2023 H2 Maths Paper 2 Q3 — Worked Solution

Part (a)

2 marks

A = \begin{pmatrix} - 1 \\ 2 \\ 5 \end{pmatrix},\quad B = \begin{pmatrix} 1 \\ - 2 \\ 8 \end{pmatrix}

We first find $\overrightarrow{AB}$ .

\overrightarrow{AB} = B - A

= \begin{pmatrix} 1 - ( - 1) \\ - 2 - 2 \\ 8 - 5 \end{pmatrix}

= \begin{pmatrix} 2 \\ - 4 \\ 3 \end{pmatrix}

Since $\overrightarrow{BC} = 2\overrightarrow{AB}$ ,

\overrightarrow{BC} = 2\begin{pmatrix} 2 \\ - 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ - 8 \\ 6 \end{pmatrix}

C = B + \overrightarrow{BC}

= \begin{pmatrix} 1 + 4 \\ - 2 - 8 \\ 8 + 6 \end{pmatrix}

= \begin{pmatrix} 5 \\ - 10 \\ 14 \end{pmatrix}

To check your answer:

\overrightarrow{AC} = C - A = \begin{pmatrix} 6 \\ - 12 \\ 9 \end{pmatrix}

, and

3\overrightarrow{AB} = \begin{pmatrix} 6 \\ - 12 \\ 9 \end{pmatrix}

, matching, since

\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AB} + 2\overrightarrow{AB} = 3\overrightarrow{AB}

✓

$\therefore$ $C$ has position vector $\begin{pmatrix} 5 \\ - 10 \\ 14 \end{pmatrix}$ .

Part (b)

2 marks

Let $D = \begin{pmatrix} 1 \\ 2 \\ d \end{pmatrix}$ . Since $\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BD} \right|$ , we first find both vectors.

\overrightarrow{AD} = D - A = \begin{pmatrix} 2 \\ 0 \\ d - 5 \end{pmatrix},\quad\overrightarrow{BD} = D - B = \begin{pmatrix} 0 \\ 4 \\ d - 8 \end{pmatrix}

Equating the squared lengths,

4 + (d - 5)^{2} = 16 + (d - 8)^{2}

\Rightarrow 4 + d^{2} - 10d + 25 = 16 + d^{2} - 16d + 64

\Rightarrow 29 - 10d = 80 - 16d

\Rightarrow 6d = 51

\Rightarrow d = \frac{17}{2}

To check your answer: at

d = 8.5

,

|AD|^{2} = 4 + (8.5 - 5)^{2} = 16.25

and

|BD|^{2} = 16 + (8.5 - 8)^{2} = 16.25

, equal ✓

$\therefore$ $d = \frac{17}{2}$ .

Part (c)

3 marks

Approach

With $D$ now fixed, $\angle\text{ADB}$ is the angle between $\overrightarrow{DA}$ and $\overrightarrow{DB}$ , found from their scalar product.

With $d = \frac{17}{2}$ , $D = \begin{pmatrix} 1 \\ 2 \\ 8.5 \end{pmatrix}$ .

\overrightarrow{DA} = A - D = \begin{pmatrix} - 2 \\ 0 \\ - 3.5 \end{pmatrix},\quad\overrightarrow{DB} = B - D = \begin{pmatrix} 0 \\ - 4 \\ - 0.5 \end{pmatrix}

\overrightarrow{DA} \cdot \overrightarrow{DB} = ( - 2)(0) + (0)( - 4) + ( - 3.5)( - 0.5)

= 1.75

\left| \overrightarrow{DA} \right| = \sqrt{4 + 0 + 12.25} = \sqrt{16.25},\quad\left| \overrightarrow{DB} \right| = \sqrt{0 + 16 + 0.25} = \sqrt{16.25}

\cos\left( \angle\text{ADB} \right) = \frac{1.75}{\sqrt{16.25} \times \sqrt{16.25}}

= \frac{1.75}{16.25}

= \frac{7}{65}

To check your answer:

\frac{7}{65} \approx 0.1077

, and

\cos^{- 1}(0.1077) \approx {83.8}^{\circ}

, a plausible angle for a triangle with

|DA| = |DB|

(isosceles, so the base angles at

A

and

B

are equal, and

\angle\text{ADB}

need not be a special angle) ✓

$\therefore$ $\angle\text{ADB} = \cos^{- 1}\left( \frac{7}{65} \right) = {83.8}^{\circ}$ (1 d.p.).

