2022 H2 Mathematics Paper 1 Question 11 Vectors II: Lines and Planes
Answers 5 x + y + 2 z = 2560. {5 x + y + 2 z = 2560.} 5 x + y + 2 z = 2560. ( 512 , 44 , − 22 ) . {\left( 512, 44, - 22 \right).} ( 512 , 44 , − 22 ) . Full solutions
(a) P Q → = O Q → − O P → = ( 200 20 − 15 ) − ( 1136 92 p ) = ( − 936 − 72 − 15 − p ) \begin{align*}
\overrightarrow{PQ} &= \overrightarrow{OQ} - \overrightarrow{OP}
\\ &= \begin{pmatrix}
200 \\
20 \\
- 15
\end{pmatrix} - \begin{pmatrix}
1136 \\
92 \\
p
\end{pmatrix}
\\ &= \begin{pmatrix}
- 936 \\
- 72 \\
- 15 - p
\end{pmatrix}
\end{align*} PQ = OQ − OP = 200 20 − 15 − 1136 92 p = − 936 − 72 − 15 − p ∣ P Q → ∣ = 939 881505 + 30 p + p 2 = 939 881505 + 30 p + p 2 = 881721 p 2 + 30 p − 216 = 0 ( p + 36 ) ( p − 6 ) = 0 p = − 36 or p = 6 \begin{gather*}
\left| \overrightarrow{PQ} \right| = 939
\\ \sqrt{881505 + 30 p + p^2} = 939
\\ 881505 + 30 p + p^2 = 881721
\\ p^2 + 30 p - 216 = 0
\\ (p + 36)(p - 6) = 0
\\ p = - 36 \quad \textrm{ or } \quad p = 6
\end{gather*} PQ = 939 881505 + 30 p + p 2 = 939 881505 + 30 p + p 2 = 881721 p 2 + 30 p − 216 = 0 ( p + 36 ) ( p − 6 ) = 0 p = − 36 or p = 6
Since
P {P} P is below
Q , {Q,} Q , p = − 36. ■ {p=- 36. \; \blacksquare} p = − 36. ■
(b) d 1 = ( 500 200 − 70 ) − ( 400 600 − 20 ) = ( 100 − 400 − 50 ) = 50 ( 2 − 8 − 1 ) d 2 = ( 600 − 340 − 50 ) − ( 400 600 − 20 ) = ( 200 − 940 − 30 ) = 10 ( 20 − 94 − 3 ) n ′ = ( 2 − 8 − 1 ) × ( 20 − 94 − 3 ) = ( − 70 − 14 − 28 ) = − 14 ( 5 1 2 ) \begin{align*}
\mathbf{d_1} &= \begin{pmatrix}
500 \\
200 \\
- 70
\end{pmatrix} - \begin{pmatrix}
400 \\
600 \\
- 20
\end{pmatrix}
\\ &= \begin{pmatrix}
100 \\
- 400 \\
- 50
\end{pmatrix}
\\ &= 50 \begin{pmatrix}
2 \\
- 8 \\
- 1
\end{pmatrix}
\\ \mathbf{d_2} &= \begin{pmatrix}
600 \\
- 340 \\
- 50
\end{pmatrix} - \begin{pmatrix}
400 \\
600 \\
- 20
\end{pmatrix}
\\ &= \begin{pmatrix}
200 \\
- 940 \\
- 30
\end{pmatrix}
\\ &= 10 \begin{pmatrix}
20 \\
- 94 \\
- 3
\end{pmatrix}
\\ \mathbf{n'} &= \begin{pmatrix}
2 \\
- 8 \\
- 1
\end{pmatrix} \times \begin{pmatrix}
20 \\
- 94 \\
- 3
\end{pmatrix}
\\ &= \begin{pmatrix}
- 70 \\
- 14 \\
- 28
\end{pmatrix}
\\ &= - 14 \begin{pmatrix}
5 \\
1 \\
2
\end{pmatrix}
\end{align*} d 1 d 2 n ′ = 500 200 − 70 − 400 600 − 20 = 100 − 400 − 50 = 50 2 − 8 − 1 = 600 − 340 − 50 − 400 600 − 20 = 200 − 940 − 30 = 10 20 − 94 − 3 = 2 − 8 − 1 × 20 − 94 − 3 = − 70 − 14 − 28 = − 14 5 1 2 r ⋅ n = a ⋅ n r ⋅ ( 5 1 2 ) = ( 400 600 − 20 ) ⋅ ( 5 1 2 ) r ⋅ ( 5 1 2 ) = 2560 \begin{align*}
\mathbf{r}\cdot\mathbf{n} &= \mathbf{a} \cdot \mathbf{n}
\\ \mathbf{r} \cdot \begin{pmatrix}
5 \\
1 \\
2
\end{pmatrix} &= \begin{pmatrix}
400 \\
600 \\
- 20
\end{pmatrix} \cdot \begin{pmatrix}
5 \\
1 \\
2
\end{pmatrix}
\\ \mathbf{r} \cdot \begin{pmatrix}
5 \\
1 \\
2
\end{pmatrix} &= 2560
\end{align*} r ⋅ n r ⋅ 5 1 2 r ⋅ 5 1 2 = a ⋅ n = 400 600 − 20 ⋅ 5 1 2 = 2560
Cartesian equation of plane is
5 x + y + 2 z = 2560 ■ 5 x + y + 2 z = 2560 \; \blacksquare 5 x + y + 2 z = 2560 ■
(c) P Q → = ( 1136 92 − 15 − ( − 36 ) ) = ( − 936 − 72 21 ) = 3 ( − 312 − 24 7 ) \begin{align*}
\overrightarrow{PQ} &= \begin{pmatrix}
1136 \\
92 \\
-15-(- 36)
\end{pmatrix}
\\ &= \begin{pmatrix}
- 936 \\
- 72 \\
21
\end{pmatrix}
