Addition, subtraction and scalar multiplication of vectors is very straight forward. Vector multiplication and vector division, however, do not make sense and therefore do not have a definition.

Vector Addition

vector arithmeticTo add two vectors together, one simply adds together the corresponding components. For example:

*** QuickLaTeX cannot compile formula:
left(begin{array}{c}3\5end{array}right)+left(begin{array}{c}4\-3end{array}right)=left(begin{array}{c}7\2end{array}right)

*** Error message:
Undefined control sequence \5.
leading text: $left(begin{array}{c}3\5

The vector that results from applying one vector followed by another by adding, i.e. {bf a}+{bf b} is the vector that points directly from the start point to the finish point. Applying the vectors the other way round, i.e. {bf b}+{bf a}, also results in the same resultant vector. This is known as the parallelogram law of vector addition. See figure.

This may also be written as overrightarrow{AB}+overrightarrow{BC}=overrightarrow{AC} if considering vectors between points A, B and C.

Vector Subtraction

vector arithmeticAs you might expect, vector subtraction is achieved by subtracting the corresponding components of the given vectors. For example:

*** QuickLaTeX cannot compile formula:
left(begin{array}{c}3\-5end{array}right)-left(begin{array}{c}-4\2end{array}right)=left(begin{array}{c}7\-7end{array}right)

*** Error message:
Undefined control sequence \2.
leading text: ...-5end{array}right)-left(begin{array}{c}-4\2

. Rather than starting from a point and moving along two vectors one after the other, consider two vectors starting from the same point, say {bf a} and {bf b}. The vector that results from subtracting {bf a} from {bf b} is the one that points directly from the end of {bf a} to the end of {bf b}. See figure. This is known as the triangle law of vector addition. It can be thought of as starting at the end of {bf a}, moving backwards along {bf a} and then forwards along {bf b} (-{bf a}+{bf b}) but written as ({bf b}-{bf a}). See Position Vectors for more on this.

Scalar Multiplication of Vectors

vector arithmeticThere is no definition for multiplying vectors together but we can multiply or divide vectors by a scalar (ie a single number). For example,

*** QuickLaTeX cannot compile formula:
2left(begin{array}{c}-1\6end{array}right)=left(begin{array}{c}-2\12end{array}right)

*** Error message:
Undefined control sequence \6.
leading text: $2left(begin{array}{c}-1\6

 or -frac{1}{3}left(begin{array}{c}3\-2end{array}right)=left(begin{array}{c}-1\frac{2}{3}end{array}right)

As we can see from the diagram, scalar multiples of vectors are all parallel.

Since scalar multiplication and vector addition is possible, it follows that any vector can be expressed as a linear combination of the standard unit vectors. For example,

*** QuickLaTeX cannot compile formula:
left(begin{array}{c}7\-5end{array}right)=left(begin{array}{c}7\0end{array}right)-left(begin{array}{c}0\5end{array}right)=7left(begin{array}{c}1\0end{array}right)-5left(begin{array}{c}0\1end{array}right)=7{bf i}-5{bf j}

*** Error message:
Undefined control sequence \0.
leading text: ...\-5end{array}right)=left(begin{array}{c}7\0

This is useful when writing vectors on a single line rather than stacked horizontally.


Examples

Example 1

Consider the following vectors:

*** QuickLaTeX cannot compile formula:
{bf a}=left(begin{array}{c}-1\3end{array}right)

*** Error message:
Undefined control sequence \3.
leading text: ${bf a}=left(begin{array}{c}-1\3

,     

*** QuickLaTeX cannot compile formula:
{bf b}=left(begin{array}{c}8\5end{array}right)

*** Error message:
Undefined control sequence \5.
leading text: ${bf b}=left(begin{array}{c}8\5

,    {bf c}=left(begin{array}{c}4\-2end{array}right).

The vector 6{bf a}+4{bf b}-9{bf c} is parallel to the vector 5{bf i}+k{bf j}. Find the value of k.

*** QuickLaTeX cannot compile formula:
begin{array}{c}6{bf a}+4{bf b}-9{bf c}&=&6left(begin{array}{c}-1\3end{array}right)+4left(begin{array}{c}8\5end{array}right)-9left(begin{array}{c}4\-2end{array}right)\&=&left(begin{array}{c}-10\56end{array}right)=-10{bf i}+56{bf j}end{array}

*** Error message:
Misplaced alignment tab character &.
leading text: $begin{array}{c}6{bf a}+4{bf b}-9{bf c}&

For this vector to be parallel to the vector 5{bf i}+k{bf j}, it must be a multiple of it. It follows that the multiple is -2 and so k=-28.

Example 2

The parallelogram ABCD has 3 of its vertices at A(-2,0), B(3,2) and C(4,-2). Find the coordinates of point D.

Again, a lot of the information for finding the solution comes from sketching the problem.

vector arithmetic

The vector that points from A to B is given by

*** QuickLaTeX cannot compile formula:
overrightarrow{AB}={bf b}-{bf a}=left(begin{array}{c}3\2end{array}right)-left(begin{array}{c}-2\0end{array}right)=left(begin{array}{c}5\2end{array}right)

*** Error message:
Undefined control sequence \2.
leading text: ...w{AB}={bf b}-{bf a}=left(begin{array}{c}3\2

Since the shape is a parallelogram,  overrightarrow{DC} is the same vector i.e.

*** QuickLaTeX cannot compile formula:
overrightarrow{DC}=left(begin{array}{c}5\2end{array}right)

*** Error message:
Undefined control sequence \2.
leading text: $overrightarrow{DC}=left(begin{array}{c}5\2

.

It follows that

*** QuickLaTeX cannot compile formula:
overrightarrow{OD}=overrightarrow{OC}-overrightarrow{DC}=left(begin{array}{c}4\-2end{array}right)-left(begin{array}{c}5\2end{array}right)=left(begin{array}{c}-1\-4end{array}right)

*** Error message:
Undefined control sequence \2.
leading text: ...\-2end{array}right)-left(begin{array}{c}5\2

Note that overrightarrow{OC} and overrightarrow{OD} are position vectors (see more on this). The coordinates of the point D are (-1,-4).

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