Pre Calculus Answers

PreCalculus fon 6.3 Notes (Day 1) Many quantities in geometry or physics can be represented by a single real number (Ex: area, time, etc.) Other quantities have both magnitude and direction. They are represented by vectors (Ex: force, velocity.) Def: P A vector is a directed line segment PQ initial point terminal point Vectors are denoted with bold-faced variables. Def: Let v be a vector with initial point P=(p. p.) and terminal point Q = (4.9). (1) v=PQ(9-P9P)=(V) This is component form. (2) The magnitude (length) of v is denoted by where =+v Sketch each vector. Find the component form and the magnitude. F Initial Point: (2,-3) Ex: Initial Point: (0,0) Terminal Point:(4, 1) Terminal Point:(2, 4) v= <2,47 M=2√5 4 4-2, 1--3 =(2,4) 141=√20 = 2√5 V2²+4² NOTE: u and v are equivalent vectors because they have the same direction and magnitude. In the second vector, the initial point is (0, 0). When this is true, v =(v)=(9.9%). The component form of the vector has the same values as the terminal point. 11. Vecos operations. A Scalar multiplication: a scalar is a constant -kv=k(vv)=(kv, kv₂) Ex Given v. find 2v. Sketch both. =(1.2) 2-(2-1,2-2) =<2,47 Note: 2v is twice as long. Ex: Given u, find -u. Sketch both. (3,4) -= (-3,-47 Note: u has the same magnitude, but opposite direction B. Vector addition (sum/resultant) (n+wa+n) = a + Ex: =-21) v = (2,3) += -2+2, 1+3 = <0,47 Sketch u, ther sketch V by- attaching it tou C. Vector subtraction u-v=u+-v = (--) Ex: u = (-2,3) u-v=ut-v-5,2> U= (-2,3) V= <3,17 -14-3-17 add the uppocite u-v: vector from the terminal point of v to the terminal point of u. It is useful to have a unit vector (vector of length 1) in the same direction as a given vector. Unit vector = (Note: the unit vector is a scalar multiple of v.) T Def: Standard Unit Vector: f(1.0) j = (0.1) horizontal vertical Ex: Convertv21+3j to component form. =2<1,0>+3<0,1> = <2,0>+<0,3> = <2,37 Ex: Therefore: vai + bj = (a,b) (Component Form) Find u + vil u21+3jand v=4i-7j. Note: the same result could have been obtained without u=<2,3> v=<4,-7> converting to component form. utv = <6,-47 = bi-4j IV. Find a unit vector in the direction of a given vector. Ex: u = -3 1/4/1=(-3)+y) // wll llull=5. -3 4 Ex: w=-4i+4/3j 1/wll=²+(43) = √64=8 V. Find the vector v with the given magnitude and the same direction as #. Ex: =3, u(5,-5) 114ll= √5²+(-5)²= √50=5√2 find the unit vector for u, then multiply it by 3 Vait vector =>=<뜰,틀> 5 5 multiply by 3:32, -32 1 Lech 6.3 (con't) Day 2 Let u be a unit vector. Then the terminal point of u is on the unit circle. (Diagram of a unit circle with a vector u) u = u = cos θ i + sin θ j Let v be a vector that makes an angle θ with the positive X-axis. v = ||v|| (cos θ i + sin θ j) v = ||v|| cos θ i + ||v|| sin θ j degrees General Form of v: v = ai + bj tan θ = sin θ / cos θ = ||v|| sin θ / ||v|| cos θ = b / a tan θ = b / a where v = ai + bj Find the magnitude and direction angle for v. Ex: v = 4(cos 20° i + sin 20° j) ||v|| = 4 θ = 20° Ex: v = 3 - √3 j v = <3, -√3> ||v|| = √(3^2 + (-√3)^2) ||v|| = 2√3 tan θ = -√3 / 3 θ = 330° Ex: v = -5i - 5j v = <-5, -5> ||v|| = √((-5)^2 + (-5)^2) ||v|| = 5√2 tan θ = -5 / -5 tan θ = 1 θ = 225° Picture sketching the vector IV III Section I v = <-2, -7> ||v|| = √53 tan θ = 7/2 θ' = 74° θ = 254° v = <-6, 2> ||v|| = 2√10 tan θ = -1/3 θ' = 19° θ = 162° Section II Find the component form of v given ||v|| and θ. Sketch v. Section IIA ||v|| = 4, θ = 90° v = ||v|| v = 4<0, 1> v = <0, 4> Section IIB ||v|| = 6, θ = 240° v = ||v|| v = 6<-1/2, -√3/2> v = <-3, -3√3> Section IIC ||v|| = 3, θ = 26° v = 3 v = 3<0.899, 0.438> v = <2.697, 1.314>