Question 15
The position vectors of points P and Q are $2\hat{\boldsymbol{\imath}}-3\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}$ and $2\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}-4\hat{\boldsymbol{k}}$ respectively.
Let O be the origin.
Determine the angle POQ.
[2 marks]Points O, P and Q are joined to form a triangle.
Determine the area of triangle POQ.
[2 marks]Both sides start at the origin
Both position vectors start at $O$, so they are two sides of the triangle and the angle between them is the angle $POQ$. Each one is traced from $O$, one axis at a time.
The orange vector product is drawn at a fifth of its length so it fits, and its direction is exact.
Turn the triangle and compare the two methods
Turn the figure and watch the angle on screen change while the true angle stays the same. Then choose a way to find the area. The last option shows the slip of forgetting the half.
Drag the picture or use the sliders to turn it. The orange arrow is drawn at a fifth of its length. When it points straight at you it shows as an orange ring at $O$.
You are looking straight down the vector product, so the triangle is in its true shape and the angle on screen is the true angle $POQ$. From any other side the shape on screen is squashed.
From this side the triangle is tilted away from you, so its shape on screen is distorted. Measure the angle with the dot product, never from a drawing.
This is Method 1 in the marking guide. Half the length of the vector product is the area of the triangle. Turn the figure to check that the orange arrow is at right angles to both $\overrightarrow{OP}$ and $\overrightarrow{OQ}$.
This is Method 2 in the marking guide. It reuses the two lengths and the angle from part a), and it gives the same area.
The length of the vector product is the area of the whole grey parallelogram, about $17.32$. The triangle is half of it, so this answer is twice the area of the triangle.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide. This is a Technology-active question, so your GDC can find the dot product, the vector product and the lengths, and I have shown the working by hand as well.
The angle $POQ$ has its vertex at $O$, so it is the angle between $\overrightarrow{OP}$ and $\overrightarrow{OQ}$. $O$ is the origin, so these are the position vectors you are given.
$$\overrightarrow{OP}=\begin{pmatrix}2\\-3\\1\end{pmatrix},\quad\overrightarrow{OQ}=\begin{pmatrix}2\\2\\-4\end{pmatrix}$$
Rearrange the dot product rule $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ for $\cos\theta$, then substitute.
$$\begin{aligned}&\overrightarrow{OP}\cdot\overrightarrow{OQ}\\&\quad=(2)(2)+(-3)(2)+(1)(-4)\\&\quad=-6\end{aligned}$$
$$\begin{aligned}\left|\overrightarrow{OP}\right|&=\sqrt{2^2+(-3)^2+1^2}\\&=\sqrt{14}\\\left|\overrightarrow{OQ}\right|&=\sqrt{2^2+2^2+(-4)^2}\\&=\sqrt{24}\end{aligned}$$
$$\cos\theta=\frac{\overrightarrow{OP}\cdot\overrightarrow{OQ}}{\left|\overrightarrow{OP}\right|\left|\overrightarrow{OQ}\right|}=\frac{-6}{\sqrt{14}\sqrt{24}}$$
The QCAA accept an equivalent numerical expression, such as the decimal value of $\cos\theta$, and this mark can be implied by your later working. They also accept a different method, such as entering the two vectors in your GDC and asking it for the angle.
The marking guide’s example decimal is $-0.32$. The value is $-0.3273\ldots$, which is $-0.33$ to two decimal places, so the guide’s $-0.32$ has been truncated. Keep the full value in your GDC for the next step.
Marker: correctly determines a numerical expression for $\cos(\theta)$
Take the inverse cosine of the unrounded value. The cosine is negative, so the angle is obtuse.
$$\begin{aligned}\theta&=\cos^{-1}\left(\frac{-6}{\sqrt{14}\sqrt{24}}\right)\\&\approx1.9043\end{aligned}$$
The QCAA accept the angle in degrees, $\theta\approx109^\circ$, and the corresponding reflex angle, $\theta\approx4.38$ or $251^\circ$. If your expression for $\cos\theta$ was wrong, you can still earn this mark on follow-through.
Rounding too early costs accuracy here. $\cos^{-1}(-0.33)\approx1.907$, which rounds to $1.91$.
Marker: determines angle POQ
$\overrightarrow{OP}$ and $\overrightarrow{OQ}$ are two sides of the triangle that start at the same corner, $O$. The length of their vector product is the area of the parallelogram they make, and the triangle is half of it.
$$A=\frac12\left|\overrightarrow{OP}\times\overrightarrow{OQ}\right|$$
Find the vector product with the rule from the formula book.
$$\mathbf{a}\times\mathbf{b}=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$$
$$\begin{aligned}&\overrightarrow{OP}\times\overrightarrow{OQ}\\&\quad=\begin{pmatrix}2\\-3\\1\end{pmatrix}\times\begin{pmatrix}2\\2\\-4\end{pmatrix}\\&\quad=\begin{pmatrix}(-3)(-4)-(1)(2)\\(1)(2)-(2)(-4)\\(2)(2)-(-3)(2)\end{pmatrix}\\&\quad=\begin{pmatrix}10\\10\\10\end{pmatrix}\end{aligned}$$
The marking guide writes this as $10\hat{\boldsymbol{\imath}}+10\hat{\boldsymbol{\jmath}}+10\hat{\boldsymbol{k}}$, and your GDC can find it too. Write down the two vectors you entered and the result.
The QCAA also accept Method 2, which reuses part a). For this mark you substitute into the area rule with the sine of the included angle.
$$\begin{aligned}A&=\frac12\left|\overrightarrow{OP}\right|\left|\overrightarrow{OQ}\right|\sin\theta\\&=\frac12\sqrt{14}\sqrt{24}\sin(1.90)\end{aligned}$$
An equivalent expression such as $\frac12(3.74)(4.90)(0.94)$ is also accepted.
Marker: Method 1: correctly determines $\overrightarrow{OP}\times\overrightarrow{OQ}$. Method 2: correctly substitutes values into suitable rule
Find the length of the vector product, then halve it.
$$\begin{aligned}A&=\tfrac12\sqrt{10^2+10^2+10^2}\\&=\tfrac12\sqrt{300}\\&=5\sqrt3\end{aligned}$$
The QCAA also accept the exact value, $A=5\sqrt3$ units². Method 2 lands on the same area, and follow-through applies if your working before this step had an error.
Check that your GDC is in radians if you use Method 2 with $\theta\approx1.90$. In degree mode, $\sin(1.90)$ gives an area of about $0.30$.
Marker: determines area
Each phrase of the question gave you something
The angle between two vectors and the area of a triangle from a vector product are both standard routines, and here the vectors arrive ready to use. Because $O$ is a corner of the triangle, the position vectors are already two of its sides, and your GDC can check every calculation.
If I had to guess where you might lose marks, it would be the half in part b). The length of the vector product, $10\sqrt3\approx17.32$, is the area of the parallelogram, and the triangle is half of that. In part a), keep the negative sign in the dot product. Dropping it gives the acute angle, about $70.9^\circ$, which is the angle between $\overrightarrow{OP}$ and $\overrightarrow{QO}$.
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 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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