Given the points A(1, 0, 4), B(2, 3, −1) and C(0, 1, − 2) , find the vector equation of the line ${L_1}$ passing through the points A and B.

The line ${L_2}$ has Cartesian equation $\frac{{x - 1}}{3} = \frac{{y + 2}}{1} = \frac{{z - 1}}{{ - 2}}$.

Show that ${L_1}$ and ${L_2}$ are skew lines.

Find the Cartesian equation of the plane ${\Pi _1}$.

(i)     Find the value of $k$.

(ii)     Find the point of intersection P of the line ${L_3}$ and the plane ${\mathit{\Pi} _2}$.

direction vector $\overrightarrow {{\rm{AB}}}  = \left( \begin{array}{c}1\\3\\ - 5\end{array} \right)$ or $\overrightarrow {{\rm{BA}}}  = \left( \begin{array}{c} - 1\\ - 3\\5\end{array} \right)$     A1

$\boldsymbol{r} = \left( \begin{array}{l}1\\0\\4\end{array} \right) + t\left( \begin{array}{c}1\\3\\ - 5\end{array} \right)$ or $\boldsymbol{r} = \left( \begin{array}{c}2\\3\\ - 1\end{array} \right) + t\left( \begin{array}{c}1\\3\\ - 5\end{array} \right)$ or equivalent     A1

Note:     Do not award final A1 unless ‘$\boldsymbol{r} = {\text{K}}$’ (or equivalent) seen.

     Allow FT on direction vector for final A1.

[2 marks]

both lines expressed in parametric form:

${L_1}$:

$x = 1 + t$

$y = 3t$

$z = 4 - 5t$

${L_2}$:

$x = 1 + 3s$

$y =  - 2 + s$     M1A1

$z =  - 2s + 1$

Notes:     Award M1 for an attempt to convert ${L_2}$ from Cartesian to parametric form.

     Award A1 for correct parametric equations for ${L_1}$ and ${L_2}$.

     Allow M1A1 at this stage if same parameter is used in both lines.

attempt to solve simultaneously for x and y:     M1

$1 + t = 1 + 3s$

$3t =  - 2 + s$

$t =  - \frac{3}{4},{\text{ }}s =  - \frac{1}{4}$     A1

substituting both values back into z values respectively gives $z = \frac{{31}}{4}$

and $z = \frac{3}{2}$ so a contradiction     R1

therefore ${L_1}$ and ${L_1}$ are skew lines     AG

[5 marks]

finding the cross product:

$\left( \begin{array}{c}1\\3\\ - 5\end{array} \right) \times \left( \begin{array}{c}3\\1\\ - 2\end{array} \right)$     (M1)

= – i  – 13j  – 8k     A1

Note:     Accept i  + 13j  + 8k

$- 1(0) - 13(1) - 8( - 2) = 3$     (M1)

$\Rightarrow  - x - 13y - 8z = 3$ or equivalent     A1

[4 marks]

(i)     $(\cos \theta  = )\frac{{\left( \begin{array}{c}k\\1\\ - 1\end{array} \right) \bullet \left( \begin{array}{l}1\\1\\0\end{array} \right)}}{{\sqrt {{k^2} + 1 + 1}  \times \sqrt {1 + 1} }}$     M1

Note:     Award M1 for an attempt to use angle between two vectors formula.

$\frac{{\sqrt 3 }}{2} = \frac{{k + 1}}{{\sqrt {2({k^2} + 2)} }}$     A1

obtaining the quadratic equation

$4{(k + 1)^2} = 6({k^2} + 2)$     M1

${k^2} - 4k + 4 = 0$

${(k - 2)^2} = 0$

$k = 2$     A1

Note:     Award M1A0M1A0 if $\cos 60^\circ$ is used $(k = 0{\text{ or }}k =  - 4)$.

(ii)     $r = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}2\\1\\ - 1\end{array} \right)$

substituting into the equation of the plane ${\Pi _2}$:

$3 + 2\lambda  + \lambda  = 12$     M1

$\lambda  = 3$     A1

point P has the coordinates:

(9, 3, –2)   A1

Notes:     Accept 9i  + 3j  – 2k and $\left( \begin{array}{l}9\\3\\- 2\end{array} \right)$.

     Do not allow FT if two values found for k.

[7 marks]