C4 June 2011 Q6

EdexcelOld spec14 marks3D Lines & Planes

6. With respect to a fixed origin \(O\), the lines \(l_1\) and \(l_2\) are given by the equations\[l_1:\ \mathbf{r} = \begin{pmatrix} 6 \\ -3 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}, \qquad l_2:\ \mathbf{r} = \begin{pmatrix} -5 \\ 15 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix},\]where \(\lambda\) and \(\mu\) are scalar parameters.

(a) Show that \(l_1\) and \(l_2\) meet and find the position vector of their point of intersection \(A\). (6)

(b) Find, to the nearest \(0.1^\circ\), the acute angle between \(l_1\) and \(l_2\). (3)

The point \(B\) has position vector \(\begin{pmatrix} 5 \\ -1 \\ 1 \end{pmatrix}\).

(d) Find the shortest distance from \(B\) to the line \(l_2\), giving your answer to 3 significant figures. (4)

Mark scheme (a)

Scheme	Marks
\(\mathbf{i}:\quad 6 - \lambda = -5 + 2\mu\) \(\mathbf{j}:\quad -3 + 2\lambda = 15 - 3\mu\) Any two equations	M1
leading to \(\lambda = 3,\ \mu = 4\)	M1 A1
\(\mathbf{r} = \begin{pmatrix} 6 \\ -3 \\ -2 \end{pmatrix} + 3\begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 7 \end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix} -5 \\ 15 \\ 3 \end{pmatrix} + 4\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 7 \end{pmatrix}\)	M1 A1
\(\mathbf{k}:\quad \text{LHS} = -2 + 3(3) = 7, \quad \text{RHS} = 3 + 4(1) = 7\) (As LHS = RHS, lines intersect)	B1
	(6)

Notes

Alternatively for B1, showing that \(\lambda = 3\) and \(\mu = 4\) both give \(\begin{pmatrix} 3 \\ 3 \\ 7 \end{pmatrix}\)

In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.

Mark scheme (b)

Scheme	Marks
\(\begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}.\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = -2 - 6 + 3 = \sqrt{14}\sqrt{14}\cos\theta \qquad (\theta \approx 110.92^\circ)\)	M1 A1
Acute angle is \(69.1^\circ\) awrt 69.1	A1
	(3)

Mark scheme (c)

Scheme	Marks
\(\mathbf{r} = \begin{pmatrix} 6 \\ -3 \\ -2 \end{pmatrix} + 1\begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ 1 \end{pmatrix} \qquad (\Rightarrow B \text{ lies on } l_1)\)	B1
	(1)

Diagram: lines l1 and l2 meeting at A at angle theta, with B on l1 and the perpendicular distance d from B to l2 — Mark scheme (d)

Scheme	Marks
Let \(d\) be shortest distance from \(B\) to \(l_2\)
\(\overrightarrow{AB} = \begin{pmatrix} 5 \\ -1 \\ 1 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 7 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ -6 \end{pmatrix}\)	M1
\(\left\|\overrightarrow{AB}\right\| = \sqrt{\left(2^2 + (-4)^2 + (-6)^2\right)} = \sqrt{56}\) awrt 7.5	A1
\(\dfrac{d}{\sqrt{56}} = \sin\theta\)	M1
\(d = \sqrt{56}\sin 69.1^\circ \approx 6.99\) awrt 6.99	A1
	(4)
	(14 marks)