Question 12
Consider the vertices $A$, $B$ and $C$ of the rectangular prism as shown.
State the coordinates of $A$, $B$ and $C$.
[1 mark] b)Determine a unit vector, $\hat{\boldsymbol{n}}$, that is normal to the plane containing $A$, $B$ and $C$.
[3 marks] c)Verify that $\hat{\boldsymbol{n}}$ is perpendicular to $\overrightarrow{AB}$.
[2 marks] d)Determine the Cartesian equation of the plane that contains $A$, $B$ and $C$.
[2 marks]Three corners pin down a plane
Read each corner by walking from $O$ along the axes. Then the cross product of two edges of the triangle gives a normal, and one corner fixes where the plane sits.
The orange arrow is $\hat{\boldsymbol{n}}$ drawn to scale, one unit long. The shaded triangle is only part of the plane, which carries on forever in every direction.
Choose two vectors in the plane and turn the prism
Pick two sides of the triangle, then swap their order. Turn the prism until the shaded triangle is edge-on and watch the orange unit normal stand at right angles to it. The last option shows the slip of crossing the position vectors $\boldsymbol{a}$ and $\boldsymbol{c}$.
Drag the picture or use the sliders to turn it. The orange arrow is $\hat{\boldsymbol{n}}$ drawn to scale, one unit long, with a right-angle mark against the first vector in the cross product.
These are the two vectors the QCAA sample response uses. Their cross product has length $7$, so dividing by $7$ gives the unit normal. The dot product with $\overrightarrow{AB}$ is $0$, and all three corners give $12$ on the right of $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{a}\cdot\boldsymbol{n}$.
This pair gives exactly the same cross product. Any two sides of the triangle work here, and the QCAA accept any two vectors in the plane, such as $\overrightarrow{BC}$.
Swapping the order reverses the cross product, so $\hat{\boldsymbol{n}}$ points out of the other side of the triangle. It is still a unit normal. The marking guide lists the reversed normal $-\left(\begin{smallmatrix}6\\2\\3\end{smallmatrix}\right)$ as an accepted answer in part b). The Cartesian equation is the earlier one multiplied by $-1$, which describes the same plane.
$\boldsymbol{a}$ and $\boldsymbol{c}$ are position vectors. They start at $O$, so they lie in the grey triangle $OAC$, and their cross product is normal to that triangle instead. The check from part c) catches it, because $\hat{\boldsymbol{n}}\cdot\overrightarrow{AB}$ is not zero. The three corners also don’t give the same value of $\boldsymbol{r}\cdot\boldsymbol{n}$, so no plane with this normal contains $A$, $B$ and $C$. Turn the prism to see the grey triangle face-on.
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 axes are marked $1$, $3$ and $2$, so the prism is $1$ unit along the $x$-axis (towards you), $3$ along the $y$-axis and $2$ up the $z$-axis. Every coordinate of a corner is either $0$ or one of those lengths. Walk from $O$ along the axes to each corner.
To reach $A$, go $1$ along $x$ and $3$ along $y$, staying on the floor. To reach $B$, go $3$ along $y$ and $2$ up. To reach $C$, go $1$ along $x$ and $2$ up.
The one mark is for all three sets of coordinates. “State” means no working is needed.
Marker: correctly states the coordinates of $A$, $B$ and $C$
A normal to the plane is perpendicular to every vector that lies in it. Start with two vectors in the plane. Each one is end minus start, using the position vectors $\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{c}$ of the three corners.
$$\begin{aligned}\overrightarrow{AB}&=\boldsymbol{b}-\boldsymbol{a}\\&=\begin{pmatrix}0\\3\\2\end{pmatrix}-\begin{pmatrix}1\\3\\0\end{pmatrix}\\&=\begin{pmatrix}-1\\0\\2\end{pmatrix}\end{aligned}$$
$$\begin{aligned}\overrightarrow{AC}&=\boldsymbol{c}-\boldsymbol{a}\\&=\begin{pmatrix}1\\0\\2\end{pmatrix}-\begin{pmatrix}1\\3\\0\end{pmatrix}\\&=\begin{pmatrix}0\\-3\\2\end{pmatrix}\end{aligned}$$
The QCAA accept any two vectors in the plane, such as $\overrightarrow{BC}=\left(\begin{smallmatrix}1\\-3\\0\end{smallmatrix}\right)$, in any vector form, such as $\overrightarrow{AB}=-\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{k}}$. This mark can be implied by your later working. From here to part d), follow-through applies if an earlier answer was wrong.
Marker: determines 2 vectors in the plane containing $A$, $B$ and $C$
The cross product of two vectors is perpendicular to both of them, so $\overrightarrow{AB}\times\overrightarrow{AC}$ is normal to the plane. The formula book writes the rule with $\boldsymbol{a}$ and $\boldsymbol{b}$, but those letters are already the position vectors of $A$ and $B$, so I have used $\boldsymbol{u}$ and $\boldsymbol{v}$.
$$\boldsymbol{u}\times\boldsymbol{v}=\begin{pmatrix}u_2v_3-u_3v_2\\u_3v_1-u_1v_3\\u_1v_2-u_2v_1\end{pmatrix}$$
With $\boldsymbol{u}=\overrightarrow{AB}$ and $\boldsymbol{v}=\overrightarrow{AC}$:
$$\begin{aligned}\boldsymbol{n}&=\overrightarrow{AB}\times\overrightarrow{AC}\\&=\begin{pmatrix}-1\\0\\2\end{pmatrix}\times\begin{pmatrix}0\\-3\\2\end{pmatrix}\\&=\begin{pmatrix}(0)(2)-(2)(-3)\\(2)(0)-(-1)(2)\\(-1)(-3)-(0)(0)\end{pmatrix}\\&=\begin{pmatrix}6\\2\\3\end{pmatrix}\end{aligned}$$
Sign slips usually happen in the middle row, which is $u_3v_1-u_1v_3$. You can check your normal before moving on: $\boldsymbol{n}\cdot\overrightarrow{AB}$ and $\boldsymbol{n}\cdot\overrightarrow{AC}$ should both be $0$.
