Challenging +1.2 This is a standard Further Maths question on finding the shortest distance between skew lines. While it requires converting between line forms, finding direction vectors, and applying the skew lines distance formula (using scalar triple product), these are well-practiced techniques at FM level. The 8 marks reflect multiple computational steps rather than conceptual difficulty. Slightly above average due to being FM content and requiring careful vector manipulation.
The line \(L_1\) has equation
$$\frac{x-2}{3} = \frac{y+4}{8} = \frac{4z-5}{5}$$
The line \(L_2\) has equation
$$\left(\mathbf{r} - \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}\right) \times \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} = \mathbf{0}$$
Find the shortest distance between the two lines, giving your answer to three significant figures.
[8 marks]
Question 11:
11 | Obtains a position vector
of a point on L .
1 | AO2.5 | B1 | 2 3 2 12
r = −4 +µ 8 = −4 +µ 32
1
5 4 5 4 5 4 5
−2 2
r = 0 +λ 1
2
3 3
2 −2 4
−4 − 0 = −4
5 4 3 −7 4
12 2 91
32 × 1 = −26
5 3 −52
4
91
−4
� �.�−26�
7
− −52 43
4
=
1 3 √ 6 9 =√56.197659...=5.18 3sf
Obtains a direction vector
for L
1
ISW | AO1.1b | B1
Obtains a position vector
of a point on L and a
2
direction vector for L
2 | AO2.5 | B1
Obtains a correct vector
between the two lines.
Follow through from their
position vectors | AO1.1b | B1F
Calculates the vector
product of their two
direction vectors
or
calculates the scalar
product of the general
vector between the lines
with both direction
vectors to obtain a pair of
simultaneous equations | AO3.1a | M1
Obtains the correct vector
product
or
obtains a correct pair of
simultaneous equations | AO1.1b | A1
Uses their vector product
or the solutions to their
simultaneous equations
to calculate the shortest
distance between the two
lines | AO3.1a | M1
Obtains the correct
shortest distance.
Allow awrt 5.18
Accept exact answer | AO1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
The line $L_1$ has equation
$$\frac{x-2}{3} = \frac{y+4}{8} = \frac{4z-5}{5}$$
The line $L_2$ has equation
$$\left(\mathbf{r} - \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}\right) \times \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} = \mathbf{0}$$
Find the shortest distance between the two lines, giving your answer to three significant figures.
[8 marks]
\hfill \mbox{\textit{AQA Further Paper 2 2019 Q11 [8]}}