C4 January 2010 Q4

EdexcelOld spec12 marks3D Lines & Planes

4. The line \(l_1\) has vector equation \[\mathbf{r} = \begin{pmatrix}-6\\4\\-1\end{pmatrix} + \lambda\begin{pmatrix}4\\-1\\3\end{pmatrix}\]

and the line \(l_2\) has vector equation \[\mathbf{r} = \begin{pmatrix}-6\\4\\-1\end{pmatrix} + \mu\begin{pmatrix}3\\-4\\1\end{pmatrix}\]

where \(\lambda\) and \(\mu\) are parameters.

The lines \(l_1\) and \(l_2\) intersect at the point \(A\) and the acute angle between \(l_1\) and \(l_2\) is \(\theta\).

(a) Write down the coordinates of \(A\). (1)

(b) Find the value of \(\cos\theta\). (3)

The point \(X\) lies on \(l_1\) where \(\lambda = 4\).

(d) Find the vector \(\overrightarrow{AX}\). (2)

(e) Hence, or otherwise, show that \(\left|\overrightarrow{AX}\right| = 4\sqrt{26}\). (2)

The point \(Y\) lies on \(l_2\). Given that the vector \(\overrightarrow{YX}\) is perpendicular to \(l_1\),

(f) find the length of \(AY\), giving your answer to 3 significant figures. (3)

Scheme	Marks
\(A{:}\ (-6, 4, -1)\) Accept vector forms	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(\begin{pmatrix}4\\-1\\3\end{pmatrix}.\begin{pmatrix}3\\-4\\1\end{pmatrix} = 12 + 4 + 3 = \sqrt{4^2 + (-1)^2 + 3^2}\sqrt{3^2 + (-4)^2 + 1^2}\cos\theta\)	M1 A1
\(\cos\theta = \dfrac{19}{26}\) awrt 0.73	A1
	(3)

Scheme	Marks
\(X{:}\ (10, 0, 11)\) Accept vector forms	B1
	(1)

Mark scheme (d)

Scheme	Marks
\(\overrightarrow{AX} = \begin{pmatrix}10\\0\\11\end{pmatrix} - \begin{pmatrix}-6\\4\\-1\end{pmatrix}\) Either order	M1
\(= \begin{pmatrix}16\\-4\\12\end{pmatrix}\) cao	A1
	(2)

Mark scheme (e)

Scheme	Marks
\(\left\|\overrightarrow{AX}\right\| = \sqrt{16^2 + (-4)^2 + 12^2}\)	M1
\(= \sqrt{416} = \sqrt{16 \times 26} = 4\sqrt{26}\ \ \ast\) Do not penalise if consistent incorrect signs in (d)	A1
	(2)

Scheme	Marks
Use of correct right angled triangle	M1
\(\dfrac{\left\|\overrightarrow{AX}\right\|}{d} = \cos\theta\)	M1
\(d = \dfrac{4\sqrt{26}}{\frac{19}{26}} \approx 27.9\) awrt 27.9	A1
	(3)
	(12 marks)