Vector Arithmetic

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:

\left(\begin{array}{c}3\\5\end{array}\right)+\left(\begin{array}{c}4\\-3\end{array}\right)=\left(\begin{array}{c}7\\2\end{array}\right)

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:

\left(\begin{array}{c}3\\-5\end{array}\right)-\left(\begin{array}{c}-4\\2\end{array}\right)=\left(\begin{array}{c}7\\-7\end{array}\right). 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,

2\left(\begin{array}{c}-1\\6\end{array}\right)=\left(\begin{array}{c}-2\\12\end{array}\right)  or -\frac{1}{3}\left(\begin{array}{c}3\\-2\end{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,

\left(\begin{array}{c}7\\-5\end{array}\right)=\left(\begin{array}{c}7\\0\end{array}\right)-\left(\begin{array}{c}0\\5\end{array}\right)=7\left(\begin{array}{c}1\\0\end{array}\right)-5\left(\begin{array}{c}0\\1\end{array}\right)=7{\bf i}-5{\bf j}

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

Examples

Consider the following vectors:

{\bf a}=\left(\begin{array}{c}-1\\3\end{array}\right),     {\bf b}=\left(\begin{array}{c}8\\5\end{array}\right),    {\bf c}=\left(\begin{array}{c}4\\-2\end{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.

\begin{array}{lll}6{\bf a}+4{\bf b}-9{\bf c}&=&6\left(\begin{array}{c}-1\\3\end{array}\right)+4\left(\begin{array}{c}8\\5\end{array}\right)-9\left(\begin{array}{c}4\\-2\end{array}\right)\\&=&\left(\begin{array}{c}-10\\56\end{array}\right)\\&=&-10{\bf i}+56{\bf j}\end{array}

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.

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

\overrightarrow{AB}={\bf b}-{\bf a}=\left(\begin{array}{c}3\\2\end{array}\right)-\left(\begin{array}{c}-2\\0\end{array}\right)=\left(\begin{array}{c}5\\2\end{array}\right)

Since the shape is a parallelogram,  \overrightarrow{DC} is the same vector i.e. \overrightarrow{DC}=\left(\begin{array}{c}5\\2\end{array}\right).

It follows that

\overrightarrow{OD}=\overrightarrow{OC}-\overrightarrow{DC}=\left(\begin{array}{c}4\\-2\end{array}\right)-\left(\begin{array}{c}5\\2\end{array}\right)=\left(\begin{array}{c}-1\\-4\end{array}\right)

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

Videos

https://youtu.be/8_ldVD15iN0

Navigating along various known vectors to obtain unknown vectors and showing that three points are colinear.

https://youtu.be/5FcaEAkUMDs

Finding expressions for various vectors running along the perimeter of a trapezium.

https://youtu.be/IHFOnS4Gt1k

A complicated vector arithmetic question involving the position vector of a point with unknown location.

AS Maths Vectors

A2 Maths Vectors