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]