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

To 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. is the vector that points directly from the start point to the finish point. Applying the vectors the other way round, i.e. , 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 if considering vectors between points , and .

## Vector Subtraction

As 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 and . The vector that results from subtracting from is the one that points directly from the end of to the end of . See figure. This is known as the **triangle law **of vector addition. It can be thought of as starting at the end of , moving backwards along and then forwards along () but written as (). See Position Vectors for more on this.

## Scalar Multiplication of Vectors

There 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

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

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

, .

The vector is parallel to the vector . Find the value of .

*** 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 , it must be a multiple of it. It follows that the multiple is and so .

The parallelogram has 3 of its vertices at , and . Find the coordinates of point .

Show Solution

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

The vector that points from to 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, 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 and are position vectors (see more on this). The coordinates of the point are .

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