2023 H2 Mathematics Paper 2 Question 3 Vectors II: Lines and Planes
Answers
( 5 − 10 14 ) . {{\begin{pmatrix}
5 \\
- 10 \\
14
\end{pmatrix}.}} 5 − 10 14 .
d = 17 2 . {{d = \frac{17}{2}.}} d = 2 17 .
83. 8 ∘ . {{83.8^\circ.}} 83. 8 ∘ .
( 7 72 7 36 241 36 ) . {{\begin{pmatrix}
\frac{7}{72} \\
\frac{7}{36} \\
\frac{241}{36}
\end{pmatrix}.}} 72 7 36 7 36 241 . Full solutions
(a)
B C → = 2 A B → = 2 ( O B → − O A → ) = 2 ( ( 1 − 2 8 ) − ( − 1 2 5 ) ) = ( 4 − 8 6 ) \begin{align*}
\overrightarrow{BC} &= 2 \overrightarrow{AB} \\
&= 2 \left( \overrightarrow{OB} - \overrightarrow{OA} \right) \\
&= 2 \left( \begin{pmatrix}
1 \\
- 2 \\
8
\end{pmatrix} - \begin{pmatrix}
- 1 \\
2 \\
5
\end{pmatrix} \right) \\
&= \begin{pmatrix}
4 \\
- 8 \\
6
\end{pmatrix}
\end{align*} BC = 2 A B = 2 ( OB − O A ) = 2 1 − 2 8 − − 1 2 5 = 4 − 8 6 O C → − O B → = ( 4 − 8 6 ) O C → = ( 4 − 8 6 ) + ( 1 − 2 8 ) = ( 5 − 10 14 ) ■ \begin{align*}
\overrightarrow{OC} - \overrightarrow{OB} &= \begin{pmatrix}
4 \\
- 8 \\
6
\end{pmatrix} \\
\overrightarrow{OC} &= \begin{pmatrix}
4 \\
- 8 \\
6
\end{pmatrix} + \begin{pmatrix}
1 \\
- 2 \\
8
\end{pmatrix} \\
&= \begin{pmatrix}
5 \\
- 10 \\
14
\end{pmatrix} \; \blacksquare
\end{align*} OC − OB OC = 4 − 8 6 = 4 − 8 6 + 1 − 2 8 = 5 − 10 14 ■
(b)
∣ A D → ∣ = ∣ B D → ∣ ∣ O D → − O A → ∣ = ∣ O D → − O B → ∣ ∣ ( 1 2 d ) − ( − 1 2 5 ) ∣ = ∣ ( 1 2 d ) − ( 1 − 2 8 ) ∣ ∣ ( 2 0 d − 5 ) ∣ = ∣ ( 0 4 d − 8 ) ∣ 2 2 + 0 2 + ( d − 5 ) 2 = 0 2 + 4 2 + ( d − 8 ) 2 \begin{align*}
\left| \overrightarrow{AD} \right| &= \left| \overrightarrow{BD} \right| \\
\left| \overrightarrow{OD} - \overrightarrow{OA} \right| &= \left| \overrightarrow{OD} - \overrightarrow{OB} \right| \\
\left| \begin{pmatrix}
1 \\
2 \\
d
\end{pmatrix} - \begin{pmatrix}
- 1 \\
2 \\
5
\end{pmatrix} \right| &= \left| \begin{pmatrix}
1 \\
2 \\
d
\end{pmatrix} - \begin{pmatrix}
1 \\
- 2 \\
8
\end{pmatrix} \right| \\
\left| \begin{pmatrix}
2 \\
0 \\
d - 5
\end{pmatrix} \right| &= \left| \begin{pmatrix}
0 \\
4 \\
d - 8
\end{pmatrix} \right| \\
\sqrt{2^2 + 0^2 + (d - 5)^2} &= \sqrt{ 0^2 + 4^2 + (d - 8)^2} \\
\end{align*} A D O D − O A 1 2 d − − 1 2 5 2 0 d − 5 2 2 + 0 2 + ( d − 5 ) 2 = B D = O D − OB = 1 2 d − 1 − 2 8 = 0 4 d − 8 = 0 2 + 4 2 + ( d − 8 ) 2 d 2 − 10 d + 29 = d 2 − 16 d + 80 − 10 d + 29 = − 16 d + 80 6 d = 51 d = 17 2 ■ \begin{align*}
d^2 - 10 d + 29 &= d^2 - 16 d + 80
