Find the coordinates of S.

The vector product $\overrightarrow {{\text{PQ}}}  \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}   { - 13} \\   7 \\   m \end{array}} \right)$. Find the value of m .

Hence calculate the area of parallelogram PQRS.

Find the Cartesian equation of the plane, ${\prod _1}$ , containing the parallelogram PQRS.

Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane ${\prod _1}$ .

Hence find the point on the plane that is closest to the origin.

A second plane, ${\prod _2}$ , has equation x − 2y + z = 3. Calculate the angle between the two planes.

$\overrightarrow {{\text{PQ}}}  = \left( {\begin{array}{*{20}{c}}   { - 1} \\   { - 1} \\   3 \end{array}} \right)$ , $\overrightarrow {{\text{SR}}}  = \left( {\begin{array}{*{20}{c}}   {0 - x} \\   {5 - y} \\   {1 - z} \end{array}} \right)$     (M1)

point S = (1, 6, −2)     A1

[2 marks]

$\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}   { - 1} \\   { - 1} \\   3 \end{array}} \right)$$\overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}   2 \\   4 \\   1 \end{array}} \right)$     A1

$\overrightarrow {{\text{PQ}}}  \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}   { - 13} \\   7 \\   { - 2} \end{array}} \right)$

m = −2     A1

[2 marks]

area of parallelogram PQRS $= \left| {\overrightarrow {{\text{PQ}}}  \times \overrightarrow {{\text{PS}}} } \right| = \sqrt {{{( - 13)}^2} + {7^2} + {{( - 2)}^2}}$     M1

$= \sqrt {222}  = 14.9$     A1

[2 marks]

equation of plane is −13x + 7y − 2z = d     M1A1

substituting any of the points given gives d = 33

−13x + 7y − 2z = 33     A1

[3 marks]

equation of line is $\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}   0 \\   0 \\   0 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}   { - 13} \\   7 \\   { - 2} \end{array}} \right)$     A1

Note: To get the A1 must have $\boldsymbol{r} =$ or equivalent.

[1 mark]

$169\lambda + 49\lambda + 4\lambda = 33$     M1

$\lambda = \frac{{33}}{{222}}{\text{ }}( = 0.149…)$     A1

closest point is $\left( { - \frac{{143}}{{74}},\frac{{77}}{{74}}, - \frac{{11}}{{37}}} \right){\text{ }}\left( { = ( - 1.93,{\text{ 1.04,  - 0.297)}}} \right)$     A1

[3 marks]

angle between planes is the same as the angle between the normals     (R1)

$\cos \theta = \frac{{ - 13 \times 1 + 7 \times - 2 - 2 \times 1}}{{\sqrt {222} \times \sqrt 6 }}$     M1A1

$\theta = 143^\circ$ (accept $\theta = 37.4^\circ$ or 2.49 radians or 0.652 radians)     A1

[4 marks]

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form $\boldsymbol{r} =$ ...

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...