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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