Position Vectors

Position Vectors

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

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.

A Level AQA Edexcel OCR

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}

A LevelAQAEdexcelOCR

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.

A LevelAQAEdexcelOCR

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}

A LevelAQAEdexcelOCR

Example Questions

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}

2\boldsymbol{a}-4\boldsymbol{b}=\dbinom{2\times7}{2\times-2}-\dbinom{4\times-3}{4\times-1}=\dbinom{14}{-4}-\dbinom{-12}{-4}=\dbinom{14-(-12)}{-4-(-4)}=\dbinom{26}{0}
\overrightarrow{XY}=\boldsymbol{y}-\boldsymbol{x}=(-3\boldsymbol{i}+4\boldsymbol{j})-(5\boldsymbol{i}-\boldsymbol{j})=-8\boldsymbol{i}+5\boldsymbol{j}

 

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}

Related Topics

MME

Vector Basics

A Level

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

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