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>
Pre Calculus Answers
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