Show that \(L\) is not perpendicular to \(\Pi \).

Given that \(L\) lies in the plane \(\Pi \), find the value of \(p\) and the value of \(q\).

(i)     Show that \(p =  - 2\).

(ii)     If \(L\) intersects \(\Pi \) at \(z =  - 1\), find the value of \(q\).

EITHER

n \( = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right)\) and d \( = \left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\)     A1A1

and n \( \ne \) kd     R1

OR

n \( \times \) d \( = \left( {\begin{array}{*{20}{c}} { - 5} \\ {3p - 1} \\ {2 - p} \end{array}} \right)\)     M1A1

the vector product is non-zero for \(p \in \mathbb{R}\)     R1

THEN

\(L\) is not perpendicular to \(\Pi \)     AG

[3 marks]

METHOD 1

\((2 + p\lambda ) + (q + 2\lambda ) + 3(1 + \lambda ) = 9\)    M1

\((q + 5) + (p + 5)\lambda  = 9\)    (A1)

\(p =  - 5\) and \(q = 4\)     A1A1

METHOD 2

direction vector of line is perpendicular to plane, so

\(\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 0\)    M1

\(p =  - 5\)    A1

\((2,{\text{ }}q,{\text{ }}1)\) is common to both \(L\) and \(\Pi \)

either \(\left( {\begin{array}{*{20}{c}} 2 \\ q \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 9\) or by substituting into \(x + y + 3z = 9\)     M1

\(q = 4\)    A1

[4 marks]

(i)     METHOD 1

\(\alpha \) is the acute angle between n and L

if \(\sin \theta  = \frac{1}{{\sqrt {11} }}\) then \(\cos \alpha  = \frac{1}{{\sqrt {11} }}\)     (M1)(A1)

attempting to use \(\cos \alpha  = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\) or \(\sin \alpha  = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\)     M1

\(\frac{{p + 5}}{{\sqrt {11}  \times \sqrt {{p^2} + 5} }} = \frac{1}{{\sqrt {11} }}\)    A1A1

\({(p + 5)^2} = {p^2} + 5\)    M1

\(10p =  - 20\) (or equivalent)     A1

\(p =  - 2\)    AG

METHOD 2

\(\alpha \) is the angle between n and L

if \(\sin \theta  = \frac{1}{{\sqrt {11} }}\) then \(\sin \alpha  = \frac{{\sqrt {10} }}{{\sqrt {11} }}\)     (M1)A1

attempting to use \(\sin \alpha  = \frac{{\left| {n \times d} \right|}}{{\left| n \right|\left| d \right|}}\)     M1

\(\frac{{\sqrt {{{( - 5)}^2} + {{(3p - 1)}^2} + {{(2 - p)}^2}} }}{{\sqrt {11}  \times \sqrt {{p^2} + 5} }} = \frac{{\sqrt {10} }}{{\sqrt {11} }}\)    A1A1

\({p^2} - p + 3 = {p^2} + 5\)    M1

\( - p + 3 = 5\) (or equivalent)     A1

\(p =  - 2\)    AG

(ii)     \(p =  - 2\) and \(z =  - 1 \Rightarrow \frac{{x - 2}}{{ - 2}} = \frac{{y - q}}{2} =  - 2\)     (A1)

\(x = 6\) and \(y = q - 4\)     (A1)

this satisfies \(\Pi \) so \(6 + q - 4 - 3 = 9\)     M1

\(q = 10\)    A1

[11 marks]

Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate \(\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\), with some scripts seen involving needlessly convoluted arguments.

Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate \(\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\), with some scripts seen involving needlessly convoluted arguments.

Part (c) often proved problematic, as some candidates unsurprisingly used the sine (or cosine) of an incorrect angle, and few consequent marks were then available. Some good clear solutions were seen, occasionally complete with diagrams in the cases of the thoughtful candidates who were able to ‘work through’ the question rather than just apply a standard vector result.