Lecture Note
University
California State UniversityCourse
MATH 150B | Calculus IIPages
5
Academic year
2023
yung dump
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12.2 Vectors 25 Def.The vector is a directed line segment AB with Terminal point initial point A Terminal point B instituel point A13 length IAB/ Two vectors are equal if they have D the same length and direction y c F R " regardless the initial point" K Vectors are usually written in lower case and boldface : U,V, w x or = with arrows: , v / 33 z Let = PQ, there is only one 62(m)%,2.) i= PQ = (Un,Ve,43) directed line segment = PQ whose initial point is the origin. 0 It is the representative of in standard y standard position and is the vector position vector X we normally use to represent Since V is a three- dimensional vector with initial point at origin and terminal point (v1, V21 V3), the component form of is V = V1, U2, V3 In two dimension = = Xz- X1, Y2-Y, Zz-Z1> is in plane The magnitude or length of V=PQ is lvl= v1 + V2 + V32 = (x2-x)2 (y2-yi) (21-21) Note: The only vector with length 0 is zero vector TO = or o'= This vector is also the only vector with no direction.
Exp Find [as the component form 26 lbs the length of the vector with initial point p(-2,5,2) and terminal point Q(-4,3,3) [as The standard position vector V = PQ has component form V = = = = 1b) lvl= = Vecfor Algebra Operations Let = (41,421 43) and V [I] Addition: u i+v= scalar 2 Scalar multiplication: ku = where to scale KEIR 75 is y y is =K41,42> is = V2 V1 3 42 x X u1 The parallelogram law of vector addition it zup 3x/ The difference u-v -vj 1.3.7 is Note + (-v) +(-v) Proof:/kuil= =
Exp Let -2,4,0> and V=L3,1,5> 27 [a) Find the component of 24+3V +3 =
Unit Vectors 28 A vector is of length 1 is called a unit vector. The standard unit vectors are =(1,0,0>, , K= Any vector v = can be written as a linear combination of the standard unit vectors : V = = + LO,V2, 0> + LO,O,V3> = v,8,1,0,0> + V2 + (0,0,1) = V1 + V2 j + V3 K the K- component of V the i-component of the j- component of 2 Pa(xiy,2.) OP, = X1 i + ys + 2, K OP2 = X2 i + y2 j + 22 K 0 j P p2= (x2-x1) + (Y2-Y,) i + (22-2) RK y X If V # 0 then I V I 7 O and I V 1 I = = 1 This means that V is a unit vector in the direction. of v. /vl V So called the direction of V . is Exp Find a unit vector is in the direction of the vector from L, (2,1,2) to P2 (4,3, 1) is = P P2= = (4-2) it (3-1)3 + (1-2) = 21 + 2j- - K /ii = 4+4+1 = 9 = 3 vector it = u 1 = 31+3j-1 is the direction of u . The unit ul
Exp Let V = 31-4j - be a velocity vector. 29 Express v as a product of its speed times a unit vector in the direction of motion. Speed is the magnitude (length) of v given by 101= 9+ 16 = V25 = 5 The unit vector V = 3 - has the same direction as V. /V/ 5 5 So 4 V = 31 4) = 5 (1/4 5 5 length Direction of motion "speed" Summary : we can express any non zero vector i interms of it's length and direction by writing v=lvl Exp Express the vector 21 + -2K as a product of its length and direction. length = 4+1+4 = 9 = 3 Direction = 2; 3 + 13-3 21 + J -2 K =3(3i+33-37) = + K Midpoint of a line Segment: The midpoint M of the line segment joining the points P,(x,y,2) and P2 (X2, 12,22) is the point f2(X,1,2,2) E,+2) Exp: Find to the midpoint of the line segment joining P,(-1,1,5) , 22(2,5,0) Midpoint is (1/3,) 2 the direction of PiP2 is =Pipz = (2+1) (5-1)j + (0-5) = 31+4J-5 I it = 9+16+25 = 55=5V2 The direction is = 3 + 4 J - I K /vi 5V2 5V2 V2
Vectors: Definition, Properties, and Operations
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