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Date	May 2019	Marks available	1	Reference code	19M.2.AHL.TZ2.H_11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Show that	Question number	H_11	Adapted from	N/A

Question

The plane П₁ contains the points P(1, 6, −7) , Q(0, 1, 1) and R(2, 0, −4).

The Cartesian equation of the plane П₂ is given by $x - 3 y - z = 3$ $x - 3y - z = 3$ .

The Cartesian equation of the plane П₃ is given by $a x + b y + c z = 1$ $ax + by + cz = 1$ .

Consider the case that П₃ contains $L$ $L$ .

Find the Cartesian equation of the plane containing P, Q and R.

[6]

Given that П₁ and П₂ meet in a line $L$ $L$ , verify that the vector equation of $L$ $L$ can be given by r $= ⎛ ⎜ ⎜ ⎝ \begin{matrix} \frac{5}{4} 0 - \frac{7}{4} \end{matrix} ⎞ ⎟ ⎟ ⎠ + λ ⎛ ⎜ ⎜ ⎝ \begin{matrix} \frac{1}{2} 1 - \frac{5}{2} \end{matrix} ⎞ ⎟ ⎟ ⎠$ $= \left( {\begin{array}{*{20}{c}} {\frac{5}{4}} \\ 0 \\ { - \frac{7}{4}} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} {\frac{1}{2}} \\ 1 \\ { - \frac{5}{2}} \end{array}} \right)$ .

[3]

Given that П₃ is parallel to the line $L$ $L$ , show that $a + 2 b - 5 c = 0$ $a + 2b - 5c = 0$ .

[1]

Show that $5 a - 7 c = 4$ $5a - 7c = 4$ .

[2]

d.i.

Given that П₃ is equally inclined to both П₁ and П₂, determine two distinct possible Cartesian equations for П₃.

[7]

d.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

for example

$- - \to PQ = ⎛ ⎜ ⎝ \begin{matrix} - 1 - 5 8 \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ { - 5} \\ 8 \end{array}} \right)$ , $- - \to PR = ⎛ ⎜ ⎝ \begin{matrix} 1 - 6 3 \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{PR}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { - 6} \\ 3 \end{array}} \right)$ A1A1

$- - \to PQ \times - - \to PR$ $\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PR}}}$ = 33i + 11j + 11k (M1)A1

r.n = a.n

$33 x + 11 y + 11 z = ⎛ ⎜ ⎝ \begin{matrix} 011 \end{matrix} ⎞ ⎟ ⎠ \cdot ⎛ ⎜ ⎝ \begin{matrix} 331111 \end{matrix} ⎞ ⎟ ⎠ = 22$ $33x + 11y + 11z = \left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 1 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {33} \\ {11} \\ {11} \end{array}} \right) = 22$ (M1)

$\Rightarrow 3 x + y + z = 2$ $\Rightarrow 3x + y + z = 2$ or equivalent A1

METHOD 2

assume plane can be written as $a x + b y + c z = 1$ $ax + by + cz = 1$ M1

substituting each set of coordinates gives the system of equations:

$a + 6 b - 7 c = 1$ $a + 6b - 7c = 1$

$0 a + b + c = 1$ $0a + b + c = 1$

$2 a + 0 b - 4 c = 1$ $2a + 0b - 4c = 1$ A1

solving by GDC (M1)

$a = \frac{3}{2}$ $a = \frac{3}{2}$ , $b = \frac{1}{2}$ $b = \frac{1}{2}$ , $C = \frac{1}{2}$ $C = \frac{1}{2}$ A1A1A1

$\Rightarrow \frac{3}{2} x + \frac{1}{2} y + \frac{1}{2} z = 1$ $\Rightarrow \frac{3}{2}x + \frac{1}{2}y + \frac{1}{2}z = 1$ or equivalent

[6 marks]

METHOD 1

substitution of equation of line into both equations of planes M1

$3 (\frac{5}{4} + \frac{λ}{2}) + (- \frac{7}{4} - \frac{5 λ}{2}) = 2$ $3\left( {\frac{5}{4} + \frac{\lambda }{2}} \right) + \left( { - \frac{7}{4} - \frac{{5\lambda }}{2}} \right) = 2$ A1

$(\frac{5}{4} + \frac{λ}{2}) - 3 λ - (- \frac{7}{4} - \frac{5 λ}{2}) = 3$ $\left( {\frac{5}{4} + \frac{\lambda }{2}} \right) - 3\lambda - \left( { - \frac{7}{4} - \frac{{5\lambda }}{2}} \right) = 3$ A1

METHOD 2

adding Π₁ and Π₂ gives $4 x - 2 y = 5$ $4x - 2y = 5$ M1

given $y = λ \Rightarrow x = \frac{5}{4} + \frac{λ}{2}$ $y = \lambda \Rightarrow x = \frac{5}{4} + \frac{\lambda }{2}$ A1

