Question 14
The origin, O, is joined to points $\text{A}(1,2,5)$ and $\text{B}(-3,4,0)$ to form triangle OAB.
Point C is the point on OB such that AC is perpendicular to OB, as shown.
Given the length of side OB is 5 units, show that the vector projection of $\overrightarrow{\text{OA}}$ on $\overrightarrow{\text{OB}}$ is
$$\frac{1}{5}\begin{pmatrix}-3\\4\\0\end{pmatrix}$$
Use your result from Question 14a) to determine the length of OC.
[1 mark] c)Determine the length of side OA.
[1 mark] d)Use Pythagoras’ theorem to determine the length of AC.
[1 mark] e)Use your result from Question 14d) to determine the area of triangle OAB.
[1 mark] f)Use a vector product method to verify your result from Question 14e).
[2 marks]The projection gives you the height, and the height gives you the area
Watch A drop straight down onto OB. The orange piece is the projection, and the drop is the height of the triangle.
The magnitude of OA × OB is the area of the whole parallelogram. The triangle is half of it, which is where the ½ comes from.
Build the projection one piece at a time
This view looks straight at the plane of the triangle, so OB runs across the page. Here $\mathbf{a}=\overrightarrow{\text{OA}}$ and $\mathbf{b}=\overrightarrow{\text{OB}}$. Step through the four pieces of $(\mathbf{a}\cdot\hat{\mathbf{b}})\,\hat{\mathbf{b}}$, then change the angle $\theta$ to see how each piece responds.
Divide $\overrightarrow{\text{OB}}$ by its length to get $\hat{\mathbf{b}}$. It points the same way as OB but has length 1, so it carries direction only. The question gives you $\left|\overrightarrow{\text{OB}}\right|=5$ to save you a step.
The scalar projection $\mathbf{a}\cdot\hat{\mathbf{b}}$ is a number, not a vector. It tells you how far the shadow of A reaches along OB, measured from O. In this question it is the length OC.
Multiply that number by $\hat{\mathbf{b}}$ and you get a vector along OB with that length. This is the vector projection, $\overrightarrow{\text{OC}}$, and it is what part a) asks for, so your answer needs components.
Take the vector projection away from $\mathbf{a}$ and what is left is perpendicular to OB. Its length is AC, the height of the triangle in part e).
The angle is more than 90°, so the shadow falls behind O. The scalar projection is negative and the vector projection points away from B.
At 90° there is no shadow, so both projections are zero.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Keep exact surds as you go. They make part f) easy to check.
The vector projection of $\mathbf{a}$ on $\mathbf{b}$ is $(\mathbf{a}\cdot\hat{\mathbf{b}})\,\hat{\mathbf{b}}$. You are told $\left|\overrightarrow{\text{OB}}\right|=5$, so the unit vector divides by 5.
$$\text{proj}=\left(\begin{pmatrix}1\\2\\5\end{pmatrix}\cdot\frac{1}{5}\begin{pmatrix}-3\\4\\0\end{pmatrix}\right)\frac{1}{5}\begin{pmatrix}-3\\4\\0\end{pmatrix}$$
Marker: correctly substitutes into a suitable rule that represents the vector projection
This is a “show that”, so the scalar product has to be visible.
$$\begin{aligned}\text{proj}&=\frac{1}{25}\big[(1)(-3)+(2)(4)+(5)(0)\big]\begin{pmatrix}-3\\4\\0\end{pmatrix}\\&=\frac{1}{25}(5)\begin{pmatrix}-3\\4\\0\end{pmatrix}=\frac{1}{5}\begin{pmatrix}-3\\4\\0\end{pmatrix}\end{aligned}$$
Marker: determines vector projection having demonstrated accurate calculation of the scalar product
C is where the perpendicular from A meets OB, so $\overrightarrow{\text{OC}}$ is the projection from part a). Its length is a fifth of $\left|\overrightarrow{\text{OB}}\right|$.
$$\text{OC}=\frac{1}{5}\times5$$
Marker: determines length of side OC
OA is the magnitude of $\overrightarrow{\text{OA}}$.
$$\text{OA}=\sqrt{1^2+2^2+5^2}$$
Marker: correctly determines the length of side OA
Triangle OCA has its right angle at C, so OA is the hypotenuse.
$$\begin{aligned}\text{OC}^2+\text{AC}^2&=\text{OA}^2\\\text{AC}^2&=30-1=29\end{aligned}$$
Marker: determines length of AC
Take OB as the base. AC is perpendicular to it, so AC is the height.
$$\text{Area}=\frac{1}{2}bh=\frac{1}{2}\times5\times\sqrt{29}$$
Marker: determines area of triangle OAB
Find the vector product of the two sides from O.
$$\begin{aligned}\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}&=\begin{pmatrix}(2)(0)-(5)(4)\\(5)(-3)-(1)(0)\\(1)(4)-(2)(-3)\end{pmatrix}=\begin{pmatrix}-20\\-15\\10\end{pmatrix}\end{aligned}$$
$\overrightarrow{\text{OB}}\times\overrightarrow{\text{OA}}=\begin{pmatrix}20\\15\\-10\end{pmatrix}$ is also accepted, since it has the same magnitude. This mark can be implied by your area calculation.
Marker: correctly calculates suitable vector product
The magnitude of the vector product is the area of the parallelogram. Halve it for the triangle.
$$\begin{aligned}\text{Area}&=\frac{1}{2}\left|\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}\right|=\frac{1}{2}\sqrt{(-20)^2+(-15)^2+10^2}\\&=\frac{1}{2}\sqrt{725}=\frac{5\sqrt{29}}{2}\end{aligned}$$
Showing that both are about 13.46 is also accepted. Since $725=25\times29$, the surds match exactly.
Marker: verifies result from Question 14e) based on vector product result
Each phrase of the question gave you something
Every part names its method. You are told to use the projection, then Pythagoras, then the base and height, then the vector product. Each part also tells you which earlier result to use.
If I had to guess where you might lose marks, it would be the middle component of the vector product. It is easy to get $+15$ instead of $-15$. The sign doesn’t change the magnitude here, but it would matter in a question that asks for a normal vector. The other risk is forgetting the ½ in part f).
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 2025 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it