AS June 2018 Paper 1 Q19

AQACurrent spec8 marks3D Lines & Planes

19 A theme park has two zip wires.

Sarah models the two zip wires as straight lines using coordinates in metres.

The ends of one wire are located at \((0, 0, 0)\) and \((0, 100, -20)\)

The ends of the other wire are located at \((10, 0, 20)\) and \((-10, 100, -5)\)

(a) Use Sarah’s model to find the shortest distance between the zip wires. [7 marks]

(b) State one way in which Sarah’s model could be refined. [1 mark]

Mark scheme (a)

Scheme	Marks	AO
Finds a direction vector for the second wire. Condone one error.	M1	3.4
Writes, in terms of a parameter, the position vector (or coordinates) of one point on each of the two lines. Condone use of same parameter.	M1	3.1a
Obtains, in terms of two parameters, a correct vector between the two lines.	A1	1.1b
Sets up two scalar products for their \(\boldsymbol{r_2} - \boldsymbol{r_1}\) and their valid direction vectors.	M1	1.1a
Obtains correct parameter values.	A1	1.1b
Uses full method for required distance	M1	1.1b
Obtains correct distance to 2, 3 or 4 significant figures with correct units. Accept 1 significant figure if full method shown.	A1	3.2a

Typical solution

\[\text{Direction vector for 2}^{\text{nd}}\text{ wire} = \begin{pmatrix}10 \\ 0 \\ 20\end{pmatrix} - \begin{pmatrix}-10 \\ 100 \\ -5\end{pmatrix}\]\[\boldsymbol{r_1} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} + \lambda\begin{pmatrix}0 \\ 100 \\ -20\end{pmatrix} \quad \text{and} \quad \boldsymbol{r_2} = \begin{pmatrix}10 \\ 0 \\ 20\end{pmatrix} + \mu\begin{pmatrix}20 \\ -100 \\ 25\end{pmatrix}\]\[\boldsymbol{r_2} - \boldsymbol{r_1} = \begin{pmatrix}10 + 20\mu \\ -100\mu - 100\lambda \\ 20 + 25\mu + 20\lambda\end{pmatrix}\]\[\begin{pmatrix}10 + 20\mu \\ -100\mu - 100\lambda \\ 20 + 25\mu + 20\lambda\end{pmatrix}\cdot\begin{pmatrix}0 \\ 100 \\ -20\end{pmatrix} = 0 \quad \text{and} \quad \begin{pmatrix}10 + 20\mu \\ -100\mu - 100\lambda \\ 20 + 25\mu + 20\lambda\end{pmatrix}\cdot\begin{pmatrix}20 \\ -100 \\ 25\end{pmatrix} = 0\]\[-10000\mu - 10000\lambda - 400 - 500\mu - 400\lambda = 0 \quad \text{and}\]\[200 + 400\mu + 10000\mu + 10000\lambda + 500 + 625\mu + 500\lambda = 0\]\[11025\mu + 10500\lambda + 700 = 0 \quad \text{and} \quad 11025\mu + 10920\lambda + 420 = 0\]\[\lambda = \frac{2}{3} \quad \text{and} \quad \mu = -\frac{44}{63}\]\[\sqrt{\left(10 + 20\left(-\tfrac{44}{63}\right)\right)^2 + \left(-100\left(-\tfrac{44}{63}\right) - 100\left(\tfrac{2}{3}\right)\right)^2 + \left(20 + 25\left(-\tfrac{44}{63}\right) + 20\left(\tfrac{2}{3}\right)\right)^2}\]

= 16.7 metres

ALT 19(a)

Scheme	Marks	AO
Finds a direction vector for the second wire.	M1	3.4
Forms two equations for a perpendicular vector	M1	3.1a
Obtains two correct equations for a perpendicular vector	A1	1.1b
Obtains a correct normal vector	A1	1.1b
Finds the unit normal vector	M1	1.1a
Uses full method for required distance	M1	1.1b
Obtains correct distance to 2, 3 or 4 significant figures with correct units. Accept 1 significant figure if full method shown.	A1	3.2a

Typical solution

\[\text{Direction vector for 2}^{\text{nd}}\text{ wire} = \begin{pmatrix}10 \\ 0 \\ 20\end{pmatrix} - \begin{pmatrix}-10 \\ 100 \\ -5\end{pmatrix}\]

Let \(\begin{pmatrix}x \\ y \\ z\end{pmatrix}\) be a vector perpendicular to both wires.

\[\therefore \begin{pmatrix}0 \\ 100 \\ -20\end{pmatrix}\cdot\begin{pmatrix}x \\ y \\ z\end{pmatrix} = 0 \quad \textbf{and} \quad \begin{pmatrix}20 \\ -100 \\ 25\end{pmatrix}\cdot\begin{pmatrix}x \\ y \\ z\end{pmatrix} = 0\]\[\Rightarrow 100y - 20z = 0 \quad \textbf{and} \quad 20x - 100y + 25z = 0\]\[\Rightarrow z = 5y \quad \textbf{and} \quad x = -1.25y\]

\(\therefore\) perpendicular vector is \(\begin{pmatrix}-1.25y \\ y \\ 5y\end{pmatrix}\)

\(\Rightarrow\) unit perpendicular vector is \(\begin{pmatrix}-1.25 \\ 1 \\ 5\end{pmatrix} \div \sqrt{(-1.25)^2 + 1^2 + 5^2}\)

a vector from 1^st line to 2^nd line is \(\begin{pmatrix}10 \\ 0 \\ 20\end{pmatrix} - \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix}10 \\ 0 \\ 20\end{pmatrix}\)

\(\therefore\) distance between lines is \(\begin{pmatrix}10 \\ 0 \\ 20\end{pmatrix}\cdot\begin{pmatrix}-1.25 \\ 1 \\ 5\end{pmatrix} \div \frac{21}{4}\)

= 16.7 metres

Scheme	Marks	AO
Suggests an improvement to the model. Do not condone criticisms without refinements.	B1	3.5c
	(8 marks)