. v+2w 

Solution to Ex. 3 a. Because v=<-2,5>, you have 2v=2<-2,5> =<2(-2),2(5)> =<-4,10> b. The difference of w and v is w-v=<3-(-2), 4-5> =<5,-1> c. The sum of v and 2w is =v+2w=<-2,5> +2<3,4> =<-2,5>+<2(3),2(4)> =<-2,5>+<6,8> =<-2+6, 5+8> =<4,13> v=<-2, 5> and w=<3,4> 

Properties of Vector Addition and Scalar Multiplication 1. u + v = v + u
 2. (u + v) + w= u + (v + w) 
 3. u + 0 = u
 4. u + (-u) = 0 
 6. (c+d)u= cu + d 7. c(du) = (cd)u
 8. c(u + v) = cu + cv 9. IIcvII=IcI IIvII For all vectors u, v, and w, and for all scalars c and d, the following properties are true: 

Unit Vectors Definition: vector with magnitude of 1 It’s helpful to find unit vectors in same direction as another vector u= unit vector = v = 1 IIvII IIvII v Scalar Vector 

Ex. 4: Unit Vectors Find the unit vector in the direction of v=<-2,5> and verify that the result has a magnitude of 1 

Solution to Ex. 4 v = IIvII <-2,5> √(-2)²+(5)² = 1 √29 <-2,5> = √29 √29 < > -2 5 , Vector has a magnitude of 1 because: √ 2 √29 ( )+ 5 √29 ( )= √4 2 2 √29 +25 √29 = √29 29 =1 

Standard Unit Vectors Unit vectors <1,0> and <0,1> are referred to as standard unit vectors denoted by i=<1,0> and j=<0,1> We use these to write component form of vectors as linear combination of vectors i and j <-1,1> = -1i+1j 

Ex. 5: Standard Unit Vectors Let u= -3i+8j and v=2i-j =2u-3v =2(-3i+8J)-3(2i-j) =-6i+16j-6i+3j =-12i+19j 

Direction Angles If v=ai+bj is any vector that makes angle θ with positive x-axis, it has the same direction as u v=IIvII =IIvII(cosθ)i+IIvII(sinθ)j Direction angle is then determined by: tanθ = sinθ cosθ = IIvIIsinθ IIvIIcosθ = b a 

Ex. 6: Direction Angles Find direction angle for u=3i+3j tan θ =3/3=1 so θ=45° Find component form given IIvII=5/2 and θ=45° =5/2 =5/2 <√2/2, √2/2> =<5/2(√2/2), 5/2(√2/2)> =(5√2/4, 5√2/4) 

Dot Product u= and v= u · v= u₁v₁+u₂v₂ Ex. 7 1) <3,4> · <2,-3> =(3)(2)+(4)(-3) =6+(-12) = -6 Orthogonal Vectors Orthogonal Vectors exist when the dot product=0 This means the vector is perpendicular and normal Ex. 8 Are the vectors orthogonal? u=<-12,30> v= <5/4,1/2> =(-12)(5/4)+(30)(1/2) =-15+15=0 So yes they are. 

Properties of Dot Product 1.u · v = v · u 2.0 · v = 0 3.u · (v+w) = u · v + u · w 4.v · v = IIvII² 5.c (u · v) = cu · v = u · cv Let u, v, and w, be vectors in the plane or in space and let c be a scalar: 

Angle Between 2 Vectors u · v IIuII IIvII cosθ = angle θ ; 0 ≤ θ ≤ π Ex.9 Find the angle between u=<4,3> and v= <3,5> cosθ = u · v IIuII IIvII 1. <4,3> · <3,5> = 27 θ= arccos 27/5√34 ≈ 22.2° II<4,3>II II<3,5>II (√4²+3²)(√3²+5²) (√25)(√34)=5√34 5√34 

Work W= IIFII IIPQII magnitude Distance W= cosθ IIFII IIPQII Determine work done by lifting 2400 lbs car 5 feet with crane W=2400lbsx5ft=1200 ft-lbs To slide an object across a floor, a person pulls a rope a constant force of 25 lbs at a constant angle of 35° above horizontal. Find work done if object dragged 40 ft W=(cos35°)(25lbs)(40ft) ! W=819.5 ft-lbs 

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