Question 15
The points $O(0,0,0)$, $A(-6,2,-2)$ and $C(3,1,2)$ are represented in three-dimensional space in the diagram.
$OABC$ forms a parallelogram in three-dimensional space.
Determine the coordinates of $B$.
[1 mark]$M$ is the midpoint of $BC$.
Determine the vector that represents $\overrightarrow{OM}$.
[1 mark]$N$ divides $AM$ in the ratio $2:1$.
Determine the vector that represents $\overrightarrow{ON}$.
[2 marks] d)Use a vector method to show that $O$, $B$ and $N$ lie on a straight line.
[2 marks]Each point is reached along vectors you already know
The figure is drawn from the real coordinates. Each new point is reached by a path from $O$, and the blue legs of the path are the vectors that add up to the white arrow.
The figure uses the real coordinates from one fixed viewpoint. You can turn it yourself in the interactive below.
Slide the dividing point and test the line
Turn the figure to see the parallelogram from any side. Then slide $N$ along $AM$. The ring marks the point two-thirds of the way along $OB$, and only one position of $N$ lands on it.
Drag the picture or use the sliders to turn it. The dashed line runs through $O$ and $B$, and the ticks split $AM$ into thirds.
This is the question’s point. $\overrightarrow{ON}=\tfrac23\overrightarrow{OB}$, so $N$ is on the line through $O$ and $B$, two-thirds of the way from $O$. The vector product is the zero vector, which tells you the same thing.
This is the ratio the wrong way round. $AN:NM=1:2$ puts $N$ one-third of the way from $A$, at $(-4,2,-1)$. That point is off the line through $O$ and $B$, so part d) could not be shown.
$N$ misses the line through $O$ and $B$. The second components force $k=\tfrac23$, but then the first and third components disagree, and the vector product is not zero. Only the ratio $2:1$ puts $N$ on the line.
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 letters of $OABC$ go in order around the parallelogram, so $B$ is the corner opposite $O$. Opposite sides of a parallelogram are equal vectors, so $\overrightarrow{AB}=\overrightarrow{OC}$. Go out along $\overrightarrow{OA}$, then along a copy of $\overrightarrow{OC}$.
$$\begin{aligned}\overrightarrow{OB}&=\overrightarrow{OA}+\overrightarrow{AB}\\&=\overrightarrow{OA}+\overrightarrow{OC}\\&=\begin{pmatrix}-6\\2\\-2\end{pmatrix}+\begin{pmatrix}3\\1\\2\end{pmatrix}\\&=\begin{pmatrix}-3\\3\\0\end{pmatrix}\end{aligned}$$
The QCAA also accept the answer as a vector, such as $\overrightarrow{OB}=\left(\begin{smallmatrix}-3\\3\\0\end{smallmatrix}\right)$. The sample response in the marking guide starts with $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AC}$. That is a misprint, because the two vectors it then adds are $\overrightarrow{OA}$ and $\overrightarrow{OC}$.
Marker: correctly determines the coordinates of $B$
$M$ is the midpoint of $BC$, so each of its coordinates is the average of the matching coordinates of $B$ and $C$.
$$\begin{aligned}M&=\left(\frac{-3+3}{2},\frac{3+1}{2},\frac{0+2}{2}\right)\\&=(0,2,1)\end{aligned}$$
The QCAA also accept another way of writing $\overrightarrow{OM}$, such as $\overrightarrow{OM}=\overrightarrow{OB}+\tfrac12\overrightarrow{BC}$, and the coordinate form $(0,2,1)$. If your $B$ was wrong, follow-through marks apply here and in part c).
Marker: determines vector $\overrightarrow{OM}$
$N$ divides $AM$ in the ratio $2:1$, so $AN$ is two of the three equal parts and $N$ is $\tfrac23$ of the way from $A$ to $M$. Go out to $A$, then two-thirds of the way along $\overrightarrow{AM}$.
$$\begin{aligned}\overrightarrow{ON}&=\overrightarrow{OA}+\tfrac23\overrightarrow{AM}\\&=\overrightarrow{OA}+\tfrac23\left(\overrightarrow{OM}-\overrightarrow{OA}\right)\end{aligned}$$
The QCAA also accept the same expression in position-vector letters, $\overrightarrow{ON}=\mathbf{a}+\tfrac23(\mathbf{m}-\mathbf{a})$, and this mark can be implied by your later working. The marking guide describes this mark as using “the 1:2 ratio”, but its working uses $\tfrac23$, which is the question’s ratio of $2:1$.
