Vectors in Space11.2

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 =.

• The magnitude (length) of v• Vector addition still means you add the value in

the x place, the y place, and the z place

2 2 21 2 3 1 2 3, , v v v is v v v v

Rules

• Scalar multiplication still means you distribute the scalar over the component form

• The dot product of

• If the dot product = 0, then the vectors are orthogonal.

• The angle between two nonzero vectors u and v is

1 2 3 1 2 3, , , , ,v v v v cv cv cv cv

1 2 3 1 2 3

1 1 2 2 3 3

, , , ,

u u u u and v u u u

is u v uv uv uv

cosu vu 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• The magnitude is

3 3 6 4 4 2 0 2 2, , , ,v

2 2 20 2 2 8 2 2v

Finding the Dot Product of Two Vectors

• Find the dot product of and

• Remember the dot product is a scalar, not a vector.

0 3 2 4 2 3 0 4 3 2 2 3

0 6 6 12

, , , , ( ) ( ) ( )( )

0 3 2, , 4 2 3, ,

Find the Angle Between Two Vectors

• Find the angle between u = and v =

1 0 2 3 1 0

5 10

, , , ,cos

u vu v

1 0 2, , 3 1 0, ,

1 364 9

50cos . o

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.

• Because PR = -2PQ, the vectors are parallel.

• Therefore, the points are collinear.

5 2 4 1 6 4 3 5 2

4 2 11 1 0 4 6 10 4

, ( ), , ,

, ( ), , ,

PQ

PR

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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)