The QCAA accept any vector normal to the plane, such as $\overrightarrow{AB}\times\overrightarrow{CB}=-\left(\begin{smallmatrix}6\\2\\3\end{smallmatrix}\right)$.
Marker: determines a vector normal to the plane
A unit vector has length $1$. Divide $\boldsymbol{n}$ by its length.
$$\begin{aligned}|\boldsymbol{n}|&=\sqrt{6^2+2^2+3^2}\\&=\sqrt{49}\\&=7\end{aligned}$$
$$\hat{\boldsymbol{n}}=\frac{1}{|\boldsymbol{n}|}\,\boldsymbol{n}$$
The QCAA also accept the unit vector written as $\left(\begin{smallmatrix}6/7\\2/7\\3/7\end{smallmatrix}\right)$. If you used the reversed normal from step 3, the same division gives $-\frac{1}{7}\left(\begin{smallmatrix}6\\2\\3\end{smallmatrix}\right)$, which is also a unit vector normal to the plane.
Marker: determines a unit vector normal to the plane
Two non-zero vectors are perpendicular exactly when their dot product is $0$. Write down what you need to show, then substitute your results from part b).
Verify that $\hat{\boldsymbol{n}}\cdot\overrightarrow{AB}=0$.
$$\hat{\boldsymbol{n}}\cdot\overrightarrow{AB}=\frac{1}{7}\begin{pmatrix}6\\2\\3\end{pmatrix}\cdot\begin{pmatrix}-1\\0\\2\end{pmatrix}$$
The QCAA also accept working with $\boldsymbol{n}$ in place of $\hat{\boldsymbol{n}}$, as in $\boldsymbol{n}\cdot\overrightarrow{AB}=0$. The factor of $\frac{1}{7}$ doesn’t change whether the answer is zero.
Marker: substitutes the results from b) into $\hat{\boldsymbol{n}}\cdot\overrightarrow{AB}$
Multiply matching components and add.
$$\begin{aligned}\hat{\boldsymbol{n}}\cdot\overrightarrow{AB}&=\frac{1}{7}\left(-6+0+6\right)\\&=0\end{aligned}$$
Finish with a sentence that says what the zero shows. The sample response ends with “The required result is verified.”
If an earlier slip means your dot product can’t come out to $0$, you can still earn this mark for a correct calculation, as long as you have written the statement you were trying to show, such as $\hat{\boldsymbol{n}}\cdot\overrightarrow{AB}=0$. The printed guide says that statement must be shown “in e)”, the same words as its note for Question 11 e). Question 12 has no part e), so this is a copying slip for part c).
Marker: verifies the result by showing that the result of the calculation is 0
A plane is fixed by a normal and one point on it. If $\boldsymbol{r}$ is any point on the plane, then $\boldsymbol{r}-\boldsymbol{a}$ lies in the plane, so it is perpendicular to $\boldsymbol{n}$. Setting $(\boldsymbol{r}-\boldsymbol{a})\cdot\boldsymbol{n}=0$ gives the vector equation $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{a}\cdot\boldsymbol{n}$. Use $\boldsymbol{n}$ from part b) and the point $A$.
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}6\\2\\3\end{pmatrix}=\begin{pmatrix}1\\3\\0\end{pmatrix}\cdot\begin{pmatrix}6\\2\\3\end{pmatrix}$$
The QCAA accept $B$ or $C$ in place of $A$, as in $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{b}\cdot\boldsymbol{n}$ or $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{c}\cdot\boldsymbol{n}$.
The marking guide also gives a second method. Start from $ax+by+cz+d=0$ with the normal’s components as $a$, $b$ and $c$, then substitute the coordinates of one corner to form an equation in $d$. For this mark you need the first line below, using $A$, $B$ or $C$.
$$\begin{aligned}6(1)+2(3)+3(0)+d&=0\\d&=-12\end{aligned}$$
Marker: determines the vector equation of the plane (method 2: substitutes a vector normal to the plane and a suitable set of coordinates into the Cartesian equation of plane formula to form an equation in terms of $d$)
Expand both dot products. The left side becomes $6x+2y+3z$, and the right side is $1\times6+3\times2+0\times3=12$.
The QCAA also accept $6x+2y+3z-12=0$, which is where the second method lands with $d=-12$.
Check with a corner you didn’t use. For $B(0,3,2)$, $6(0)+2(3)+3(2)=12$, and for $C(1,0,2)$, $6(1)+2(0)+3(2)=12$.
Marker: determines the Cartesian equation of the plane
Each phrase of the question gave you something
This is the standard routine for a plane through three points, split into small pieces with the order chosen for you. Two vectors and a cross product give the normal, and one corner gives the constant in the equation. Part c) is a dot product you would do anyway as a check on part b).
If I had to guess where you might lose marks, it would be part a). The $x$-axis comes towards you in the diagram, so it is easy to swap the $x$ and $y$ coordinates or read a corner from the wrong face. The check in part c) won’t catch a wrong corner, because $\hat{\boldsymbol{n}}$ is built from your own $\overrightarrow{AB}$, so trace each corner from $O$ along the axes before you move on. The other place I would slow down is the middle row of the cross product.
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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