Find the vector equation of the line (BC).

Determine whether or not the lines (OA) and (BC) intersect.

Find the Cartesian equation of the plane Π$_1$, which passes through C and is perpendicular to $\overrightarrow {{\rm{OA}}}$.

Show that the line (BC) lies in the plane Π$_1$.

Verify that 2j + k is perpendicular to the plane Π$_2$.

Find a vector perpendicular to the plane Π$_3$.

Find the acute angle between the planes Π$_2$ and Π$_3$.

$\overrightarrow {{\rm{BC}}}$ = (i + 3j + 3k) $-$ (2i $-$ j + 2k) = $-$i + 4j + k    (A1)

r = (2i $-$ j + 2k) + $\lambda$($-$i + 4j + k)

(or r = (i + 3j + 3k) + $\lambda$($-$i + 4j + k)     (M1)A1

Note:     Do not award A1 unless r = or equivalent correct notation seen.

[3 marks]

attempt to write in parametric form using two different parameters AND equate     M1

$2\mu = 2 - \lambda$

$\mu = - 1 + 4\lambda$

$- 2\mu = 2 + \lambda$     A1

attempt to solve first pair of simultaneous equations for two parameters     M1

solving first two equations gives $\lambda = \frac{4}{9},{\text{ }}\mu = \frac{7}{9}$     (A1)

substitution of these two values in third equation     (M1)

since the values do not fit, the lines do not intersect     R1

Note:     Candidates may note that adding the first and third equations immediately leads to a contradiction and hence they can immediately deduce that the lines do not intersect.

[6 marks]

METHOD 1

plane is of the form r $\bullet$ (2i + j $-$ 2k) = d     (A1)

d = (i + 3j + 3k) $\bullet$ (2i + j $-$ 2k) = $-$1     (M1)

hence Cartesian form of plane is $2x + y - 2z = - 1$     A1

METHOD 2

plane is of the form $2x + y - 2z = d$     (A1)

substituting $(1,{\text{ }}3,{\text{ }}3)$ (to find gives $2 + 3 - 6 = - 1$)     (M1)

hence Cartesian form of plane is $2x + y - 2z = - 1$     A1

[3 marks]

METHOD 1

attempt scalar product of direction vector BC with normal to plane     M1

($-$i + 4j + k) $\bullet$ (2i + j $-$ 2k) $= - 2 + 4 - 2$

$= 0$     A1

hence BC lies in Π$_1$     AG

METHOD 2

substitute eqn of line into plane     M1

${\text{line }}r = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 2 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \\ 1 \end{array}} \right).{\text{ Plane }}{\pi _1}:2x + y - 2z = - 1$

$2(2 - \lambda ) + ( - 1 + 4\lambda ) - 2(2 + \lambda )$

$= - 1$     A1

hence BC lies in Π$_1$     AG

Note:     Candidates may also just substitute $2i - j + 2k$ into the plane since they are told C lies on ${\pi _1}$.

Note:     Do not award A1FT.

[2 marks]

METHOD 1

applying scalar product to $\overrightarrow {{\rm{OA}}}$ and $\overrightarrow {{\rm{OB}}}$     M1

(2j + k) $\bullet$ (2i + j $-$ 2k) = 0     A1

(2j + k) $\bullet$ (2i $-$ j + 2k) =0     A1

METHOD 2

attempt to find cross product of $\overrightarrow {{\rm{OA}}}$ and $\overrightarrow {{\rm{OB}}}$     M1

plane Π$_2$ has normal $\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}}$ = $-$ 8j $-$ 4k     A1

since $-$8j $-$ 4k = $-$4(2j + k), 2j + k is perpendicular to the plane Π$_2$     R1

[3 marks]

plane Π$_3$ has normal $\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OC}}}$ = 9i $-$ 8j + 5k     A1

[1 mark]

attempt to use dot product of normal vectors     (M1)

$\cos \theta = \frac{{(2j + k) \bullet (9i - 8j + 5k)}}{{\left| {2j + k} \right|\left| {9i - 8j + 5k} \right|}}$     (M1)

$= \frac{{ - 11}}{{\sqrt 5 \sqrt {170} }}\,\,\,( = - 0.377 \ldots )$     (A1)

Note:     Accept $\frac{{11}}{{\sqrt 5 \sqrt {170} }}$.   acute angle between planes $= 67.8^\circ \,\,\,{\text{(}} = 1.18^\circ )$     A1

[4 marks]