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.