AS June 2021 Paper 1 Q15

15 Two submarines are travelling on different straight lines.
The two lines are described by the equations

\[\mathbf{r} = \begin{bmatrix}2 \\ -1 \\ 4\end{bmatrix} + \lambda\begin{bmatrix}5 \\ 3 \\ -2\end{bmatrix} \quad \text{and} \quad \frac{x - 5}{4} = \frac{y}{2} = 4 - z\]

(a)

(i) Show that the two lines intersect. [3 marks]

(ii) Find the position vector of the point of intersection. [1 mark]

(b) Tracey says that the submarines will collide because there is a common point on the two lines.

Explain why Tracey is not necessarily correct. [1 mark]

(c) Calculate the acute angle between the lines\[\mathbf{r} = \begin{bmatrix}2 \\ -1 \\ 4\end{bmatrix} + \lambda\begin{bmatrix}5 \\ 3 \\ -2\end{bmatrix} \quad \text{and} \quad \frac{x - 5}{4} = \frac{y}{2} = 4 - z\]

Give your angle to the nearest \(0.1^\circ\) [3 marks]

Mark scheme (a)

Scheme	Marks	AO
(i) Forms one equation by substituting two of \(x = 2 + 5\lambda\), \(y = -1 + 3\lambda\), \(z = 4 - 2\lambda\) into the equation of the second line. Or introduces a second parameter (eg \(\mu\)) to form at least two equations in \(\lambda\) and \(\mu\), e.g. \(2 + 5\lambda = 4\mu + 5\), \(-1 + 3\lambda = 2\mu\), \(4 - 2\lambda = 4 - \mu\)	M1	1.1a
Solves their equation(s) to obtain the correct value of \(\lambda\) (or \(\mu\)).	A1	1.1b
Completes a correct argument to conclude that the lines intersect. eg shows that \(\lambda = -1\) satisfies another equation formed by substituting a different combination of \(x\), \(y\) and \(z\) Or shows that \(\lambda = -1\) and \(\mu = -2\) satisfy a third equation. Or shows that \((-3, -4, 6)\) lies on both lines.	R1	2.1
	(3)
(ii) Obtains \(\begin{bmatrix}-3 \\ -4 \\ 6\end{bmatrix}\) Accept \(-3\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}\) Condone \((-3, -4, 6)\)	B1	1.1b
	(1)

Typical solution

(i)

\[\frac{2 + 5\lambda - 5}{4} = \frac{-1 + 3\lambda}{2}\]\[\begin{aligned}2(5\lambda - 3) &= 4(3\lambda - 1) \\ 10\lambda - 12\lambda &= -4 + 6 \\ -2\lambda &= 2 \\ \lambda &= -1\end{aligned}\]\[\frac{2 + 5\lambda - 5}{4} = 4 - (4 - 2\lambda)\]\[\begin{aligned}5\lambda - 3 &= 4 \times 2\lambda \\ -3 &= 8\lambda - 5\lambda \\ -3 &= 3\lambda \\ \lambda &= -1\end{aligned}\]

same value of \(\lambda\) \(\therefore\) the lines intersect

(ii)

\[\mathbf{r} = \begin{bmatrix}2 \\ -1 \\ 4\end{bmatrix} - 1\begin{bmatrix}5 \\ 3 \\ -2\end{bmatrix} = \begin{bmatrix}-3 \\ -4 \\ 6\end{bmatrix}\]

Scheme	Marks	AO
States a correct explanation.	B1	3.2a
	(1)

Mark scheme (c)

Scheme	Marks	AO
Calculates the scalar product of the two direction vectors. Allow one error in each direction vector.	M1	1.1a
Calculates the product of the direction vector magnitudes.	M1	1.1a
Obtains correct angle. Condone 0.13 (radians). Accept \(7.61^\circ\) or better.	A1	1.1b
	(3)
	(8 marks)

Typical solution

\[\begin{aligned}&\begin{bmatrix}5 \\ 3 \\ -2\end{bmatrix}\cdot\begin{bmatrix}4 \\ 2 \\ -1\end{bmatrix} \\ &= 5 \times 4 + 3 \times 2 + (-2) \times (-1) = 28\end{aligned}\]\[\begin{aligned}&\sqrt{5^2 + 3^2 + (-2)^2} \times \sqrt{4^2 + 2^2 + (-1)^2} \\ &= \sqrt{38 \times 21} = \sqrt{798}\end{aligned}\]\[\cos\theta = \frac{28}{\sqrt{798}}\]\[\theta = 7.6^\circ \text{ (1dp)}\]