(a)     Write the vector equations of the following lines in parametric form.

$${r_1} = \left( {\begin{array}{*{20}{c}}   3 \\   2 \\   7 \end{array}} \right) + m\left( {\begin{array}{*{20}{c}}   2 \\   { - 1} \\   2 \end{array}} \right)$$

$${r_2} = \left( {\begin{array}{*{20}{c}}   1 \\   4 \\   2 \end{array}} \right) + n\left( {\begin{array}{*{20}{c}}   4 \\   { - 1} \\   1 \end{array}} \right)$$

(b)     Hence show that these two lines intersect and find the point of intersection, A.

(c)     Find the Cartesian equation of the plane $\prod$ that contains these two lines.

(d)     Let B be the point of intersection of the plane $\prod$ and the line${r} = \left( {\begin{array}{*{20}{c}}   { - 8} \\   { - 3} \\   0 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}   3 \\   8 \\   2 \end{array}} \right)$.

Find the coordinates of B.

(e)     If C is the mid-point of AB, find the vector equation of the line perpendicular to the plane $\prod$ and passing through C.

(a)     $x = 3 + 2m$

$y = 2 - m$

$z = 7 + 2m$     A1

$x = 1 + 4n$

$y = 4 - n$

$z = 2 + n$     A1

[2 marks]

(b)     $3 + 2m = 1 + 4n \Rightarrow 2m - 4n = - 2{\text{ }}({\text{i}})$

$2 - m = 4 - n \Rightarrow m - n = - 2{\text{ }}({\text{ii}})$     M1

$7 + 2m = 2 + n \Rightarrow 2m - n = - 5{\text{ }}({\text{iii}})$

$({\text{iii}}) - ({\text{ii}}) \Rightarrow m = - 3$     A1

$\Rightarrow n = - 1$     A1

Substitute in (i), –6 + 4 = –2 . Hence lines intersect.     R1

Point of intersection A is (–3, 5,1)     A1

[5 marks]

(c)     $\left| {\begin{array}{*{20}{c}}   i&j&k \\   2&{ - 1}&2 \\   4&{ - 1}&1 \end{array}} \right| = \left( {\begin{array}{*{20}{c}}   1 \\   6 \\   2 \end{array}} \right)$     M1A1

$r \cdot \left( {\begin{array}{*{20}{c}}   1 \\   6 \\   2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   3 \\   2 \\   7 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}   1 \\   6 \\   2 \end{array}} \right)$     (M1)

$r \cdot \left( {\begin{array}{*{20}{c}}   1 \\   6 \\   2 \end{array}} \right) = 29$

x + 6y + 2z = 29     A1

Note: Award M1A0 if answer is not in Cartesian form.

[4 marks]

(d)     $x =  - 8 + 3\lambda$

$y = - 3 + 8\lambda$     (M1)

$z = 2\lambda$

Substitute in equation of plane.

$- 8 + 3\lambda - 18 + 48\lambda + 4\lambda = 29$     M1

$55\lambda = 55$

$\lambda = 1$     A1

Coordinates of B are (–5, 5, 2)     A1

[4 marks]

(e)     Coordinates of C are $\left( { - 4,{\text{ 5, }}\frac{3}{2}} \right)$     (A1)

$r = \left( {\begin{array}{*{20}{c}}   { - 4} \\   5 \\   {\frac{3}{2}} \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}   1 \\   6 \\   2 \end{array}} \right)$     M1A1

Note: Award M1A0 unless candidate writes r = or $\left( {\begin{array}{*{20}{c}}   x \\   y \\   z \end{array}} \right)$

[3 marks]

Total [18 marks]

Most candidates found this question to their liking and many correct solutions were seen. In (b), some candidates solved two equations for m and n but then failed to show that these values satisfied the third equation. In (e), some candidates used an incorrect formula to determine the coordinates of the mid-point of AB .