Vectors in Space
11.2
JMerrill, 2010
Rules
• The same rules apply in 3-D space:
• The component form is found by subtracting the
coordinates of the initial point from the corresponding
coordinates of the terminal point.
• Two vectors are = iff their corresponding components
are =.
2
2
2

v
,
v
,
v
is
v

v

v

v
• The magnitude (length) of v
1 2 3
1
2
3
• Vector addition still means you add the value in the x
place, the y place, and the z place
Rules
• Scalar multiplication still means you distribute the scalar
over the component formv  v1,v 2,v3 , cv  cv1,cv 2,cv3
• The dot product of u  u1, u2 , u3 and v  u1, u2 , u3
is u  v  uv
1 1  u 2v 2  u3v 3
• If the dot product = 0, then the vectors are orthogonal.
• The angle between two nonzero vectors u and v is
u v
cos  
u v
Finding the Component Form of a Vector
• Find the component form and magnitude of
vector v having an initial point of (3,4,2) and
a terminal point of (3,6,4)
• The component form is v  3  3, 6  4, 4  2  0, 2, 2
• The magnitude is
v  02  22  2 2  8  2 2
Finding the Dot Product of Two Vectors
• Find the dot product of 0, 3, 2 and 4, 2, 3
0, 3, 2  4, 2, 3  0(4)  3(2)  (2)(3)
 0  6  6  12
• Remember the dot product is a scalar, not a
vector.
Find the Angle Between Two Vectors
• Find the angle between u = 1, 0, 2 and v = 3,1, 0
1, 0, 2  3,1, 0
u v
cos 

u v
5 10
  cos
1
3
o
 64.9
50
Parallel Vectors
• Vector w has an initial point (1,-2,0) and a terminal
point (3,2,1). Which of the following vectors is parallel
to w?
• A. u = <4,8,2>
B. v = <4,8,4>
• Put w into component form <2,4,1>
• Vector u is the answer because it is just double vector
w. U can be written as 2<2,4,1>
Using Vectors to Determine Collinear Points
• Determine whether the points P(2,-1,4),
Q(5,4,6), and R(-4,-11,0) are collinear.
• The points P,Q, and R are collinear iff the
vectors PQ and PR are parallel.
PQ  5  2, 4  ( 1), 6  4  3, 5, 2
PR  4  2, 11  (1), 0  4  6, 10, 4
• Because PR = -2PQ, the vectors are parallel.
• Therefore, the points are collinear.
Finding the Terminal Point of a Vector
• The initial point of v = <4,2,-1> is P(3,-1,6). What is
the terminal point?
• We use the initial point of P(3,-1,6) and the terminal
point of Q(q1, q2, q3)
• PQ = <q1–3, q2 -(-1), q3–6> = <4,2,-1>
• The 1st term = the 1st term, the 2nd term = the 2nd
term… So, q1–3 = 4, q2 -(-1) = 2, q3–6 = -1
• Point Q = (7, 1, 5)