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9.4 Vectors in the Plane

vector: (v or v )a mathematical object with
both magnitude and direction. Vectors are
depicted in two ways:
component form: the form <a,b> used to
denote a vector.
standard representation: the arrow from the
origin to the point (a, b)
9.4 Vectors in the Plane
magnitude of v: denoted |v|, it is the length of the
vector.
direction of v: the direction in which the arrow
points
initial point: the starting point of the arrow, also
called the “tail”
terminal point: the ending point of the arrow, also
called the “head”
Head Minus Tail Rule
If an arrow has initial point x1 , y1  and terminal
point x2 , y2  it represents the vector x2  x1, y2  y1
equivalent vectors: two arrows which represent the
same vector.
Example 1:
Show that the arrows from (2, 3) to (4, 7) and from
(-1, 1) to (1, 5) are equivalent
x2  x1 , y2  y1
x2  x1 , y2  y1
4  2,7  3
1  1,5  1
2,4
2,4
Magnitude
If v is represented by the arrow from x1 , y1  to
x2 , y2  then v  x  x    y  y 
2
2
If v = <a, b> then
1
2
2
1
v  a 2  b2
Example 2:
Find the magnitude of the vector from the point
(-2, 3) to (4, 6)
v
x2  x1 2   y2  y1 2
v
4  22  6  32
v
62  32
v  36  9
v  45
v 3 5
Vector Addition and Scalar Multiplication
Let u=<u1, u2> and v=<v1, v2> and let k be a real
number (scalar). Then the sum of the vectors is:
u  v  u1  v1 , u2  v2
The product of the scalar k and a vector is:
ku  k u1 , u2  ku1 , ku2
Example 3:
Let u=<3, 2> and let v=<-1, 3>. Find 3u + 2v
3u  2v
3 3,2  2  1,3
9,6   2,6
7,12
Unit Vectors
unit vector: a vector with magnitude 1
unit vector in the direction of v: to find the unit
vector u in the direction of v use the following
formula:
v 1
u  v
v v
Unit Vectors
standard unit vectors: the vectors i=<1, 0> and
j=<0, 1> are unit vectors in the direction of the
positive x-axis and y-axis respectively.
linear combination: the expression v=ai+bj used to
represent the vector <a, b>
Example 4:
Find a unit vector in the direction of the vector
2
2
v=<3, 4>
v  a b
2
2
v  3 4
v
u
v
v  9  16
u
3,4
5
3 4
u ,
5 5
v  25
v 5
Direction Angles
If v has the direction angle θ as measured from the
x-axis, then the component form of the vector
can be found by the following formula:
v  v cos  , v sin 
Example 5:
A basketball is shot at a 60° angle with an initial
speed of 10 m/s. Find the component form of
this vector.
v  v cos  , v sin 
v  10 cos 60,10 sin 60
v  5,5 3
9.4 HW Assignment
Pg. 603: #s 18-52 evens, 58-62 evens