Find the cosine of the angle between the two planes in the form \(\sqrt {\frac{p}{q}} \) where \(p,{\text{ }}q \in \mathbb{Z}\).

(i)     Show that \(L\) has direction \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\).

(ii)     Show that the point \({\text{A }}(1,{\text{ }}0,{\text{ }}4)\) lies on both planes.

(iii)     Write down a vector equation of \(L\).

Given the vector \(\overrightarrow {{\text{AB}}} \) is perpendicular to \(L\) find the value of \(a\) and the value of \(b\).

Show that \({\text{AB}} = 3\sqrt 2 \).

Find the coordinates of the two possible positions of \(P\).

Note:     Throughout the question condone vectors written horizontally.

angle between planes is equal to the angles between the normal to the planes     (M1)

\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { - 1} \end{array}} \right) = 18\)    (A1)

let \(\theta \) be the angle between the normal to the planes

\(\cos \theta  = \frac{{18}}{{\sqrt {18} \sqrt {26} }} = \sqrt {\frac{{18}}{{26}}} {\text{ }}\left( {{\text{or equivalent, for example }}\sqrt {\frac{{324}}{{468}}} {\text{ or }}\sqrt {\frac{9}{{13}}} } \right)\)    M1A1

[4 marks]

Note:     Throughout the question condone vectors written horizontally.

(i)     METHOD 1

\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { - 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 4} \\ 8 \\ 8 \end{array}} \right)\)    M1A1

which is a multiple of \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\)     R1AG

Note:     Allow any equivalent wording or \(\left( {\begin{array}{*{20}{c}} { - 4} \\ 8 \\ 8 \end{array}} \right) = 4\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\), do not allow \(\left( {\begin{array}{*{20}{c}} { - 4} \\ 8 \\ 8 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\).

METHOD 2

let \(z = t\) (or equivalent)

solve simultaneously to get     M1

\(y = t - 4,{\text{ }}x = 3 - 0.5t\)    A1

hence direction vector is \(\left( {\begin{array}{*{20}{c}} { - 0.5} \\ 1 \\ 1 \end{array}} \right)\)

which is a multiple of \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\)     R1AG

METHOD 3

\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = - 4 + 2 + 2 = 0\)    M1A1

\(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { - 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = - 4 + 6 - 2 = 0\)    A1

Note:     If only one scalar product is found award M0A0A0.

(ii)     \({\Pi _1}:{\text{ }}4 + 0 + 4 = 8\) and \({\Pi _2}:{\text{ }}4 + 0 - 4 = 0\)     R1

(iii)     r \( = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\)     A1A1

Note:     A1 for “r \( = \)” and a correct point on the line, A1 for a parameter and a correct direction vector.

[6 marks]

Note:     Throughout the question condone vectors written horizontally.

\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {a - 1} \\ b \\ { - 3} \end{array}} \right)\)    (A1)

\(\left( {\begin{array}{*{20}{c}} {a - 1} \\ b \\ { - 3} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = 0\)    M1

Note:     Award M0 for \(\left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = 0\).

\( - a + 1 + 2b - 6 = 0 \Rightarrow a - 2b =  - 5\)    A1

lies on \({\Pi _1}\) so \(4a + b + 1 = 8 \Rightarrow 4a + b = 7\)     M1

\(a = 1,{\text{ }}b = 3\)    A1

[5 marks]

Note:     Throughout the question condone vectors written horizontally.

\({\text{AB}} = \sqrt {{0^2} + {3^2} + {{( - 3)}^2}}  = 3\sqrt 2 \)    M1AG

[1 mark]

Note:     Throughout the question condone vectors written horizontally.

METHOD 1

\(\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{AP}}} } \right| = 3\sqrt 2 \)    (M1)

\(\overrightarrow {{\text{AP}}} = t\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\)    (A1)

\(\left| {3t} \right| = 3\sqrt 2  \Rightarrow t =  \pm \sqrt 2 \)    (M1)A1

\({\text{P}}\left( {1 - \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) and \(\left( {1 + \sqrt 2 ,{\text{ }} - 2\sqrt 2 ,{\text{ }}4 - 2\sqrt 2 } \right)\)     A1

[5 marks]

METHOD 2

let P have coordinates \((1 - \lambda ,{\text{ }}2\lambda ,{\text{ }}4 + 2\lambda )\)     M1

\(\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 3 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BP}}} = \left( {\begin{array}{*{20}{c}} { - \lambda } \\ {2\lambda - 3} \\ {3 + 2\lambda } \end{array}} \right)\)    A1

\(\cos 45^\circ  = \frac{{\overrightarrow {{\text{BA}}}  \bullet \overrightarrow {{\text{BP}}} }}{{\left| {{\text{BA}}} \right|\left| {{\text{BP}}} \right|}}\)    M1

Note:     Award M1 even if AB rather than BA is used in the scalar product.

\(\overrightarrow {{\text{BA}}}  \bullet \overrightarrow {{\text{BP}}}  = 18\)

\(\frac{1}{{\sqrt 2 }} = \frac{{18}}{{\sqrt {18} \sqrt {9{\lambda ^2} + 18} }}\)

\(\lambda  =  \pm \sqrt 2 \)    A1

\({\text{P}}\left( {1 - \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) and \(\left( {1 + \sqrt 2 ,{\text{ }} - 2\sqrt 2 ,{\text{ }}4 - 2\sqrt 2 } \right)\)     A1

Note:     Accept answers given as position vectors.

[5 marks]

This was successfully done, though some candidates lost marks unnecessarily by not giving the answer in the form requested in the question.

(b)(i) A variety of techniques were successfully used here. A common error was not to justify why the vector obtained from the vector product was in the same direction as the one given in the question.

(b)(ii) and (iii) These were well done, though too many candidates still lose a mark unnecessarily by writing the vector equation of a line as \(l = \) rather than \(r = \).

Weaker candidates found the rest of this question more difficult. Though most obtained one equation for \(a\) and \(b\) they did not take note of the fact that it was also on the given plane, which gave the second equation.

This was done successfully by the majority of candidates. Candidates need to be aware that the notation AB means the length of the line segment joining the points A and B (as in the course guide).

This proved to be a difficult question for most candidates. Those who were successful were equally split between the two approaches given in the markscheme.