PhysicsLAB Resource Lesson
Properties of Vectors

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A vector can be equivalently defined as a "directed line segment" or as a physical quantity that must include both its "magnitude and direction" to be completely described. For the purposes of this lesson, let's use the letters A, B, C … to represent vector quantities. If A represents a vector that begins at the origin (0,0) and ends are the point (3,4) we can represent it mathematically in several ways.
  • in rectangular form as A = <3,4>
  • in i, j notation as A = 3i + 4j
  • graphically as
  • in polar notation as A = (5, 53º)
When multiplying a vector by a positive scalar, only the magnitude of the vector is affected.
  • 2A = <6,8>
  • 2A = 6i + 8j
  • 2A = (10, 53º)
If the scalar is negative, then the direction is also reversed, or changed by 180º.
  • -½A = <-1.5, -2>
  • -½A = -1.5i - 2j
  • -½A = (2.5, 233º)
Refer to the following information for the next two questions.

B = <8, -15>
C = -12i + 5j
 Express vector B in polar notation.

 Express vector C in polar notation.

To add two vectors graphically we will use the "head-to-tail" method. This literally means placing the tail of the second vector at the head of the first vector and "tracing the path" of their sum, or the resultant vector.
The simplest addition is when you are asked to add vectors that are either parallel (pointing in the same direction) or anti-parallel (pointing in 180º the opposite direction). This requires that you merely add or subtract their magnitudes. Shown below are three examples where we used i,j notation illustrate this process. In this discussion we used two new vectors: D = <10, 0> and E = <4, 0>.
  • R = D + E
R = (10i + 0j) + (4i + 0j) = 14i + 0j
  • R = D - E
R = (10i + 0j) - (4i + 0j) = 6i + 0j
  • R = ½D - 2E
R = (5i + 0j) + (-8i + 0j) = -3i + 0j
      Note: subtracting 2E is
    equivalent to adding (-2E)
Now let's look at a more complicated example by graphing the sum of our original three vectors, R = A + B + C.
Notice that vector B starts at the end of vector A and vector C starts at the end of B. We are literally "tracing the path" of A + B + C. The resultant, R, equals the sum of the three vectors. It begins at the "tail" of A and ends at the "head" of C. That is, it begins at the start of vector A and ends at the end of vector C. To mathematically determine the magnitude and direction of our resultant we will have to use an x|y chart which will be discussed momentarily after we have alerted you to the commutative nature of vector addition.
R = B + A + C
R = C + B + A
Notice that the resultant, R, always starts at the beginning of the first vector (the solid xy-origin) and terminates at the end of the last vector. Also note that in each case R has the same rectangular form: <-1, -6> regardless of the order in which A, B, and C were added.
To analytically determine the sum of two or more vectors without the assistance of a head-to-tail diagram, the use of an x|y chart is recommended. This chart organizes the i,j components of each vector. Given below is an x|y chart using the information from our last example: R = A + B + C.
Our resultant can be expressed in i,j form as R = -i - 6j.

As shown in the column to the right, R can be expressed in polar form as R = (6.1, 260.5º).
To express this outcome in polar notation requires the use of the Pythagorean Theorem and the trig function tangent.
Our final discussion will be on how to calculate the components of a vector when it is initially stated in polar form. Let's start our investigation with the vector F = (50, 106.3º) which terminates in Quadrant II.
If we wanted to add F to vector 2A = <6,8> we would first need to transform F into its two rectangular components: one directed along the negative-x axis and the other directed along the positive-y axis as shown in the diagram below. Note that these directions were chosen because they form the "borders" of Quadrant II. 
To begin our conversion, we first drop dotted perpendiculars from the end of F to each axis. The x-component is calculated using the trig function cosine.
cos θ is defined as the ratio of the length of the adjacent side to the length of the hypotenuse in a right triangle; or cos θ = adj/hyp.
x = F cos θ
The y-component is calculated using the trig function sine.
sin θ is defined as the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle; or sin θ = opp/hyp.
y = F sin θ
In our case, the reference angle, θ, equals 73.5º. Recall from trig class that the reference angle is an acute angle that is always measured to the nearest x-axis, never to the y-axis. Our diagram shows us that vector F = (50, 106.3º) can now be restated in rectangular form as F = <-48, 14>.
Returning to our original problem, the sum of vector F and the vector 2A would be
R = (6i + 8j) + (-48i + 14j) = -42i + 22j
which expressed in polar form would be
R = (47.4, 152º)

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