Numbers and Operations

3.5 Vectors

A vector in a plane is defined to be an ordered pair of real numbers. A vector in space is defined as an ordered triple of real numbers. On a coordinate system, a vector is usually represented by an arrow whose initial point is the origin and whose terminal point is at the ordered pair (or triple) that named the vector. Vector quantities always have a magnitude or norm (the length of the arrow) and direction (the angle the arrow makes with the positive x-axis). Vectors are often used to represent motion or force.

All properties of two-dimensional vectors can be extended to three-dimensional vectors. We will express the properties in terms of two-dimensional vectors for convenience. If vector  is designated by (v1v2) and vector  is designated by (u1u2), vector  is designated by (uv1,uv2) and called the resultant of  and . Vector –  has the same magnitude as  but has a direction opposite that of .

On the plane, every vector  can be expressed in terms of any other two unit (magnitude 1) vectors parallel to the x - and y-axes. If vector  = (1,0) and vector  = (0,1), any vector  = ai + bj, where a and b are real numbers. A unit vector parallel to  can be determined by dividing  by its norm, denoted by  and equal to 

It is possible to determine algebraically whether two vectors are perpendicular by defining the dot product or inner product of two vectors, (v1v2) and (u1u2).

Notice that the dot product of two vectors is a real number , not a vector. Two vectors,  and , are perpendicular if and only if 


1. Let vector = (2, 3) and vector = (6, –4).

    (A) What is the resultant of  and?

    (B) What is the norm of ?

    (C) Express  in terms of  and .

    (D) Are  and  perpendicular?


(A) The resultant,  equals (6 + 2, –4 + 3) = (8, –1).

(B) The norm of 

(C)  To verify this, use the definitions of  and  = 2(1,0) + 3(0,1) = (2, 0) + (0, 3) = (2, 3) = 

(D)  = 6 · 2 + (– 4) · 3 = 12 – 12 = 0. Therefore,  and  are perpendicular because the dot product is equal to zero.

2. If = (–1, 4) and the resultant of  and  is (4,5), find .

Let  The resultant  = (–1,4) + (v1v2) = (4,5). Therefore, (–1 + v1, 4 + v2) = (4,5), which implies that –1 + v1 = 4 and 4 + v2 = 5. Thus, v1 = 5 and 


1. Suppose  Find the magnitude of 

      (A)  2

      (B)  3

      (C)  4

      (D)  5

      (E)  6

2. If  and  the resultant vector of  equals






3. A unit vector perpendicular to vector  is

      (A)  (4,3)





Answers and Explanations

1. (D) Add the components to get  The magnitude is 

2. (D)  and  so 

3. (D) All answer choices except A are unit vectors. Backsolve to find that the only one having a zero dot product with (3, –4) is .