Question 15
A flat triangular shade sail is fixed at three anchor points $A(1,\,0,\,2)$, $B(3,\,1,\,2)$ and $C(1,\,2,\,4)$. Coordinates are measured in metres from an origin $O$ at ground level, and the $z$-axis points vertically upwards.
Determine $\overrightarrow{AB}$ and $\overrightarrow{AC}$.
[1 mark] b)Determine $\overrightarrow{AB}\times\overrightarrow{AC}$.
[1 mark] c)Use your result from Question 15b) to determine the Cartesian equation of the plane that contains the shade sail.
[2 marks] d)Use your result from Question 15b) to determine the area of the shade sail.
[1 mark]The top of a straight support pole touches the shade sail.
The pole lies along the line $\boldsymbol{r}=4\hat{\boldsymbol{\imath}}+3\hat{\boldsymbol{\jmath}}+t\left(-4\hat{\boldsymbol{\imath}}-4\hat{\boldsymbol{\jmath}}+5\hat{\boldsymbol{k}}\right)$, where $t$ is a scalar.
Determine the coordinates of the point where the pole touches the shade sail.
[2 marks]The sail’s plane, piece by piece, then the pole
Each part adds one piece to the printed diagram: the side vectors, their vector product, the plane, the area. Then the pole rises along its line until it touches the sail.
Drawn in the same view as the printed diagram. The normal is drawn at a quarter of its length, and the short dashed piece of the pole is hidden under the sail.
Push the pole up until it touches the sail
Turn on the pieces from parts a) to d) to see them on the sail. Then slide $t$ to move the top of the pole along its line. The readout puts the top of the pole into $x-2y+2z$, which equals $5$ only on the plane of the sail.
It touches the sail. $14t-2=5$ gives $t=\tfrac12$, and the point is $\left(2,\,1,\,\tfrac52\right)$, 2.5 m above the ground.
At $t=0$ the line is on the ground, at $(4,\,3,\,0)$. Slide $t$ up.
The top of the pole has not reached the plane of the sail yet: $14t-2$ is less than $5$. Slide $t$ up.
Too far. $14t-2$ is more than $5$, so the top of the pole has gone through the plane of the sail. Slide $t$ back.
Very close. Use the nudge buttons.
The top is under the sail now, so it is drawn dashed.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Write each point as a column and subtract $A$ from $B$, then from $C$, row by row.
$$\begin{aligned}\overrightarrow{AB}&=\begin{pmatrix}3\\1\\2\end{pmatrix}-\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}3-1\\1-0\\2-2\end{pmatrix}\\[6pt]\overrightarrow{AC}&=\begin{pmatrix}1\\2\\4\end{pmatrix}-\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}1-1\\2-0\\4-2\end{pmatrix}\end{aligned}$$
Marker: correctly determines $\overrightarrow{AB}$ and $\overrightarrow{AC}$
Substitute $\boldsymbol{a}=\overrightarrow{AB}$ and $\boldsymbol{b}=\overrightarrow{AC}$ into the formula sheet's vector product.
$$\begin{aligned}\overrightarrow{AB}\times\overrightarrow{AC}&=\begin{pmatrix}1\times2-0\times2\\0\times0-2\times2\\2\times2-1\times0\end{pmatrix}\\&=\begin{pmatrix}2-0\\0-4\\4-0\end{pmatrix}\end{aligned}$$
The formula sheet gives $\boldsymbol{a}\times\boldsymbol{b}=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$. Order matters: $\overrightarrow{AB}$ is written first. Take each component straight from the formula, so no signs have to be remembered.
Marker: correctly determines the vector product
The vector product is perpendicular to $\overrightarrow{AB}$ and $\overrightarrow{AC}$, so use it as the normal, with $A$ as the point on the plane.
$$\begin{aligned}\boldsymbol{n}&=\begin{pmatrix}2\\-4\\4\end{pmatrix},\quad\boldsymbol{a}=\overrightarrow{OA}=\begin{pmatrix}1\\0\\2\end{pmatrix}\\[6pt]\boldsymbol{r}\cdot\boldsymbol{n}&=\boldsymbol{a}\cdot\boldsymbol{n}\\&=\begin{pmatrix}1\\0\\2\end{pmatrix}\cdot\begin{pmatrix}2\\-4\\4\end{pmatrix}\\&=(1)(2)+(0)(-4)+(2)(4)\\&=10\end{aligned}$$
Any non-zero multiple of the normal vector is accepted.
