EXN.2.AHL.TZ0.11

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

 

$\overrightarrow{\text{AB}}=\left(\begin{array}{c}0\\ 6\\ -6\end{array}\right) \left(=6\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)\right)$        A1

$\overrightarrow{\text{AC}}=\left(\begin{array}{c}-6\\ 0\\ -6\end{array}\right) \left(=6\left(\begin{array}{c}-1\\ 0\\ -1\end{array}\right)\right)$        A1

 

[2 marks]

attempts to use  $\cos \text{B}\hat{\text{A}}\text{C}=\frac{\overrightarrow{\text{AB}}·\overrightarrow{\text{AC}}}{\left|\overrightarrow{\text{AB}}\right|\left|\overrightarrow{\text{AC}}\right|}$        (M1)

$=\frac{\left(\begin{array}{c}0\\ 6\\ -6\end{array}\right)·\left(\begin{array}{c}-6\\ 0\\ -6\end{array}\right)}{\sqrt{72}\times \sqrt{72}}$        A1

$=\frac{1}{2}$        A1

so $\text{B}\hat{\text{A}}\text{C}=60°$        AG

 

[3 marks]

attempts to find a vector normal to $\Pi$        M1

for example, $\overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}}=\left(\begin{array}{c}-36\\ 36\\ 36\end{array}\right) \left(=36\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)\right)$ leading to        A1

a vector normal to $\Pi$ is $\mathit{n}=\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)$

 

EITHER

substitutes $\left(5, -2, -5\right)$ (or $\left(5, 4, -1\right)$ or $\left(-1, -2, -1\right)$) into $-x+y+z=d$ and attempts to find the value of $d$

for example, $d=-5-2+5 \left(=-2\right)$        M1

 

OR

attempts to use $\mathit{r}\mathbf{·}\mathit{n}=\mathit{a}\mathbf{·}\mathit{n}$        M1

for example, $\left(\begin{array}{c}x\\ y\\ z\end{array}\right)·\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}5\\ -2\\ 5\end{array}\right)·\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)$

 

THEN

leading to the Cartesian equation of $\Pi$ as $-x+y+z=-2$        AG

 

[3 marks]

$\mathit{r}=\left(\begin{array}{c}7\\ -4\\ -3\end{array}\right)+\lambda \left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right) \left(\lambda \in \mathrm{ℝ}\right)$        A1

 

[1 mark]

substitutes $x=7-\lambda , y=-4+\lambda , z=-3+\lambda$ into $-x+y+z=-2$        (M1)

$-\left(7-\lambda \right)+\left(-4+\lambda \right)+\left(-3+\lambda \right)=-2 \left(3\lambda =12\right)$

$\lambda =4$        A1

shows a correct calculation for finding $d_{\text{min}}$, for example, attempts to find

$\left|4\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)\right|$        M1

$d_{\text{min}}=4\sqrt{3} \left(=6.93\right)$        A1

 

[4 marks]

let the area of triangle $\text{ABC}$ be $A$

 

EITHER

attempts to find $A=\frac{1}{2}\left|\overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}}\right|$, for example       M1

$A=\frac{1}{2}\left|\left(\begin{array}{c}-36\\ 36\\ 36\end{array}\right)\right|$

 

OR

attempts to find $\frac{1}{2}\left|\overrightarrow{\text{AB}}\right|\left|\overrightarrow{\text{AC}}\right|\sin \theta$, for example       M1

$A=\frac{1}{2}\times 6\sqrt{2}\times 6\sqrt{2}\times \frac{\sqrt{3}}{2}$  (where $\sin \frac{\mathrm{\pi }}{3}=\frac{\sqrt{3}}{2}$)

 

THEN

$A=18\sqrt{3} \left(=31.2\right)$       A1

uses $V=\frac{1}{3}Ah$ where $A$ is the area of triangle $\text{ABC}$ and $h=d_{\text{min}}$       M1

 $V=\frac{1}{3}\times 18\sqrt{3}\times 4\sqrt{3}$

$=72$       A1

 

[4 marks]

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