Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)\).

Find a vector equation for \(L\).

Find the value of \(p\).

The point D has coordinates \(({q^2},{\text{ }}0,{\text{ }}q)\). Given that \(\overrightarrow {{\text{DC}}} \) is perpendicular to \(L\), find the possible values of \(q\).

correct approach     A1

 

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ { - 4} \\ { - 2} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)\)

 

\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)\)    AG     N0

[1 mark]

any correct equation in the form \(r = a + tb\) (any parameter for \(t\))

 

where \(a\) is \(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)\) and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)\)     A2     N2

 

eg\(\,\,\,\,\,\)\(r = \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right),{\text{ }}(x,{\text{ }}y,{\text{ }}z) = ( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 3 + 2t} \\ {4 - t} \\ {2 + t} \end{array}} \right)\)

 

Note:     Award A1 for the form \(a + tb\), A1 for the form \(L = a + tb\), A0 for the form \(r = b + ta\).

 

[2 marks]

METHOD 1 – finding value of parameter

valid approach     (M1)

 

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1) = (3,{\text{ }}1,{\text{ }}p)\)

 

one correct equation (not involving \(p\))     (A1)

eg\(\,\,\,\,\,\)\( - 3 + 2t = 3,{\text{ }} - 1 - 2s = 3,{\text{ }}4 - t = 1,{\text{ }}3 + s = 1\)

correct parameter from their equation (may be seen in substitution)     A1

eg\(\,\,\,\,\,\)\(t = 3,{\text{ }}s =  - 2\)

correct substitution     (A1)

 

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + 3\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}3 - ( - 2)\)

 

\(p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)\)     A1     N2

 

METHOD 2 – eliminating parameter

valid approach     (M1)

 

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1) = (3,{\text{ }}1,{\text{ }}p)\)

 

one correct equation (not involving \(p\))     (A1)

eg\(\,\,\,\,\,\)\( - 3 + 2t = 3,{\text{ }} - 1 - 2s = 3,{\text{ }}4 - t = 1,{\text{ }}3 + s = 1\)

correct equation (with \(p\))     A1

eg\(\,\,\,\,\,\)\(2 + t = p,{\text{ }}3 - s = p\)

correct working to solve for \(p\)     (A1)

eg\(\,\,\,\,\,\)\(7 = 2p - 3,{\text{ }}6 = 1 + p\)

 

\(p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)\)     A1     N2

 

[5 marks]

valid approach to find \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \)     (M1)

 

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right)\)

 

correct vector for \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) (may be seen in scalar product)     A1

 

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2} - 3} \\ { - 1} \\ {q - 5} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right)\)

 

recognizing scalar product of \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) with direction vector of \(L\) is zero (seen anywhere)     (M1)

 

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0,{\text{ }}\overrightarrow {{\text{DC}}} \bullet \overrightarrow {{\text{AC}}} = 0,{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0\)

 

correct scalar product in terms of only \(q\)     A1

eg\(\,\,\,\,\,\)\(6 - 2{q^2} - 1 + 5 - q,{\text{ }}2{q^2} + q - 10 = 0,{\text{ }}2(3 - {q^2}) - 1 + 5 - q\)

correct working to solve quadratic     (A1)

eg\(\,\,\,\,\,\)\((2q + 5)(q - 2),{\text{ }}\frac{{ - 1 \pm \sqrt {1 - 4(2)( - 10)} }}{{2(2)}}\)

\(q =  - \frac{5}{2},{\text{ }}2\)     A1A1     N3

 

[7 marks]

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