Question 7
Given $\boldsymbol{a}=(3n+2)\hat{i}+2\hat{j}$, $\boldsymbol{b}=(n-2)\hat{j}$ and $\boldsymbol{a}\times\boldsymbol{b}=(1-2n)\hat{k}$, the possible values of $n$ are
Not quite. These solve $(n+5)(3n-1)=0$, which expands to $3n^2+14n-5$. The equation you need is $3n^2-2n-5=0$, which factorises as $(3n-5)(n+1)$. The 5 and the 1 have swapped places. Check $n=-5$: $(3n+2)(n-2)=(-13)(-7)=91$, but $1-2n=11$. In 2022, 9.0% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2022, 63.6% of Queensland students chose (B).
Not quite. These are the right numbers with the wrong signs. The quadratic factorises as $(3n-5)(n+1)=0$, so $n=\tfrac53$ or $n=-1$. Check $n=1$: $(3n+2)(n-2)=(5)(-1)=-5$, but $1-2n=-1$. In 2022, 11.8% of Queensland students chose (C).
Not quite. These solve $(n-5)(3n+1)=0$, which expands to $3n^2-14n-5$. The right factors are $(3n-5)(n+1)$, so the 3 has gone into the wrong bracket. Check $n=5$: $(3n+2)(n-2)=(17)(3)=51$, but $1-2n=-9$. In 2022, 13.8% of Queensland students chose (D).
Drag the picture or use the sliders to turn it. The blue arrows $\boldsymbol{a}$ and $\boldsymbol{b}$ both lie in the $xy$-plane. The black arrow is $\boldsymbol{a}\times\boldsymbol{b}$, and the orange ring marks $(1-2n)\hat{k}$. The scale changes with $n$ so that everything fits.
The black arrow ends exactly on the orange ring, so this value of $n$ works.
With $n=2$, $\boldsymbol{b}$ is the zero vector, so $\boldsymbol{a}\times\boldsymbol{b}=\boldsymbol{0}$. The ring is at $-3$, so this value fails.
With $n=-\tfrac23$, $\boldsymbol{a}=2\hat{j}$ is parallel to $\boldsymbol{b}$, so $\boldsymbol{a}\times\boldsymbol{b}=\boldsymbol{0}$. The ring is at $\tfrac73$, so this value fails.
The black arrow misses the orange ring, so this value of $n$ fails. Slide until the tip lands on the ring. It happens twice.
Write each vector as a column. Neither vector has a $\hat{k}$ part, and $\boldsymbol{b}$ has no $\hat{i}$ part, so put zeros in those places.
$$\boldsymbol{a}=\begin{pmatrix}3n+2\\2\\0\end{pmatrix}$$
$$\boldsymbol{b}=\begin{pmatrix}0\\n-2\\0\end{pmatrix}$$
Use the cross product formula from the formula book. The zeros clear the first two rows.
$$\boldsymbol{a}\times\boldsymbol{b}=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}=$$
$$\begin{pmatrix}2(0)-0(n-2)\\0(0)-(3n+2)(0)\\(3n+2)(n-2)-2(0)\end{pmatrix}$$
$$=\begin{pmatrix}0\\0\\(3n+2)(n-2)\end{pmatrix}$$
Set the $\hat{k}$ parts equal and solve the quadratic. Your calculator’s solve function does this in one line.
$$\begin{aligned}(3n+2)(n-2)&=1-2n\\3n^2-4n-4&=1-2n\\3n^2-2n-5&=0\\(3n-5)(n+1)&=0\end{aligned}$$
So $n=-1$ or $n=\tfrac53$. Try both on the left.
Both vectors lie in the $xy$-plane, so only the $\hat{k}$ row of the cross product survives.
$$\begin{pmatrix}3n+2\\2\\0\end{pmatrix}\times\begin{pmatrix}0\\n-2\\0\end{pmatrix}$$
$$=\begin{pmatrix}0\\0\\(3n+2)(n-2)\end{pmatrix}$$
$$\begin{aligned}(3n+2)(n-2)&=1-2n\\3n^2-2n-5&=0\\(3n-5)(n+1)&=0\end{aligned}$$
$$n=-1\ \text{or}\ n=\tfrac53$$
Check $n=-1$: $\boldsymbol{a}=-\hat{i}+2\hat{j}$ and $\boldsymbol{b}=-3\hat{j}$, so $\boldsymbol{a}\times\boldsymbol{b}=(-1)(-3)\hat{k}=3\hat{k}$, and $1-2(-1)=3$.
How Queensland students answered in 2022
From the QCAA Specialist Mathematics subject report for 2022. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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