(a)     Show that a Cartesian equation of the line, \({l_1}\), containing points A(1, −1, 2) and B(3, 0, 3) has the form \(\frac{{x - 1}}{2} = \frac{{y + 1}}{1} = \frac{{z - 2}}{1}\).

(b)     An equation of a second line, \({l_2}\), has the form \(\frac{{x - 1}}{1} = \frac{{y - 2}}{2} = \frac{{z - 3}}{1}\). Show that the lines \({l_1}\) and \({l_2}\) intersect, and find the coordinates of their point of intersection.

(c)     Given that direction vectors of \({l_1}\) and \({l_2}\) are d\(_1\) and d\(_2\) respectively, determine d\(_1 \times \) d\(_2\).

(d)     Show that a Cartesian equation of the plane, \(\prod \), that contains \({l_1}\) and \({l_2}\) is \( - x - y + 3z = 6\).

(e)     Find a vector equation of the line \({l_3}\) which is perpendicular to the plane \(\prod \) and passes through the point T(3, 1, −4).

(f)     (i)     Find the point of intersection of the line \({l_3}\) and the plane \(\prod \).

         (ii)     Find the coordinates of \({{\text{T}}}'\), the reflection of the point T in the plane \(\prod \).

         (iii)     Hence find the magnitude of the vector \(\overrightarrow {{\text{TT}}'} \).

(a)     identifies a direction vector e.g. \(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}}
  2 \\
  1 \\
  1
\end{array}} \right)\) or \(\overrightarrow {{\text{BA}}}  = \left( {\begin{array}{*{20}{c}}
  { - 2} \\
  { - 1} \\
  { - 1}
\end{array}} \right)\)     A1

identifies the point (1, –1, 2)     A1

line \({l_1}:{\text{ }}\frac{{x - 1}}{2} = \frac{{y + 1}}{1} = \frac{{z - 2}}{1}\)     AG

[2 marks]

 

(b)     \(r = \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  2
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  2 \\
  1 \\
  1
\end{array}} \right)\)     \(r = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  3
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  1
\end{array}} \right)\)

\(1 + 2\lambda = 1 + \mu ,{\text{ }} - 1 + \lambda = 2 + 2\mu ,{\text{ }}2 + \lambda = 3 + \mu \)     (M1)

equating two of the three equations gives \(\lambda = - 1\) and \(\mu = - 2\)     A1A1

check in the third equation

satisfies third equation therefore the lines intersect     R1

therefore coordinates of intersection are (–1, –2, 1)     A1

[5 marks]

 

(c)     d\(_1\) = 2i + j + k, d\(_2\) = i + 2j + k     A1

d\(_1\) \( \times \) d\(_2\) \( = \left| {\begin{array}{*{20}{c}}
  \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
  2&1&1 \\
  1&2&1
\end{array}} \right| = \) –i \(-\) j \(+\) 3k     M1A1

Note: Accept scalar multiples of above vectors.

 

[3 marks]

 

(d)     equation of plane is \( - x - y + 3z = k\)     M1A1

contains (1, 2, 3) \(\left( {{\text{or }}( - 1,{\text{ }} - 2,{\text{ }}1){\text{ or }}(1,{\text{ }} - 1,{\text{ }}2)} \right){\text{ }}\therefore k = - 1 - 2 + 3 \times 3 = 6\)     A1

\( - x - y + 3z = 6\)     AG

[3 marks]

 

(e)     direction vector of the perpendicular line is \(\left( {\begin{array}{*{20}{c}}
  { - 1} \\
  { - 1} \\
  3
\end{array}} \right)\)     (M1)

\(r = \left( {\begin{array}{*{20}{c}}
  3 \\
  1 \\
  { - 4}
\end{array}} \right) + m\left( {\begin{array}{*{20}{c}}
  { - 1} \\
  { - 1} \\
  3
\end{array}} \right)\)     A1

Note: Award A0 if r omitted.

 

[2 marks]

 

(f)     (i)     find point where line meets plane

\( - (3 - m) - (1 - m) + 3( - 4 + 3m) = 6\)     M1

m = 2     A1

point of intersection is (1, –1, 2)     A1

 

(ii)     for \({\text{T}}', m = 4\)     (M1)

so \({\text{T}}' = \) (\( - 1\), \( - 3\), \(8\))     A1

 

(iii)     \(\overrightarrow {{\text{TT}'}} = \sqrt {{{(3 + 1)}^2} + {{(1 + 3)}^2} + {{( - 4 - 8)}^2}} \)     (M1)

\( = \sqrt {176} \,\,\,\,\,( = 4\sqrt {11} )\)     A1

[7 marks]

 

Total [22 marks]

This question was done very well by many students. The common errors were using the same variable for line 2 and in stating the vectors in b) were not parallel and therefore the lines did intersect. Many students did not check the solution in order to establish this.

When required to give the equation of the line in e) many did not state it as an equation, let alone a vector equation. 

The difference between position vectors and coordinates was not clear on many papers.

In f) many used inefficient techniques that were time consuming to find the point of reflection.