Working with 3D coordinates (SQA National 5 Maths): Revision Note

The diagram shows a triangular prism $\mathrm{ABCDEF}$, relative to the coordinate axes.

• $\mathrm{AB}=\mathrm{BC}$.

• $\mathrm{CD}=10\mathrm{units}$.

• Edges $\mathrm{AF},\mathrm{BE}$ and $\mathrm{CD}$ are parallel to the $y$-axis.

a) Write down the coordinates of $\mathrm{E}$ and $\mathrm{D}$.

b) Write the vector $\overrightarrow{\mathrm{BD}}$ in component form.

Answer:

Part (a)

Because ABCDEF is a triangular prism, the three edges $\mathrm{AF},\mathrm{BE}$ and $\mathrm{CD}$ all have the same length (10 units)

Because $\mathrm{BE}$ is parallel to the $y$-axis

• $\mathrm{E}$ will have the same $x$ and $z$ coordinates as $\mathrm{B}$

• Only the $y$ coordinate will change (increase by 10)

$\mathrm{E}\left(6,10,7\right)$

Because $\mathrm{AB}=\mathrm{BC}$, triangle $\mathrm{ABC}$ is isosceles

• This means it is symmetrical around its perpendicular height

Then because $\mathrm{CD}$ is parallel to the $y$-axis

• $\mathrm{D}$ will have the same $x$ and $z$ coordinates as $\mathrm{C}$

• Only the $y$ coordinate will change (increase by 10)

$\mathrm{D}\left(9,10,0\right)$

Part (b)

To find vector $\overrightarrow{\mathrm{BD}}$ in component form

• subtract the coordinates for the first point ($\mathrm{B}$) from the coordinates for the second point ($\mathrm{D}$)

$\overrightarrow{\mathrm{BD}}=\left(\begin{array}{c}9-6\\ 10-0\\ 0-7\end{array}\right)$

$\overrightarrow{\mathrm{BD}}=\left(\begin{array}{c}3\\ 10\\ -7\end{array}\right)$