3D Vectors

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

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 Level

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 all work the same with 3D vectors as with 2D vectors.

A Level

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.

Example:

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 $\sqrt{(x_{1}-x_{2})^2 +(y_{1}-y_{2})^2 +(z_{1}-z_{2})^2}$

Example:

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 $(2 \text{ dp})$

A Level
A Level

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 Level

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 $\overrightarrow{AB}$:

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

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 $(2 \text{ dp})$

A Level

A Level

A Level

A Level

A Level

You May Also Like...

A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99

A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99

Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99