$z = 2 - y - 3 x = - \frac{7}{4} - \frac{5 λ}{2}$ $z = 2 - y - 3x = - \frac{7}{4} - \frac{{5\lambda }}{2}$ A1

⇒r $= ⎛ ⎜ ⎜ ⎝ \begin{matrix} \frac{5}{4} 0 - \frac{7}{4} \end{matrix} ⎞ ⎟ ⎟ ⎠ + λ ⎛ ⎜ ⎜ ⎝ \begin{matrix} \frac{1}{2} 1 - \frac{5}{2} \end{matrix} ⎞ ⎟ ⎟ ⎠$ $= \left( {\begin{array}{*{20}{c}} {\frac{5}{4}} \\ 0 \\ { - \frac{7}{4}} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} {\frac{1}{2}} \\ 1 \\ { - \frac{5}{2}} \end{array}} \right)$ AG

METHOD 3

n₁ × n₂ = $⎛ ⎜ ⎝ \begin{matrix} 24 - 10 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 10} \end{array}} \right)$ A1

$= 4 ⎛ ⎜ ⎜ ⎝ \begin{matrix} \frac{1}{2} 1 - \frac{5}{2} \end{matrix} ⎞ ⎟ ⎟ ⎠$ $= 4\left( {\begin{array}{*{20}{c}} {\frac{1}{2}} \\ 1 \\ { - \frac{5}{2}} \end{array}} \right)$ R1

common point $\frac{5}{4} - 3 (0) - (- \frac{7}{4}) = 3$ $\frac{5}{4} - 3\left( 0 \right) - \left( { - \frac{7}{4}} \right) = 3$ and $- 3 (\frac{5}{4}) - 0 - (- \frac{7}{4}) = - 2$ $- 3\left( {\frac{5}{4}} \right) - 0 - \left( { - \frac{7}{4}} \right) = - 2$ A1

[3 marks]

normal to П₃ is perpendicular to direction of $L$ $L$

$\Rightarrow ⎛ ⎜ ⎝ \begin{matrix} a b c \end{matrix} ⎞ ⎟ ⎠ \cdot ⎛ ⎜ ⎝ \begin{matrix} 12 - 5 \end{matrix} ⎞ ⎟ ⎠ = 0$ $\Rightarrow \left( {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 5} \end{array}} \right) = 0$ A1

⇒ $a + 2 b - 5 c = 0$ $a + 2b - 5c = 0$ AG

[1 mark]

substituting $⎛ ⎜ ⎜ ⎝ \begin{matrix} \frac{5}{4} 0 - \frac{7}{4} \end{matrix} ⎞ ⎟ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} {\frac{5}{4}} \\ 0 \\ { - \frac{7}{4}} \end{array}} \right)$ into П₃: M1

$\frac{5 a}{4} - \frac{7 c}{4} = 1$ $\frac{{5a}}{4} - \frac{{7c}}{4} = 1$ A1

$5 a - 7 c = 4$ $5a - 7c = 4$ AG

[2 marks]

d.i.

attempt to find scalar products for П₁ and П₃, П₂ and П₃.

and equating M1

$\frac{3 a + b + C}{\sqrt{11} \sqrt{a^{2} + b^{2} + c^{2}}} = \frac{a - 3 b - c}{\sqrt{11} \sqrt{a^{2} + b^{2} + c^{2}}}$ $\frac{{3a + b + C}}{{\sqrt {11} \sqrt {{a^2} + {b^2} + {c^2}} }} = \frac{{a - 3b - c}}{{\sqrt {11} \sqrt {{a^2} + {b^2} + {c^2}} }}$ M1

Note: Accept $3 a + b + c = a - 3 b - c$ $3a + b + c = a - 3b - c$ .

$\Rightarrow a + 2 b + c = 0$ $\Rightarrow a + 2b + c = 0$ A1

attempt to solve $a + 2 b + c = 0$ $a + 2b + c = 0$ , $a + 2 b - 5 c = 0$ $a + 2b - 5c = 0$ , $5 a - 7 c = 4$ $5a - 7c = 4$ M1

$\Rightarrow a = \frac{4}{5}, b = - \frac{2}{5}, c = 0$ $\Rightarrow a = \frac{4}{5}{\text{,}}\,\,b = - \frac{2}{5}{\text{,}}\,\,c = 0$ A1

hence equation is $\frac{4 x}{5} - \frac{2 y}{5} = 1$ $\frac{{4x}}{5} - \frac{{2y}}{5} = 1$

for second equation:

$\frac{3 a + b + C}{\sqrt{11} \sqrt{a^{2} + b^{2} + c^{2}}} = - \frac{a - 3 b - c}{\sqrt{11} \sqrt{a^{2} + b^{2} + c^{2}}}$ $\frac{{3a + b + C}}{{\sqrt {11} \sqrt {{a^2} + {b^2} + {c^2}} }} = - \frac{{a - 3b - c}}{{\sqrt {11} \sqrt {{a^2} + {b^2} + {c^2}} }}$ (M1)