AS June 2020 Paper 1 Q13

AQACurrent spec9 marks3D Lines & Planes

13 Line \(l_1\) has equation

\[\frac{x - 2}{3} = \frac{1 - 2y}{4} = -z\]

and line \(l_2\) has equation

\[\mathbf{r} = \begin{bmatrix}-7 \\ 4 \\ -2\end{bmatrix} + \mu\begin{bmatrix}12 \\ a + 3 \\ 2b\end{bmatrix}\]

(a) In the case when \(l_1\) and \(l_2\) are parallel, show that \(a = -11\) and find the value of \(b\). [4 marks]

(b) In a different case, the lines \(l_1\) and \(l_2\) intersect at exactly one point, and the value of \(b\) is 3

Find the value of \(a\). [5 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains the correct direction vector for line \(l_1\).	B1	1.1b
Selects a method to find \(a\) and \(b\) by equating (multiples of) their \(l_1\) direction vector and \(\begin{bmatrix}12 \\ a + 3 \\ 2b\end{bmatrix}\).	M1	3.1a
Equates all components of their vectors and finds the multiplier ‘\(p\)’.	M1	1.1a
Shows correctly that \(a = -11\) and obtains \(b = -2\).	R1	2.1

Typical solution

direction of \(l_1\) is \(\begin{bmatrix}3 \\ -2 \\ -1\end{bmatrix}\)

\[\begin{bmatrix}12 \\ a + 3 \\ 2b\end{bmatrix} = p\begin{bmatrix}3 \\ -2 \\ -1\end{bmatrix}\]\[12 = 3p\]\[p = 4\]\[a + 3 = -2p \quad \text{and} \quad 2b = -p\]\[a + 3 = -8 \quad \text{and} \quad 2b = -4\]\[a = -11 \quad \text{and} \quad b = -2\]

Mark scheme (b)

Scheme	Marks	AO
Selects a method to find the intersection of \(l_1\) and \(l_2\) by equating vector equations of the two lines or by substituting components of \(l_2\) in the Cartesian equation of \(l_1\).	M1	3.1a
Forms an equation in \(\mu\), or writes at least two simultaneous equations in \(\mu\) and another parameter. Allow one arithmetic error.	M1	1.1a
Obtains the correct value of \(\mu\).	A1	1.1b
Forms an equation in \(a\) by substituting their value of \(\mu\) into an appropriate equation, from the intersection of the two lines.	M1	1.1a
Obtains the correct value of \(a\).	A1	1.1b
	(9 marks)

Typical solution

\[\frac{x - 2}{3} = \frac{y - \frac{1}{2}}{-2} = \frac{z - 0}{-1}\]\[r = \begin{bmatrix}2 \\ \frac{1}{2} \\ 0\end{bmatrix} + \lambda\begin{bmatrix}3 \\ -2 \\ -1\end{bmatrix}\]\[\begin{bmatrix}-7 + 12\mu \\ 4 + \mu(a + 3) \\ -2 + 6\mu\end{bmatrix} = \begin{bmatrix}2 + 3\lambda \\ 0.5 - 2\lambda \\ -\lambda\end{bmatrix}\]\[\begin{aligned}-7 + 12\mu &= 2 + 3\lambda \\ 4 + \mu a + 3\mu &= 0.5 - 2\lambda \\ -2 + 6\mu &= -\lambda\end{aligned}\]\[-7 + 12\mu = 2 + 3(2 - 6\mu)\]\[\mu = \tfrac{1}{2}\]\[\lambda = 2 - 6 \times \tfrac{1}{2} = -1\]\[4 + \tfrac{1}{2}a + 3 \times \tfrac{1}{2} = \tfrac{1}{2} - 2 \times -1\]\[a = -6\]