# Position Vectors

## Position Vectors Revision

**Position Vectors**

A **position vector** describes where a point lies, in relation to the **origin**, \boldsymbol{O}

This allows you to write other vectors in terms of position vectors.

Make sure you are happy with the following topics before continuing.

**Working with Position Vectors**

For a point A, the **position vector** is denoted as \overrightarrow{OA}

Likewise the point B is denoted as \overrightarrow{OB}

Writing points as **position vectors** then allows you to describe vectors between points as position vectors.

If we let, \boldsymbol{a}=\overrightarrow{OA} and \boldsymbol{b}=\overrightarrow{OB}

Then, \overrightarrow{AB}=-\overrightarrow{OA}+\overrightarrow{OB}=-\boldsymbol{a}+\boldsymbol{b}=\boldsymbol{b}-\boldsymbol{a}

**Describing vectors using \boldsymbol{i} and \boldsymbol{j} Units**

- A vector with a
**magnitude**of 1 unit is called a**unit vector** - The
**standard unit vectors**are \boldsymbol{i} and \boldsymbol{j}, which are in the direction of the x-axis and y-axis respectively. - These standard unit vectors are used to express the
**horizontal**and**vertical**position of the end of a vector compared to its start point

Using the diagram on the right, we can see the position vector of point A=3\boldsymbol{i}+5\boldsymbol{j} and the position vector of point B=5\boldsymbol{i}+2\boldsymbol{j}

To write \overrightarrow{AB} in terms of **standard unit vectors**, we can just **add **and **subtract **the \boldsymbol{i} and \boldsymbol{j} components of \boldsymbol{a} and \boldsymbol{b} separately.

Thus the vector \overrightarrow{AB}=-\boldsymbol{a}+\boldsymbol{b}=-(3\boldsymbol{i}+5\boldsymbol{j})+(5\boldsymbol{i}+2\boldsymbol{j})=2\boldsymbol{i}-3\boldsymbol{j}

This means that to go from A to B you go 2 units to the right and 3 units down.

**Column Vectors**

**Column vectors** are another way of expressing vectors.

If A=4\boldsymbol{i}-3\boldsymbol{j}, then the column vector for A is \dbinom{4}{-3}

To **add **and **subtract **column vectors from one another you simply add or subtract the **top row** and then add or subtract the **bottom row**. For example, if A=\dbinom{5}{-2} and B=\dbinom{1}{3}, then A+B=\dbinom{5+1}{-2+3}=\dbinom{6}{1}

**Multiplying **a column vector by a **scalar **is also straight forward, you just need to **multiply both the top and bottom numbers by the scalar**. For instance, if C=\dbinom{-3}{4}, 3C=\dbinom{3\times-3}{3\times4}=\dbinom{-9}{12}

## Position Vectors Example Questions

**Question 1: **Point K has the coordinates (-1,3).

Give the position vector of point K in:

a) Standard unit vector form

b) Column vector form

**[2 marks]**

If A=(x,y), then A=x\boldsymbol{i}+y\boldsymbol{j}=\dbinom{x}{y}

So,

a) K=-\boldsymbol{i}+3\boldsymbol{j}

b) K=\dbinom{-1}{3}

**Question 2: **\boldsymbol{a} has a position vector of \dbinom{7}{-2} and \boldsymbol{b} has a position vector of \dbinom{-3}{-1}, what is 2\boldsymbol{a}-4\boldsymbol{b}?

Give your answer as a column vector.

**[2 marks]**

**Question 3: **Points X and Y have position vectors 5\boldsymbol{i}-\boldsymbol{j} and -3\boldsymbol{i}+4\boldsymbol{j} respectively. Point Z lies on the line XY, such that XZ:ZY=2:3.

Calculate the position vector of Z, giving your answer in standard unit form.

**[4 marks] **

We know that point Z lies on the line XY, such that XZ:ZY=2:3, so \overrightarrow{XZ}=\dfrac{2}{5}\overrightarrow{XY}

Therefore we can calculate the position of point Z from X:

\overrightarrow{XZ}=\dfrac{2}{5}(-8\boldsymbol{i}+5\boldsymbol{j})=-\dfrac{16}{5}\boldsymbol{i}+2\boldsymbol{j}Finally, add the position vector of \overrightarrow{XZ} to the position vector of X to get the position vector of Z (from the origin):

Z = 5\boldsymbol{i}-\boldsymbol{j}-\dfrac{16}{5}\boldsymbol{i}+2\boldsymbol{j}=\dfrac{9}{5}\boldsymbol{i}+\boldsymbol{j}