\\ - 10 d + 29 &= - 16 d + 80
\\ 6 d &= 51
\\ d &= \frac{17}{2} \; \blacksquare
\end{align*} d 2 − 10 d + 29 − 10 d + 29 6 d d = d 2 − 16 d + 80 = − 16 d + 80 = 51 = 2 17 ■
(c)
D A → = O A → − O D → = ( − 1 2 5 ) − ( 1 2 17 2 ) = ( − 2 0 − 7 2 ) D B → = O B → − O D → = ( 1 − 2 8 ) − ( 1 2 17 2 ) = ( 0 − 4 − 1 2 ) \begin{align*}
\overrightarrow{DA} &= \overrightarrow{OA} - \overrightarrow{OD} \\
&= \begin{pmatrix}
- 1 \\
2 \\
5
\end{pmatrix} - \begin{pmatrix}
1 \\
2 \\
\frac{17}{2}
\end{pmatrix} \\
&= \begin{pmatrix}
- 2 \\
0 \\
- \frac{7}{2}
\end{pmatrix} \\
\overrightarrow{DB} &= \overrightarrow{OB} - \overrightarrow{OD} \\
&= \begin{pmatrix}
1 \\
- 2 \\
8
\end{pmatrix} - \begin{pmatrix}
1 \\
2 \\
\frac{17}{2}
\end{pmatrix} \\
&= \begin{pmatrix}
0 \\
- 4 \\
- \frac{1}{2}
\end{pmatrix}
\end{align*} D A D B = O A − O D = − 1 2 5 − 1 2 2 17 = − 2 0 − 2 7 = OB − O D = 1 − 2 8 − 1 2 2 17 = 0 − 4 − 2 1 D A → ⋅ D B → = ∣ D A → ∣ ∣ D B → ∣ cos ∠ A D B ( − 2 0 − 7 2 ) ⋅ ( 0 − 4 − 1 2 ) = ∣ ( − 2 0 − 7 2 ) ∣ ∣ ( 0 − 4 − 1 2 ) ∣ cos ∠ A D B cos ∠ A D B = 7 4 1 2 65 ( 1 2 65 ) \begin{gather*}
\overrightarrow{DA} \cdot \overrightarrow{DB} = \left| \overrightarrow{DA} \right| \left| \overrightarrow{DB} \right| \cos \angle ADB \\
\begin{pmatrix}
- 2 \\
0 \\
- \frac{7}{2}
\end{pmatrix} \cdot \begin{pmatrix}
0 \\
- 4 \\
- \frac{1}{2}
\end{pmatrix} = \left| \begin{pmatrix}
- 2 \\
0 \\
- \frac{7}{2}
\end{pmatrix} \right| \left| \begin{pmatrix}
0 \\
- 4 \\
- \frac{1}{2}
\end{pmatrix} \right| \cos \angle ADB \\
\cos \angle ADB = \frac{\frac{7}{4}}{\frac{1}{2} \sqrt{65}\left(\frac{1}{2} \sqrt{65}\right)} \\
\end{gather*} D A ⋅ D B = D A D B cos ∠ A D B − 2 0 − 2 7 ⋅ 0 − 4 − 2 1 = − 2 0 − 2 7 0 − 4 − 2 1 cos ∠ A D B cos ∠ A D B = 2 1 65 ( 2 1 65 ) 4 7 ∠ A D B = cos − 1 ( 7 4 1 2 65 ( 1 2 65 ) ) = 83. 8 ∘ ( 1 dp ) ■ \begin{align*}
\angle ADB &= \cos^{-1} \left( \frac{\frac{7}{4}}{\frac{1}{2} \sqrt{65}\left(\frac{1}{2} \sqrt{65}\right)} \right) \\
&= 83.8^\circ \; (\textrm{1 dp}) \; \blacksquare
\end{align*} ∠ A D B = cos − 1 ( 2 1 65 ( 2 1 65 ) 4 7 ) = 83. 8 ∘ ( 1 dp ) ■
(d)
Let M {{M}} M A B . {{AB.}} A B .
Since ∣ A D → ∣ = ∣ B D → ∣ , {{| \overrightarrow{AD} | = | \overrightarrow{BD} |,}} ∣ A D ∣ = ∣ B D ∣ , D {{D}} D M {{M}} M A B . {{AB.}} A B .
Hence the centre of the circle, P , {{P,}} P , M D . {{MD.}} M D .
