Question 11
$OABC$ is a triangular-based pyramid, as shown.
Use a vector method to determine the area of the shaded face of the pyramid.
[4 marks]Two sides of the face give its area
The shaded face is the triangle $ABC$, which does not touch $O$. Each side is traced as a path from $A$, one axis at a time, and the two sides then build a parallelogram.
Each side is end minus start. The orange vector product is drawn at a fifth of its length so it fits, and its direction is exact.
Choose two sides and turn the pyramid
Pick two sides of the shaded face that start at the same corner, then swap their order. Turn the pyramid to see the orange vector product stick straight out of the face. The last option shows the slip of crossing the position vectors $\mathbf{a}$ and $\mathbf{b}$.
Drag the picture or use the sliders to turn it. The orange arrow is drawn at a fifth of its length. The booklet’s sketch is not to scale: with the real coordinates you see the shaded face from the front when the $x$-axis points away from you, so the figure starts on that side.
These are the two sides the QCAA sample response uses. Both start at $A$, and half the length of their vector product is the area of the shaded face.
A different corner gives a different pair of sides, yet the vector product and the area are exactly the same. In fact any two sides of the triangle give the same length of vector product, even two that don’t share a corner.
Swapping the order reverses the vector product, so the arrow now points out of the other side of the face. Its length is unchanged, so the area is too. The QCAA accept $-21\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}+6\hat{\boldsymbol{k}}$.
These are the position vectors of $A$ and $B$. They are sides of the grey triangle $OAB$, so this gives the area of that face, about $7.43$. The shaded face needs vectors along its own sides, such as $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide.
The shaded face is the triangle $ABC$. You need two of its sides as vectors. $O$ is the origin, so the position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are just the coordinates, and each side is end minus start.
$$\begin{aligned}\overrightarrow{AB}&=\mathbf{b}-\mathbf{a}\\&=\begin{pmatrix}-1\\2\\-2\end{pmatrix}-\begin{pmatrix}1\\2\\5\end{pmatrix}\\&=\begin{pmatrix}-2\\0\\-7\end{pmatrix}\end{aligned}$$
$$\begin{aligned}\overrightarrow{AC}&=\mathbf{c}-\mathbf{a}\\&=\begin{pmatrix}0\\5\\2\end{pmatrix}-\begin{pmatrix}1\\2\\5\end{pmatrix}\\&=\begin{pmatrix}-1\\3\\-3\end{pmatrix}\end{aligned}$$
The QCAA accept any two sides of the triangle in any vector form, such as $\overrightarrow{BA}=2\hat{\boldsymbol{\imath}}+7\hat{\boldsymbol{k}}$ or $\overrightarrow{BC}=\left(\begin{smallmatrix}1\\3\\4\end{smallmatrix}\right)$.
Marker: correctly determines the two vectors representing any two sides of the triangle $ABC$
Take the vector product of the two sides. The formula book writes the rule with $\mathbf{a}$ and $\mathbf{b}$, but those letters are already the position vectors here, so I have used $\mathbf{u}$ and $\mathbf{v}$.
$$\mathbf{u}\times\mathbf{v}=\begin{pmatrix}u_2v_3-u_3v_2\\u_3v_1-u_1v_3\\u_1v_2-u_2v_1\end{pmatrix}$$
$$\begin{aligned}&\overrightarrow{AB}\times\overrightarrow{AC}\\&\quad=\begin{pmatrix}-2\\0\\-7\end{pmatrix}\times\begin{pmatrix}-1\\3\\-3\end{pmatrix}\\&\quad=\begin{pmatrix}(0)(-3)-(-7)(3)\\(-7)(-1)-(-2)(-3)\\(-2)(3)-(0)(-1)\end{pmatrix}\\&\quad=\begin{pmatrix}21\\1\\-6\end{pmatrix}\end{aligned}$$
This is Paper 2, so your GDC can find the vector product. Write down the two vectors you entered and the result. The QCAA also accept $-21\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}+6\hat{\boldsymbol{k}}$, which is the same product in the other order. Follow-through applies if a side from step 1 was wrong, and this mark can be implied by your later working.
The QCAA also accept a different route that skips the vector product. For this mark you find the lengths of two sides, such as $\left|\overrightarrow{AB}\right|=\sqrt{53}\approx7.28$, $\left|\overrightarrow{AC}\right|=\sqrt{19}\approx4.36$ or $\left|\overrightarrow{BC}\right|=\sqrt{26}\approx5.10$.
Marker: determines cross product of two vectors representing any two sides of the triangle $ABC$
The length of $\overrightarrow{AB}\times\overrightarrow{AC}$ is the area of the parallelogram with sides $AB$ and $AC$. The triangle is half of that parallelogram.
$$\text{Area}=\frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|$$
This mark can be implied by your later working. On the lengths route, this mark is for one angle of the triangle, found with the dot product. For example, $\overrightarrow{AB}\cdot\overrightarrow{AC}=2+0+21=23$ gives
$$\begin{aligned}\cos A&=\frac{23}{\sqrt{53}\sqrt{19}}\\A&\approx43.55^\circ\end{aligned}$$
and then the area is $\frac12\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|\sin A$. The QCAA list $\angle A\approx43.55^\circ$ or $0.76$, $\angle B\approx36.08^\circ$ or $0.63$, and $\angle C\approx100.37^\circ$ or $1.75$.
Marker: uses vector expression to represent the area of a triangle
Find the length of the vector product, then halve it.
$$\begin{aligned}\text{Area}&=\tfrac12\sqrt{21^2+1^2+(-6)^2}\\&=\tfrac12\sqrt{478}\end{aligned}$$
The QCAA also accept $10.9$ or $\dfrac{\sqrt{478}}{2}$. The lengths route lands on the same value: $\frac12\sqrt{53}\sqrt{19}\sin43.55^\circ\approx10.93$.
Marker: determines area
Each phrase of the question gave you something
This is the standard vector product method for the area of a triangle, applied once. The coordinates are given as small integers, and your GDC can check the vector product.
If I had to guess where you might lose marks, it would be crossing the position vectors $\mathbf{a}$ and $\mathbf{b}$ straight from the diagram. That finds the area of the face $OAB$, about $7.43$, because $\mathbf{a}$ and $\mathbf{b}$ are sides that start at $O$. The other places to watch are the middle row of the vector product, $u_3v_1-u_1v_3$, and the $\frac12$ at the end.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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