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]