Find the vector \(\overrightarrow {{\text{CA}}}  \times \overrightarrow {{\text{CB}}} \).

Find an exact value for the area of the triangle ABC.

Show that the Cartesian equation of the plane \({\Pi _1}\), containing the triangle ABC, is \(2x + 3y - 4z = 2\).

A second plane \({\Pi _2}\) is defined by the Cartesian equation \({\Pi _2}:4x - y - z = 4\). \({L_1}\) is the line of intersection of the planes \({\Pi _1}\) and \({\Pi _2}\).

Find a vector equation for \({L_1}\).

Find the value of \(\alpha \) if all three planes contain \({L_1}\).

Find conditions on \(\alpha \) and \(\beta \) if the plane \({\Pi _3}\) does not intersect with \({L_1}\).

\(\overrightarrow {{\rm{CA}}}  = \left( \begin{array}{c}1\\ - 2\\ - 1\end{array} \right)\)     (A1)

\(\overrightarrow {{\rm{CB}}}  = \left( \begin{array}{c}2\\0\\1\end{array} \right)\)     (A1)

Note:     If \(\overrightarrow {{\text{AC}}} \) and \(\overrightarrow {{\text{BC}}} \) found correctly award (A1) (A0).

\(\overrightarrow {{\rm{CA}}}  \times \overrightarrow {{\rm{CB}}}  = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ - 2}&{ - 1}\\2&0&1\end{array}} \right|\)     (M1)

\(\left( \begin{array}{c} - 2\\ - 3\\4\end{array} \right)\)     A1

[4 marks]

METHOD 1

\(\frac{1}{2}\left| {\overrightarrow {{\text{CA}}}  \times \overrightarrow {{\text{CB}}} } \right|\)     (M1)

\( = \frac{1}{2}\sqrt {{{( - 2)}^2} + {{( - 3)}^2} + {4^2}} \)     (A1)

\( = \frac{{\sqrt {29} }}{2}\)     A1

METHOD 2

attempt to apply \(\frac{1}{2}\left| {{\text{CA}}} \right|\left| {{\text{CB}}} \right|\sin C\)     (M1)

\({\text{CA.CB}} = \sqrt 5 .\sqrt 6 \cos C \Rightarrow \cos C = \frac{1}{{\sqrt {30} }} \Rightarrow \sin C = \frac{{\sqrt {29} }}{{\sqrt {30} }}\)     (A1)

\({\text{area}} = \frac{{\sqrt {29} }}{2}\)     A1

[3 marks]

METHOD 1

r.\(\left( \begin{array}{c} - 2\\ - 3\\4\end{array} \right) = \left( \begin{array}{l}1\\0\\0\end{array} \right) \bullet \left( \begin{array}{c} - 2\\ - 3\\4\end{array} \right)\)     M1A1

\( \Rightarrow  - 2x - 3y + 4z =  - 2\)     A1

\( \Rightarrow 2x + 3y - 4z = 2\)     AG

METHOD 2

\( - 2x - 3y + 4z = d\)

substituting a point in the plane     M1A1

\({\text{d}} =  - 2\)     A1

\( \Rightarrow  - 2x - 3y + 4z =  - 2\)

\( \Rightarrow 2x + 3y - 4z = 2\)     AG

Note:     Accept verification that all 3 vertices of the triangle lie on the given plane.

[3 marks]

METHOD 1

\(\left| {\begin{array}{*{20}{c}}{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}}\\2&3&{ - 4}\\4&{ - 1}&{ - 1}\end{array}} \right| = \left( \begin{array}{c} - 7\\ - 14\\ - 14\end{array} \right)\)    M1A1

\({\mathbf{n}} = \left( \begin{array}{l}1\\2\\2\end{array} \right)\)

\(z = 0 \Rightarrow y = 0,{\text{ }}x = 1\)     (M1)(A1)

\({L_1}:{\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\)     A1

Note:     Do not award the final A1 if \(\mathbf{r} =\) is not seen.

METHOD 2

eliminate 1 of the variables, eg x     M1

\( - 7y + 7z = 0\)     (A1)

introduce a parameter     M1

\( \Rightarrow z = \lambda \),

\(y = \lambda {\text{, }}x = 1 + \frac{\lambda }{2}\)     (A1)

\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent     A1

Note:     Do not award the final A1 if \(\mathbf{r} =\) is not seen.

METHOD 3

\(z = t\)     M1

write x and y in terms of \(t \Rightarrow 4x - y = 4 + t,{\text{ }}2x + 3y = 2 + 4t\) or equivalent     A1

attempt to eliminate x or y     M1

\(x,{\text{ }}y,{\text{ }}z\) expressed in parameters

\( \Rightarrow z = t\),

\(y = t,{\text{ }}x = 1 + \frac{t}{2}\)     A1

\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + t \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent     A1

Note:     Do not award the final A1 if \(\mathbf{r} =\) is not seen.

[5 marks]

METHOD 1

direction of the line is perpendicular to the normal of the plane

\(\left( \begin{array}{c}16\\\alpha \\ - 3\end{array} \right) \bullet \left( \begin{array}{l}1\\2\\2\end{array} \right) = 0\)     M1A1

\(16 + 2\alpha  - 6 = 0 \Rightarrow \alpha  =  - 5\)     A1

METHOD 2

solving line/plane simultaneously

\(16(1 + \lambda ) + 2\alpha \lambda  - 6\lambda  = \beta \)     M1A1

\(16 + (10 + 2\alpha )\lambda  = \beta \)

\( \Rightarrow \alpha  =  - 5\)     A1

METHOD 3

\(\left| {\begin{array}{*{20}{c}}2&3&{ - 4}\\4&{ - 1}&{ - 1}\\{16}&\alpha &{ - 3}\end{array}} \right| = 0\)     M1

\(2(3 + \alpha ) - 3( - 12 + 16) - 4(4\alpha  + 16) = 0\)     A1

\( \Rightarrow \alpha  =  - 5\)     A1

METHOD 4

attempt to use row reduction on augmented matrix     M1

to obtain \(\left( {\begin{array}{*{20}{c}}2&3&{ - 4}\\0&{ - 1}&1\\0&0&{\alpha  + 5}\end{array}\left| \begin{array}{c}2\\0\\\beta  - 16\end{array} \right.} \right)\)     A1

\( \Rightarrow \alpha  =  - 5\)     A1

[3 marks]

\(\alpha  =  - 5\)     A1

\(\beta  \ne 16\)     A1

[2 marks]

Part a) proved an easy start, though a few (weaker) candidates still believe \(\overrightarrow {{\text{CA}}} \) to be \(\overrightarrow {{\text{OC}}}  - \overrightarrow {{\text{OA}}} \).

Part b) was an easy 3 marks and incorrect answers were rare.

Part c) was answered well, though reasoning sometimes seemed sparse, especially given that this was a ‘show that’ question.

Part d) proved more challenging, despite being a very standard question. Many candidates gained only 2 marks, either through correctly calculating the direction vector, or by successfully eliminating one of the variables. A number of clear fully correct solutions were seen, though the absence of ‘r  =’ is still prevalent, and candidates might be reminded of the correct form for the vector equation of a line.

Part e) proved a puzzle for most, though an attempt to use row reduction on an augmented matrix seemed to be the choice way for most successful candidates.

Only the very best were able to demonstrate a complete understanding of intersecting planes and thus answer part f) correctly.