Vector Subtraction

Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract \(\mathbf{\text{B}}\) from \(\mathbf{\text{A}}\), written \(\mathbf{\text{A}}–\mathbf{\text{B}}\), we must first define what we mean by subtraction.

The negative of a vector \(\mathbf{\text{B}}\)is defined to be \(\mathbf{\text{–B}}\); that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in the figure below.

In other words, \(\mathbf{\text{B}}\) has the same length as \(\mathbf{\text{–B}}\), but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.

Two vectors are shown. One of the vectors is labeled as vector in north east direction. The other vector is of the same magnitude and is in the opposite direction to that of vector B. This vector is denoted as negative B.

The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So \(\mathbf{\text{B}}\) is the negative of \(\mathbf{\text{–B}}\); it has the same length but opposite direction.

The subtraction of vector \(\mathbf{\text{B}}\) from vector \(\mathbf{\text{A}}\) is then simply defined to be the addition of \(\mathbf{\text{–B}}\) to \(\mathbf{\text{A}}\). Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.

\(\mathbf{\text{A – B = A + }\left(\text{–B}\right)\text{.}}\)

This is analogous to the subtraction of scalars (where, for example, \(\text{5 – 2 = 5 + }\left(\text{–2}\right)\)). Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates.

Example: Subtracting Vectors Graphically: A Woman Sailing a Boat

A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction \(\text{66.0º}\) north of east from her current location, and then travel 30.0 m in a direction \(\text{112º}\) north of east (or \(\text{22.0º}\) west of north).

If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.

A vector of magnitude twenty seven point five meters is shown. It is inclined to the horizontal at an angle of sixty six degrees. Another vector of magnitude thirty point zero meters is shown. It is inclined to the horizontal at an angle of one hundred and twelve degrees.

Strategy

We can represent the first leg of the trip with a vector \(\mathbf{\text{A}}\), and the second leg of the trip with a vector \(\mathbf{\text{B}}\). The dock is located at a location \(\mathbf{\text{A}}+\mathbf{\text{B}}\). If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance \(B\) (30.0 m) in the direction \(180º–112º=68º\) south of east.

We represent this as \(\mathbf{\text{–B}}\), as shown below. The vector \(\mathbf{\text{–B}}\) has the same magnitude as \(\mathbf{\text{B}}\) but is in the opposite direction. Thus, she will end up at a location \(\mathbf{\text{A}}+\left(\mathbf{\text{–B}}\right)\), or \(\mathbf{\text{A}}–\mathbf{\text{B}}\).

A vector labeled negative B is inclined at an angle of sixty-eight degrees below a horizontal line. A dotted line in the reverse direction inclined at one hundred and twelve degrees above the horizontal line is also shown.

We will perform vector addition to compare the location of the dock, \(\text{A }\text{+ }\mathbf{B}\), with the location at which the woman mistakenly arrives, \(\text{A + }\left(\text{–B}\right)\).

Solution

(1) To determine the location at which the woman arrives by accident, draw vectors \(\mathbf{\text{A}}\) and \(\mathbf{\text{–B}}\).

(2) Place the vectors head to tail.

(3) Draw the resultant vector \(\mathbf{R}\).

(4) Use a ruler and protractor to measure the magnitude and direction of \(\mathbf{R}\).

Vectors A and negative B are connected in head to tail method. Vector A is inclined with horizontal with positive slope and vector negative B with a negative slope. The resultant of these two vectors is shown as a vector R from tail of A to the head of negative B. The length of the resultant is twenty three point zero meters and has a negative slope of seven point five degrees.

In this case, \(R=\text{23}\text{.}\text{0 m}\) and\(\theta =7\text{.}\text{5º}\) south of east.

(5) To determine the location of the dock, we repeat this method to add vectors \(\mathbf{\text{A}}\) and \(\mathbf{\text{B}}\). We obtain the resultant vector \(\mathbf{\text{R}}\text{'}\):

A vector A inclined at sixty six degrees with horizontal is shown. From the head of this vector another vector B is started. Vector B is inclined at one hundred and twelve degrees with the horizontal. Another vector labeled as R prime from the tail of vector A to the head of vector B is drawn. The length of this vector is fifty two point nine meters and its inclination with the horizontal is shown as ninety point one degrees. Vector R prime is equal to the sum of vectors A and B.

In this case \(R\text{ = 52.9 m}\) and \(\theta =\text{90.1º}\) north of east.

We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip.

Discussion

Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition.

Vector Subtraction

Vector Subtraction

Example: Subtracting Vectors Graphically: A Woman Sailing a Boat

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