Show that \(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right)\) .

The line \({L_1}\) may be represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  3 \\
  { - 3} \\
  8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
  1 \\
  { - 2} \\
  3
\end{array}} \right)\) .

(i)     What information does the vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
8
\end{array}} \right)\) give about \({L_1}\) ?

(ii)    Write down another vector representation for \({L_1}\) using \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
8
\end{array}} \right)\) .

The point \({\text{T}}( - 1{\text{, }}5{\text{, }}p)\) lies on \({L_1}\) .

Find the value of \(p\) .

The point T also lies on \({L_2}\) with equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 3}\\
9\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
q
\end{array}} \right)\) .

Show that \(q = - 3\) .

Let \(\theta \) be the obtuse angle between \({L_1}\) and \({L_2}\) . Calculate the size of \(\theta \) .

evidence of correct approach     A1

e.g. \(\overrightarrow {{\rm{PQ}}}  = \overrightarrow {{\rm{OQ}}} - \overrightarrow {{\rm{OP}}} \) , \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
8
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
2\\
{ - 1}\\
5
\end{array}} \right)\)

\(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right)\)     AG     N0

[1 mark]

(i) correct description     R1     N1

e.g. reference to \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
8
\end{array}} \right)\) being the position vector of a point on the line, a vector to the line, a point on the line.

(ii) any correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\)    A2     N2

where \({\boldsymbol{a}}\) is \(\left( {\begin{array}{*{20}{c}}
  3 \\
  { - 3} \\
  8
\end{array}} \right)\) , and \({\boldsymbol{b}}\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
  1 \\
  { - 2} \\
  3
\end{array}} \right)\)

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  3 \\
  { - 3} \\
  8
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  { - 1} \\
  2 \\
  { - 3}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  {3 + 2s} \\
  { - 3 - 4s} \\
  {8 + 6s}
\end{array}} \right)\)

[3 marks]

one correct equation     (A1)

e.g. \(3 + s = - 1\) , \( - 3 - 2s = 5\)

\(s = - 4\)     A1

\(p = - 4\)     A1     N2

[3 marks]

one correct equation     A1

e.g. \( - 3 + t = - 1\) , \(9 - 2t = 5\)

\(t = 2\)     A1

substituting \(t = 2\)

e.g. \(2 + 2q = - 4\) , \(2q =  - 6\)     A1

\(q = - 3\)     AG     N0

[3 marks]

choosing correct direction vectors \(\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
{ - 3}
\end{array}} \right)\)     (A1)(A1)

finding correct scalar product and magnitudes     (A1)(A1)(A1)

scalar product \((1)(1) + ( - 2)( - 2) + ( - 3)(3)\)     \(( = - 4)\)

magnitudes \(\sqrt {{1^2} + {{( - 2)}^2} + {3^2}} \) \( = \sqrt {14} \) , \(\sqrt {{1^2} + {{( - 2)}^2} + {{( - 3)}^2}} \) \( = \sqrt {14} \)

evidence of substituting into scalar product     M1

e.g. \(\cos \theta  = \frac{{ - 4}}{{3.741 \ldots  \times 3.741 \ldots }}\)

\(\theta  = 1.86\) radians (or \(107^\circ \))    A1     N4

[7 marks]

Most candidates answered part (a) easily.

For part (b), a number of candidates stated that the vector was a "starting point," which misses the idea that it is a position vector to some point on the line.

Parts (c) and (d) proved accessible to many.

Parts (c) and (d) proved accessible to many.

For part (e), a surprising number of candidates chose incorrect vectors. Few candidates seemed to have a good conceptual understanding of the vector equation of a line.