## 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 (*v*_{1}, *v*_{2}) and vector is designated by (*u*_{1}, *u*_{2}), vector is designated by (*u*_{1 }+ *v*_{1},*u*_{2 }+ *v*_{2}) 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, (*v*_{1}, *v*_{2}) and (*u*_{1}, *u*_{2}).

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) + (*v*_{1}, *v*_{2}) = (4,5). Therefore, (–1 + *v*_{1}, 4 + *v*_{2}) = (4,5), which implies that –1 + *v*_{1} = 4 and 4 + *v*_{2} = 5. Thus, *v*_{1} = 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 .