Marker: correctly uses $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{a}\cdot\boldsymbol{n}$ with a point on the plane
Write $\boldsymbol{r}$ as the column of $x$, $y$ and $z$, expand the dot product, then divide every term by 2.
$$\begin{aligned}\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}2\\-4\\4\end{pmatrix}&=10\\2x-4y+4z&=10\end{aligned}$$
Any equivalent form of the Cartesian equation, such as $x-2y+2z-5=0$, is accepted, so the mark is already earned at $2x-4y+4z=10$.
Marker: correctly determines the Cartesian equation of the plane
Halve the magnitude of the vector product.
$$\begin{aligned}\text{Area}&=\tfrac12\left|\begin{pmatrix}2\\-4\\4\end{pmatrix}\right|\\&=\tfrac12\sqrt{2^2+(-4)^2+4^2}\\&=\tfrac12\sqrt{36}\\&=\tfrac12\times6\end{aligned}$$
$\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|$ is the area of the parallelogram with sides $\overrightarrow{AB}$ and $\overrightarrow{AC}$, and the triangle is half of it.
Marker: correctly determines the area
Write the line as column vectors, substitute each row into $x-2y+2z=5$ and solve for $t$.
$$\begin{aligned}\boldsymbol{r}&=\begin{pmatrix}4\\3\\0\end{pmatrix}+t\begin{pmatrix}-4\\-4\\5\end{pmatrix}\\&=\begin{pmatrix}4-4t\\3-4t\\5t\end{pmatrix}\end{aligned}$$
$$\begin{aligned}(4-4t)-2(3-4t)+2(5t)&=5\\4-4t-6+8t+10t&=5\\14t-2&=5\\t&=\tfrac12\end{aligned}$$
Follow-through applies in part b) from incorrect vectors in part a), in parts c) and d) from an incorrect vector product, and in part e) from an incorrect plane.
Marker: correctly substitutes the equation of the line into the equation of the plane to determine the value of $t$
Substitute $t=\tfrac12$ back into the line.
$$\boldsymbol{r}=\begin{pmatrix}4-4\left(\tfrac12\right)\\3-4\left(\tfrac12\right)\\5\left(\tfrac12\right)\end{pmatrix}=\begin{pmatrix}2\\1\\\tfrac52\end{pmatrix}$$
Marker: correctly determines the coordinates of the point
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Dropping the minus sign on the $\hat{\boldsymbol{\jmath}}$ term in b). You get $2\hat{\boldsymbol{\imath}}+4\hat{\boldsymbol{\jmath}}+4\hat{\boldsymbol{k}}$. The dot product check catches it: $(2,4,4)\cdot(2,1,0)=8\ne0$. Carried into c), it gives $2x+4y+4z=10$, which $B$ does not satisfy ($6+4+8=18$).
- Using $\overrightarrow{AB}$ instead of $\overrightarrow{OA}$ as $\boldsymbol{a}$ in c). $\overrightarrow{AB}$ is a direction, not a position. $\overrightarrow{AB}\cdot\boldsymbol{n}=0$, which gives $x-2y+2z=0$, a parallel plane through the origin. Working out $\overrightarrow{AC}\times\overrightarrow{AB}=-2\hat{\boldsymbol{\imath}}+4\hat{\boldsymbol{\jmath}}-4\hat{\boldsymbol{k}}$ is not an error: it is a non-zero multiple of $\boldsymbol{n}$ and leads to the same plane.
- Forgetting the $\tfrac12$ in d). That gives $6$ m$^2$, the parallelogram. Forgetting the square root gives $\tfrac12\times36=18$ m$^2$.
- Expanding $-2(3-4t)$ as $-6-8t$ in e). You get $-2t-2=5$, so $t=-\tfrac72$ and the point $\left(18,\,17,\,-\tfrac{35}{2}\right)$. A negative $z$ (below ground) should tell you something went wrong.
- Halving only the left side of $2x-4y+4z=10$. Using $2x-4y+4z=5$ gives $28t-4=5$ and $t=\tfrac{9}{28}$. Also common: stopping at $t=\tfrac12$ without finding the coordinates, which loses the last mark.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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