Homework Questions
Chapter 6
Section 6.1 Vectors
Vectors
• These are directed line segments
Initial Point vs. Terminal Point
• Let u be the vector from
R(-4, 2) to S(-1, 6) and v be the
vector from Q(0, 0) to P(3, 4).
• Prove u=v
Use the distance formula
d  ( x  x)  ( y  y )
2
d  (1  4)  (6  2)
2
d 5
• Therefore, RS=QP, so u=v
d  ( x  x) 2  ( y  y ) 2
2
2
d  (3  0)  (4  0)
2
d 5
2
Standard Position
• Starts at the origin
• Component Form - <v1, v2>
• Also called the position of the vector of
the point (v1, v2)
Plot the vector
• <2, 3>
If not in Standard Position…
• V1 = x2-x1
• V2 = y2-y1
OR
• V = < x2-x1, y2-y1>
Find the component form:
• P = (-3, 4)
• Q = (4, 9)
• R = (-2, 5)
• S = (2, -8)
• Find PQ
• Find QR
• Find RS
Magnitude
• This really just means length
• Distance formula
v  v1  v 2
2
2
OR
( x2  x1 ) 2  ( y 2  y1 ) 2
Find the magnitude
• Of RS from before
• Find
v of PS
Vector Addition and Scalar Mult
• u+v = <u1, u2> + <v1, v2>
= < u1+ v1, u2 + v2>
• ku = k<u1, u2>
<ku1, ku2>
Solve
• u = <-1, 3>
• v = <4, 7>
• Find u+v
• Find 3u
• Find 2u+(-1)v
Direction Angles
v = <|v|cos×, |v|sin×>
v
|v|
Finding the components of a vector:
1. Find the components of the vector v with
direction angle 115° and magnitude 6.
Examples
2. Find the components of the vector v with
direction angle 120° and magnitude 10.
Examples
3. Find the components of the vector v with
direction angle 45° and magnitude 8.
Examples
4. Find the components of the vector v with
direction angle 210° and magnitude 24.
Finding Direction Angles
5.
r = <4, -5>
Finding Direction Angles
6. p = 3i + 7j
7. p = -6i + 2j
8. <-3, -8>
HOMEWORK
• Pg. 511 (1-35 odd)
• Remember, you can check in the back
of the book to make sure you are
doing these correctly! Come with
questions ready! 