O M → = O A → + O B → 2 = ( − 1 2 5 ) + ( 1 − 2 8 ) 2 = ( 0 0 13 2 ) \begin{align*}
\overrightarrow{OM} &= \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} \\
&= \frac{\begin{pmatrix}
- 1 \\
2 \\
5
\end{pmatrix} + \begin{pmatrix}
1 \\
- 2 \\
8
\end{pmatrix}}{2} \\
&= \begin{pmatrix}
0 \\
0 \\
\frac{13}{2}
\end{pmatrix}
\end{align*} OM = 2 O A + OB = 2 − 1 2 5 + 1 − 2 8 = 0 0 2 13 M D → = O D → − O M → = ( 1 2 17 2 ) − ( 0 0 13 2 ) = ( 1 2 2 ) \begin{align*}
\overrightarrow{MD} &= \overrightarrow{OD} - \overrightarrow{OM} \\
&= \begin{pmatrix}
1 \\
2 \\
\frac{17}{2}
\end{pmatrix} - \begin{pmatrix}
0 \\
0 \\
\frac{13}{2}
\end{pmatrix} \\
&= \begin{pmatrix}
1 \\
2 \\
2
\end{pmatrix}
\end{align*} M D = O D − OM = 1 2 2 17 − 0 0 2 13 = 1 2 2 l M D : r = ( 1 2 17 2 ) + λ ( 1 2 2 ) {l_{MD}: \mathbf{r} = \begin{pmatrix}
1 \\
2 \\
\frac{17}{2}
\end{pmatrix} + \lambda \begin{pmatrix}
1 \\
2 \\
2
\end{pmatrix}} l M D : r = 1 2 2 17 + λ 1 2 2 O P → = ( 1 + λ 2 + 2 λ 17 2 + 2 λ ) \begin{equation}
\overrightarrow{OP} = \begin{pmatrix}
1 + \lambda \\
2 + 2 \lambda \\
\frac{17}{2} + 2 \lambda
\end{pmatrix}
\end{equation} OP = 1 + λ 2 + 2 λ 2 17 + 2 λ ∣ D P → ∣ = ∣ A P → ∣ ∣ O P → − O D → ∣ = ∣ O A → − O P → ∣ ∣ ( 1 + λ 2 + 2 λ 17 2 + 2 λ ) − ( 1 2 17 2 ) ∣ = ∣ ( 1 + λ 2 + 2 λ 17 2 + 2 λ ) − ( − 1 2 5 ) ∣ ∣ ( λ 2 λ 2 λ ) ∣ = ∣ ( 2 + λ 2 λ 7 2 + 2 λ ) ∣ λ 2 + ( 2 λ ) 2 + ( 2 λ ) 2 = ( 2 + λ ) 2 + ( 2 λ ) 2 + ( 7 2 + 2 λ ) 2 9 λ 2 = 65 4 + 18 λ + 9 λ 2 \begin{gather*}
| \overrightarrow{DP} | = \left| \overrightarrow{AP} \right| \\
\left| \overrightarrow{OP} - \overrightarrow{OD} \right| = \left| \overrightarrow{OA} - \overrightarrow{OP} \right| \\
\left| \begin{pmatrix}
1 + \lambda \\
2 + 2 \lambda \\
\frac{17}{2} + 2 \lambda
\end{pmatrix} - \begin{pmatrix}
1 \\
2 \\
\frac{17}{2}
\end{pmatrix} \right| = \left| \begin{pmatrix}
1 + \lambda \\
2 + 2 \lambda \\
\frac{17}{2} + 2 \lambda
\end{pmatrix} - \begin{pmatrix}
- 1 \\
2 \\
5
\end{pmatrix} \right| \\
\left| \begin{pmatrix}
λ \\
2 λ \\
2 λ
\end{pmatrix} \right| = \left| \begin{pmatrix}
2 + λ \\
2 λ \\
\frac{7}{2} + 2 λ
\end{pmatrix} \right| \\
\sqrt{ λ^2 + (2 λ)^2 + (2 λ)^2 } = \sqrt{ (2 + λ)^2 + (2 λ)^2 + \left(\frac{7}{2} + 2 λ\right)^2 } \\
9 λ^2 = \frac{65}{4} + 18 λ + 9 λ^2
\end{gather*} ∣ D P ∣ = A P OP − O D = O A − OP 1 + λ 2 + 2 λ 2 17 + 2 λ − 1 2 2 17 = 1 + λ 2 + 2 λ 2 17 + 2 λ − − 1 2 5 λ 2 λ 2 λ = 2 + λ 2 λ 2 7 + 2 λ λ 2 + ( 2 λ ) 2 + ( 2 λ ) 2 = ( 2 + λ ) 2 + ( 2 λ ) 2 + ( 2 7 + 2 λ ) 2 9 λ 2 = 4 65 + 18 λ + 9 λ 2 18 λ + 65 4 = 0 18 λ = − 65 4 λ = − 65 72 \begin{gather*}
18 λ + \frac{65}{4} = 0
\\ 18 λ = - \frac{65}{4}
\\ λ = - \frac{65}{72}
\end{gather*} 18 λ + 4 65 = 0 18 λ = − 4 65 λ = − 72 65 Substituting λ = − 65 72 {{λ = - \frac{65}{72}}} λ = − 72 65 ( 1 ) , {{(1),}} ( 1 ) ,
O P → = ( 7 72 7 36 241 36 ) ■ \begin{align*}
\overrightarrow{OP} &= \begin{pmatrix}
\frac{7}{72} \\
\frac{7}{36} \\
\frac{241}{36}
\end{pmatrix} \; \blacksquare
\end{align*} OP = 72 7 36 7 36 241 ■
Question Commentary The first three parts are pretty standard, but the last part is novel and challenging.
For the O Levels A Math syllabus we learnt in coordinate geometry that the centre of circles can be found by
finding the perpendicular bisector of two points on the circle.
However, for a 3D question like this one, finding the
perpendicular bisector is not easy and we will have to observe the earlier result in part (b).
Thereafter, we can use magnitudes to find the final answer as shown above.
An alternative method could be to equate P = ( x , y , z ) {{P = (x,y,z)}} P = ( x , y , z ) ∣ D P → ∣ = ∣ A P → ∣ = ∣ B P → ∣ . {{\left| \overrightarrow{DP} \right| = \left| \overrightarrow{AP} \right| = \left| \overrightarrow{BP} \right|. }} D P = A P = BP .