Marker: determines an expression for $\overrightarrow{ON}$ involving $\overrightarrow{OM}$ and the $2:1$ ratio
Expand the bracket and collect the two vectors, then substitute.
$$\begin{aligned}\overrightarrow{ON}&=\tfrac13\overrightarrow{OA}+\tfrac23\overrightarrow{OM}\\&=\frac13\begin{pmatrix}-6\\2\\-2\end{pmatrix}+\frac23\begin{pmatrix}0\\2\\1\end{pmatrix}\\&=\begin{pmatrix}-2\\[4pt]\dfrac23\\[8pt]-\dfrac23\end{pmatrix}+\begin{pmatrix}0\\[4pt]\dfrac43\\[8pt]\dfrac23\end{pmatrix}\end{aligned}$$
The QCAA also accept the coordinate form $(-2,2,0)$. Using $\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=\left(\begin{smallmatrix}6\\0\\3\end{smallmatrix}\right)$ straight away gives the same answer, since $\left(\begin{smallmatrix}-6\\2\\-2\end{smallmatrix}\right)+\tfrac23\left(\begin{smallmatrix}6\\0\\3\end{smallmatrix}\right)=\left(\begin{smallmatrix}-2\\2\\0\end{smallmatrix}\right)$.
Marker: determines vector $\overrightarrow{ON}$
Write down what you need to show before you calculate anything. Three points lie on a straight line when two vectors joining them are parallel and share a point. Both $\overrightarrow{ON}$ and $\overrightarrow{OB}$ start at $O$, so the requirement is that one is a multiple of the other.
$$\overrightarrow{ON}=k\,\overrightarrow{OB}$$
A second method uses the vector product, which is the zero vector for parallel vectors. Then the requirement is $\overrightarrow{ON}\times\overrightarrow{OB}=\mathbf{0}$, and the QCAA also accept $\left|\overrightarrow{ON}\times\overrightarrow{OB}\right|=0$.
This mark can be implied by your later working, and the QCAA accept an equivalent requirement. The marking guide’s example of one is equal gradients for $ON$ and $OM$. That looks like a misprint for $ON$ and $OB$, since $M$ is not on the line.
Marker: correctly communicates a requirement that $O$, $B$ and $N$ lie on a straight line
Use your answers to parts a) and c). Every component of $\overrightarrow{ON}$ is $\tfrac23$ of the matching component of $\overrightarrow{OB}$.
$$\begin{pmatrix}-2\\2\\0\end{pmatrix}=\frac23\begin{pmatrix}-3\\3\\0\end{pmatrix}$$
The QCAA also accept $\overrightarrow{OB}=\tfrac32\overrightarrow{ON}$. For the vector product method, use the rule from the formula book.
$$\mathbf{a}\times\mathbf{b}=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$$
$$\begin{aligned}&\overrightarrow{ON}\times\overrightarrow{OB}\\&\quad=\begin{pmatrix}-2\\2\\0\end{pmatrix}\times\begin{pmatrix}-3\\3\\0\end{pmatrix}\\&\quad=\begin{pmatrix}(2)(0)-(0)(3)\\(0)(-3)-(-2)(0)\\(-2)(3)-(2)(-3)\end{pmatrix}\\&\quad=\begin{pmatrix}0\\0\\0\end{pmatrix}\end{aligned}$$
The vector product is $\mathbf{0}$, so the vectors are parallel, and with the common point $O$ the three points lie on a straight line.
Marker: correctly demonstrates that the requirement has been met
Each phrase of the question gave you something
Each part is one routine vector skill, and each answer feeds the next part. The coordinates are small integers, and the thirds that appear in part c) cancel to whole numbers.
If I had to guess where you might lose marks, it would be part a). The order of the letters in $OABC$ puts $B$ opposite $O$, so $\overrightarrow{OB}$ is a diagonal and you add $\overrightarrow{OA}$ and $\overrightarrow{OC}$. Subtracting them gives a point that won’t line up with $N$ in part d). In part c), check which end the ratio starts from: $2:1$ from $A$ means two-thirds of the way along $AM$ from $A$. In part d), finish with a sentence that names the common point $O$.
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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