Question 10
A plane is represented by the equation $x-2z=5$. A vector normal to this plane is
Not quite. The $-2$ belongs to $z$, and there is no $y$ term, so the middle component is $0$. The $5$ on the right is a constant. It moves the plane but never goes into the normal. This vector has $(1,-2,5)\cdot(2,0,1)=7\neq0$. In 2022, 6.0% of Queensland students chose (A).
Not quite. The coefficient of $z$ is $-2$, so the sign matters. This vector has $(1,0,2)\cdot(2,0,1)=4\neq0$, so it isn’t perpendicular to the plane. In 2022, 6.8% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2022, 79.9% of Queensland students chose (C).
Not quite. Rewriting as $x-2z-5=0$ and reading off $1,\ -2,\ -5$ puts the $-2$ in the $y$ slot and uses the constant as a component. This vector has $(1,-2,-5)\cdot(2,0,1)=-3\neq0$. In 2022, 7.0% of Queensland students chose (D).
Drag the picture or use the sliders to turn it. The blue sheet is part of the plane. The two thin black arrows from lie in the plane, along and . The orange arrow is the option you are testing. Every orange arrow is drawn the same length, so only its direction matters.
Both dot products are $0$, so this vector is perpendicular to two different directions in the plane. It is a normal. Turn the picture and it stays at right angles to the sheet.
This vector is perpendicular to $(0,1,0)$ but not to $(2,0,1)$, so it leans out of the perpendicular. Turn the picture until the sheet is edge-on to see the lean.
Neither dot product is $0$, so this vector isn’t perpendicular to the plane.
A plane written as $ax+by+cz=d$ has normal vector $\mathbf{n}=(a,b,c)$. The constant $d$ only shifts the plane, so it never appears in the normal.
There is no $y$ term, so its coefficient is $0$. Write the equation with every term showing.
$$1x+0y-2z=5\ \Rightarrow\ \mathbf{n}=\begin{pmatrix}1\\0\\-2\end{pmatrix}$$
Check it against two directions that lie in the plane. Points $(5,0,0)$ and $(5,1,0)$ are on the plane, so $(0,1,0)$ is one direction. Points $(5,0,0)$ and $(7,0,1)$ give $(2,0,1)$.
$$\begin{pmatrix}1\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}0\\1\\0\end{pmatrix}=0$$
$$\begin{pmatrix}1\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}2\\0\\1\end{pmatrix}=2-2=0$$
Write the equation with the $y$ term showing, then read the coefficients of $x$, $y$ and $z$.
$$1x+0y-2z=5\ \Rightarrow\ \mathbf{n}=\begin{pmatrix}1\\0\\-2\end{pmatrix}$$
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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