Part (d)

5 marks

Approach

The centre $P$ of the circle through $A$ , $B$ , $D$ lies in the plane $\text{ABD}$ and is equidistant from all three points. This gives three linear equations: $|PA|^{2} = |PB|^{2}$ , $|PA|^{2} = |PD|^{2}$ , and $P$ lying in the plane of $A$ , $B$ , $D$ .

Let $P = (x,y,z)$ .

Equating $|PA|^{2} = |PB|^{2}$ ,

(x + 1)^{2} + (y - 2)^{2} + (z - 5)^{2} = (x - 1)^{2} + (y + 2)^{2} + (z - 8)^{2}

Expanding (the $x^{2},y^{2},z^{2}$ terms cancel),

2x - 4y - 10z + 30 = - 2x + 4y - 16z + 69

\Rightarrow 4x - 8y + 6z = 39\quad\text{...(1)}

Equating $|PA|^{2} = |PD|^{2}$ , using $D = (1,2,8.5)$ (the $y$ -terms cancel since $A$ and $D$ share $y = 2$ ),

(x + 1)^{2} + (z - 5)^{2} = (x - 1)^{2} + (z - 8.5)^{2}

\Rightarrow 2x - 10z + 26 = - 2x - 17z + 73.25

\Rightarrow 4x + 7z = 47.25\quad\text{...(2)}

For the third equation, $P$ lies in the plane of $A$ , $B$ , $D$ , whose normal is $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AD}$ , where $\overrightarrow{AD} = D - A = \begin{pmatrix} 2 \\ 0 \\ 3.5 \end{pmatrix}$ .

\mathbf{n} = \begin{pmatrix} 2 \\ - 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 2 \\ 0 \\ 3.5 \end{pmatrix}

= \begin{pmatrix} ( - 4)(3.5) - (3)(0) \\ (3)(2) - (2)(3.5) \\ (2)(0) - ( - 4)(2) \end{pmatrix}

= \begin{pmatrix} - 14 \\ - 1 \\ 8 \end{pmatrix}

Using point $A$ , the plane equation $\mathbf{n} \cdot (P - A) = 0$ is

- 14(x + 1) - (y - 2) + 8(z - 5) = 0

\Rightarrow 14x + y - 8z + 52 = 0\quad\text{...(3)}

To check your answer: substituting

B = (1, - 2,8)

into (3):

14(1) - 2 - 8(8) + 52 = 0

✓; substituting

D = (1,2,8.5)

:

14(1) + 2 - 8(8.5) + 52 = 0

✓

Solving (1), (2), (3) simultaneously: from (2), $x = \frac{47.25 - 7z}{4}$ ; from (3), $y = - 52 - 14x + 8z$ . Substituting both into (1),

4x - 8( - 52 - 14x + 8z) + 6z = 39

\Rightarrow 116x - 58z = - 377

Substituting $x = \frac{47.25 - 7z}{4}$ ,

29(47.25 - 7z) - 58z = - 377

\Rightarrow - 261z = - 1747.25

\Rightarrow z = \frac{1747.25}{261} = \frac{241}{36}

Back-substituting into (2), $x = \frac{7}{72}$ , and into (3), $y = \frac{7}{36}$ .

To check your answer: with

P = \left( \frac{7}{72},\frac{7}{36},\frac{241}{36} \right)

, computing

|PA|

,

|PB|

,

|PD|

all give exactly

\frac{65}{24}

(the circumradius), confirming

P

is equidistant from

A

,

B

,

D

✓

$\therefore$ $P = \left( \frac{7}{72},\ \frac{7}{36},\ \frac{241}{36} \right)$ .

2023 H2 Maths Paper 2, Question 3

Part (a)

Part (b)

Part (c)

Part (d)

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