\\ &= 3 \begin{pmatrix}
- 312 \\
- 24 \\
7
\end{pmatrix}
\end{align*} PQ = 1136 92 − 15 − ( − 36 ) = − 936 − 72 21 = 3 − 312 − 24 7 l P Q : r = ( 1136 92 − 36 ) + λ ( − 312 − 24 7 ) l_{PQ}: \mathbf{r} = \begin{pmatrix}
1136 \\
92 \\
- 36
\end{pmatrix} + \lambda \begin{pmatrix}
- 312 \\
- 24 \\
7
\end{pmatrix} l PQ : r = 1136 92 − 36 + λ − 312 − 24 7
Substituting equation of
l {l} l into equation of plane,
( 1136 − 312 λ 92 − 24 λ − 36 + 7 λ ) ⋅ ( 5 1 2 ) = 2560 5 ( 1136 − 312 λ ) + ( 92 − 24 λ ) + 2 ( − 36 + 7 λ ) = 2560 λ = 2 \begin{gather*}
\begin{pmatrix}
1136 - 312 \lambda \\
92 - 24 \lambda \\
- 36 + 7 \lambda
\end{pmatrix} \cdot \begin{pmatrix}
5 \\
1 \\
2
\end{pmatrix} = 2560
\\ 5(1136 - 312 \lambda) + (92 - 24 \lambda) + 2(- 36 + 7 \lambda) = 2560
\\ \lambda = 2
\end{gather*} 1136 − 312 λ 92 − 24 λ − 36 + 7 λ ⋅ 5 1 2 = 2560 5 ( 1136 − 312 λ ) + ( 92 − 24 λ ) + 2 ( − 36 + 7 λ ) = 2560 λ = 2
Let
X {X} X denote the point of intersection.
Substituting
λ = 2 {\lambda = 2} λ = 2 into equation of line,
O X → = ( 1136 92 − 36 ) + 2 ( − 312 − 24 7 ) = ( 512 44 − 22 ) \begin{align*}
\overrightarrow{OX} &= \begin{pmatrix}
1136 \\
92 \\
- 36
\end{pmatrix} + 2 \begin{pmatrix}
- 312 \\
- 24 \\
7
\end{pmatrix}
\\ &= \begin{pmatrix}
512 \\
44 \\
- 22
\end{pmatrix}
\end{align*} OX = 1136 92 − 36 + 2 − 312 − 24 7 = 512 44 − 22
Hence the coordinates of the point where the pipeline meets the rock is
( 512 , 44 , − 22 ) ■ {\left( 512, 44, - 22 \right) \; \blacksquare} ( 512 , 44 , − 22 ) ■
(d) The horizontal can be modelled as a plane with normal vector
( 0 0 1 ) {\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}} 0 0 1
Let
θ {\theta} θ denote the angle between the pipeline and the horizontal
sin θ = ∣ d ⋅ n ∣ ∣ d ∣ ∣ n ∣ = ∣ ( − 312 − 24 7 ) ⋅ ( 0 0 1 ) ∣ ∣ ( − 312 − 24 7 ) ∣ ∣ ( 0 0 1 ) ∣ = 7 97969 θ = 1. 3 ∘ (1 dp) ■ \begin{align*}
\sin \theta &= \frac{\left| \mathbf{d} \cdot \mathbf{n} \right|}{\left| \mathbf{d} \right|\left| \mathbf{n} \right|}
\\ &= \frac{\left| \begin{pmatrix}
- 312 \\
- 24 \\
7
\end{pmatrix} \cdot \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix} \right|}{\left| \begin{pmatrix}
- 312 \\
- 24 \\
7
\end{pmatrix} \right|\left| \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix} \right|}
\\ &= \frac{7}{\sqrt{97969}}
\\ \theta &= 1.3^\circ \textrm{ (1 dp)} \; \blacksquare
\end{align*} sin θ θ = ∣ d ∣ ∣ n ∣ ∣ d ⋅ n ∣ = − 312 − 24 7 0 0 1 − 312 − 24 7 ⋅ 0 0 1 = 97969 7 = 1. 3 ∘ (1 dp) ■ Question Commentary
Like in 2019 paper 1 question 12 , 2018 paper 2 question 3 and many others,
this vectors question is set in a real world context. The trick will then be to translate the information provided
into the vectors/lines/planes concepts that we learn in the syllabus.
For part (a), the length comes from the concept of vector magnitude. Part (b) involves finding the normal vector of the plane
by taking a cross/vector product. Part (c) involves finding the point of intersection between a line (the pipeline) and
a plane (the rock).
Finally, part (d) involves finding the angle between a line/vector and a plane. This part could be tricky
as it may not be immediately clear what the horizontal refers to, but it is the plane with the z {z} z