3D Vectors

A LevelAQAEdexcelOCR

3D Vectors Revision

3D Vectors

Working with 3D vectors is mostly similar to 2D vectors, however the calculations can be more complicated.

3D vectors introduces another unit vector, \boldsymbol{\textcolor{blue}{k}}, which corresponds to the \textcolor{blue}{z}-axis.

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

A LevelAQAEdexcelOCR

Unit Vectors in Three Dimensions

Three dimensions inevitably introduces a third axis, the z-axis. We can visualise the z axis by imagining that the x-axis and y-axis lie flat on a surface and then the z-axis points upwards, as shown by the diagram on the right.

When working with vectors \boldsymbol{k} is the unit vector in the direction of the z-axis.

Three dimensional vectors follow the same form as two dimensional vectors, they are written as: x\boldsymbol{i}+y\boldsymbol{j}+z\boldsymbol{k}=\begin{pmatrix} x \\ y \\ z \end{pmatrix}

Addition, subtraction, multiplying with scalars and showing two vectors are parallel (Vector Basics) all work the same with 3D vectors as with 2D vectors.

A LevelAQAEdexcelOCR

Pythagoras in Three Dimensions

To find the distance of any 3D point from the origin we can use a variation of Pythagoras’ theorem:

Distance of point (x,y,z) from the origin =\sqrt{x^2+y^2+z^2}

Therefore, finding the magnitude of any 3D vector can be found in the same way as finding the magnitude of a 2D vector.


Find the magnitude of the vector \boldsymbol{s}=3\boldsymbol{i}-5\boldsymbol{j}+2\boldsymbol{k} to 2 decimal places.

|\boldsymbol{s}|=\sqrt{3^2+(-5)^2+2^2}=\sqrt{38}=6.16 units (2\text{ dp})


We can also use a Pythagoras-based formula to find the distance between two points in three dimensions:

Distance between points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is \sqrt{(x_{1}-x_{2})^2 +(y_{1}-y_{2})^2 +(z_{1}-z_{2})^2}


The position vector of point O is \begin{pmatrix} 4 \\ 6 \\ -3 \end{pmatrix} and the position vector of point P is \begin{pmatrix} 1 \\ -5 \\ 4 \end{pmatrix}.

Find |\overrightarrow{OP}| to 2 decimal places.

|\overrightarrow{OP}|=\sqrt{(4-1)^2+(6-(-5))^2+((-3)-4)^2}=\sqrt{9+121+49}=13.38 units (2 \text{ dp})

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Example: Problems in 3D

Point A has a position vector of -4\boldsymbol{i}+3\boldsymbol{j}+6\boldsymbol{k} and point B has a position vector of 2\boldsymbol{i}-5\boldsymbol{j}+3\boldsymbol{k}. Point C lies on the vector line AB and divides the line in the ratio 4:3. Find the position vector of C.


It is often helpful to split 3D problems into multiple parts and draw 2D diagrams to help you visualise them.

1) We first need to find the vector \overrightarrow{AB}:

\begin{aligned} \overrightarrow{AB} &= \overrightarrow{OB}-\overrightarrow{OA} \\ &= (2-(-4))\boldsymbol{i}+(-5-3)\boldsymbol{j}+(3-6)\boldsymbol{k} \\ &=6\boldsymbol{i}-8\boldsymbol{j}-3\boldsymbol{k} \end{aligned}

2) We are told that C divides \overrightarrow{AB} in the ratio 4:3, so C is \dfrac{4}{7} of the way along \overrightarrow{AB}.

Thus, \overrightarrow{AC}=\dfrac{4}{7}\overrightarrow{AB}=\dfrac{4}{7}(6\boldsymbol{i}-8\boldsymbol{j}-3\boldsymbol{k})=\dfrac{24}{7}\boldsymbol{i}-\dfrac{32}{7}\boldsymbol{j}-\dfrac{12}{7}\boldsymbol{k}

3) Finally, to find the position vector of C we need to add the vector \overrightarrow{AC} to \overrightarrow{OA}.

\begin{aligned} \overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC} &=(-4\boldsymbol{i}+3\boldsymbol{j}+6\boldsymbol{k})+(\dfrac{24}{7}\boldsymbol{i}-\dfrac{32}{7}\boldsymbol{j}-\dfrac{12}{7}\boldsymbol{k}) \\ &= -\dfrac{4}{7}\boldsymbol{i}-\dfrac{11}{7}\boldsymbol{j}+\dfrac{30}{7}\boldsymbol{k} \end{aligned}
A LevelAQAEdexcelOCR

3D Vectors Example Questions

Using a variation of Pythagoras’ theorem:


a)  |\overrightarrow{OA}|=\sqrt{5^2+(-2)^2+1^2}=\sqrt{30}=5.48 units (2 \text{ dp})


b) |\overrightarrow{OB}|=\sqrt{(-2)^2+(-6)^2+7^2}=\sqrt{89}=9.43 units (2 \text{ dp})


c) |\overrightarrow{OC}|=\sqrt{4^2+9^2+(-3)^2}=\sqrt{106}=10.30 units (2 \text{ dp})

\begin{aligned} 3\boldsymbol{y}+2\boldsymbol{z}-\boldsymbol{x} &=3(-2\boldsymbol{i}-7\boldsymbol{j}-\boldsymbol{k})+2(6\boldsymbol{i}+9\boldsymbol{j}+3\boldsymbol{k})-(3\boldsymbol{i}+8\boldsymbol{j}-4\boldsymbol{k})\\ &=(-6+12-3)\boldsymbol{i}+(-21+18-8)\boldsymbol{j}+(-3+6+4)\boldsymbol{k}\\ &=3\boldsymbol{i}-11\boldsymbol{j}+7\boldsymbol{k} \end{aligned}

We need to first find \overrightarrow{MN}:


\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\begin{pmatrix} 5 \\ 7 \\ -5 \end{pmatrix}-\begin{pmatrix} 3 \\ -3 \\ -1 \end{pmatrix}=\begin{pmatrix} 2 \\ 10 \\ -4 \end{pmatrix}


We can see that \begin{pmatrix} 2 \\ 10 \\ -4 \end{pmatrix} \times \dfrac{1}{2}=\begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix}


Thus, 2\overrightarrow{OL}=\overrightarrow{MN}, so \overrightarrow{OL} is parallel to \overrightarrow{MN} because they are scalar multiples of each other.

First, we need to find the position vector of \overrightarrow{AB}:



We know that the point X divides the line AB in the ratio 3:1, so \overrightarrow{AX}=\dfrac{3}{4}\overrightarrow{AB}=-\dfrac{3}{4}\boldsymbol{i}+\dfrac{3}{2}\boldsymbol{j}-6\boldsymbol{k}


Then to find the distance between the point X and the origin, we need to find the position vector of \overrightarrow{OX}:


\begin{aligned}\overrightarrow{OX}=\overrightarrow{OA}+\overrightarrow{AX} &= (2\boldsymbol{i}-6\boldsymbol{j}+5\boldsymbol{k})+(-\dfrac{3}{4}\boldsymbol{i}+\dfrac{3}{2}\boldsymbol{j}-6\boldsymbol{k}) \\ &=\dfrac{5}{4}\boldsymbol{i}-\dfrac{9}{2}\boldsymbol{j}-\boldsymbol{k} \end{aligned}


Finally, |\overrightarrow{OX}|=\sqrt{(\dfrac{5}{4})^2+(-\dfrac{9}{2})^2+(-1)^2}=4.78 units (2 \text{ dp})


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