# 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 $(2 \text{ dp})$

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