SAT SUBJECT TEST MATH LEVEL 2
PART 2
REVIEW OF MAJOR TOPICS
CHAPTER 3
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 (v1, v2) and vector 
is designated by (u1, u2), vector 
is designated by (u1 + v1,u2 + v2) 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, (v1, v2) and (u1, u2).
Notice that the dot product of two vectors is a real number , not a vector. Two vectors, and , are perpendicular if and only if 
EXAMPLES
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?
SOLUTIONS
(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) + (v1, v2) = (4,5). Therefore, (–1 + v1, 4 + v2) = (4,5), which implies that –1 + v1 = 4 and 4 + v2 = 5. Thus, v1 = 5 and 
EXERCISES
1. Suppose 
Find the magnitude of 
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
2. If 
and 
the resultant vector of 
equals
(A) 
(B) 
(C) 
(D) 
(E) 
3. A unit vector perpendicular to vector 
is
(A) (4,3)
(B) 
(C) 
(D) 
(E) 
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 
.