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Date	May 2019	Marks available	3	Reference code	19M.1.AHL.TZ1.H_11
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Verify	Question number	H_11	Adapted from	N/A

Question

Two distinct lines, ${l_1}$ and ${l_2}$ , intersect at a point ${\text{P}}$ . In addition to ${\text{P}}$ , four distinct points are marked out on ${l_1}$ and three distinct points on ${l_2}$ . A mathematician decides to join some of these eight points to form polygons.

The line ${l_1}$ has vector equation r₁ $= \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 1 \end{array}} \right)$ , $\lambda \in \mathbb{R}$ and the line ${l_2}$ has vector equation r₂ $= \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 2 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}} 5 \\ 6 \\ 2 \end{array}} \right)$ , $\mu \in \mathbb{R}$ .

The point ${\text{P}}$ has coordinates (4, 6, 4).

The point ${\text{A}}$ has coordinates (3, 4, 3) and lies on ${l_1}$ .

The point ${\text{B}}$ has coordinates (−1, 0, 2) and lies on ${l_2}$ .

Find how many sets of four points can be selected which can form the vertices of a quadrilateral.

[2]

a.i.

Find how many sets of three points can be selected which can form the vertices of a triangle.

[4]

a.ii.

Verify that ${\text{P}}$ is the point of intersection of the two lines.

[3]

Write down the value of $\lambda$ corresponding to the point ${\text{A}}$ .

[1]

Write down $\overrightarrow {{\text{PA}}}$ and $\overrightarrow {{\text{PB}}}$ .

[2]

Let ${\text{C}}$ be the point on ${l_1}$ with coordinates (1, 0, 1) and ${\text{D}}$ be the point on ${l_2}$ with parameter $\mu = - 2$ .

Find the area of the quadrilateral ${\text{CDBA}}$ .

[8]

Markscheme

appreciation that two points distinct from ${\text{P}}$ need to be chosen from each line M1

${}^4{C_2} \times {}^3{C_2}$

=18 A1

[2 marks]

a.i.

EITHER

consider cases for triangles including ${\text{P}}$ or triangles not including ${\text{P}}$ M1

$3 \times 4 + 4 \times {}^3{C_2} + 3 \times {}^4{C_2}$ (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

consider total number of ways to select 3 points and subtract those with 3 points on the same line M1

${}^8{C_3} - {}^5{C_3} - {}^4{C_3}$ (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

56−10−4

THEN

= 42 A1

[4 marks]

a.ii.

METHOD 1

substitution of (4, 6, 4) into both equations (M1)

$\lambda = 3$ and $\mu = 1$ A1A1

(4, 6, 4) AG

METHOD 2

attempting to solve two of the three parametric equations M1

$\lambda = 3$ and $\mu = 1$ A1

check both of the above give (4, 6, 4) M1AG

Note: If they have shown the curve intersects for all three coordinates they only need to check (4,6,4) with one of " $\lambda$ " or " $\mu$ ".

[3 marks]

$\lambda = 2$ A1

[1 mark]

$\overrightarrow {{\text{PA}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ { - 2} \\ { - 1} \end{array}} \right)$ , $\overrightarrow {{\text{PB}}} = \left( {\begin{array}{*{20}{c}} { - 5} \\ { - 6} \\ { - 2} \end{array}} \right)$ A1A1

Note: Award A1A0 if both are given as coordinates.

[2 marks]

METHOD 1

area triangle ${\text{ABP}} = \frac{1}{2}\left| {\overrightarrow {{\text{PB}}} \times \overrightarrow {{\text{PA}}} } \right|$ M1

$\left( { = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} { - 5} \\ { - 6} \\ { - 2} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} { - 1} \\ { - 2} \\ { - 1} \end{array}} \right)} \right|} \right) = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} 2 \\ { - 3} \\ 4 \end{array}} \right)} \right|$ A1

$= \frac{{\sqrt {29} }}{2}$ A1

EITHER

$\overrightarrow {{\text{PC}}} = 3\overrightarrow {\,{\text{PA}}}$ , $\overrightarrow {{\text{PD}}} = 3\overrightarrow {\,{\text{PB}}}$ (M1)

area triangle ${\text{PCD}} = 9 \times$ area triangle ${\text{ABP}}$ (M1)A1

$= \frac{{9\sqrt {29} }}{2}$ A1

${\text{D}}$ has coordinates (−11, −12, −2) A1

area triangle ${\text{PCD}} = \frac{1}{2}\left| {\overrightarrow {{\text{PD}}} \times \overrightarrow {{\text{PC}}} } \right| = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} { - 15} \\ { - 18} \\ { - 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 6} \\ { - 3} \end{array}} \right)} \right|$ M1A1

Note: A1 is for the correct vectors in the correct formula.

$= \frac{{9\sqrt {29} }}{2}$ A1

THEN

area of ${\text{CDBA}} = \frac{{9\sqrt {29} }}{2} - \frac{{\sqrt {29} }}{2}$

$= 4\sqrt {29}$ A1

METHOD 2

${\text{D}}$ has coordinates (−11, −12, −2) A1

area $= \frac{1}{2}\left| {\overrightarrow {{\text{CB}}} \times \overrightarrow {{\text{CA}}} } \right| + \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{BD}}} } \right|$ M1

Note: Award M1 for use of correct formula on appropriate non-overlapping triangles.

Note: Different triangles or vectors could be used.

$\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} { - 2} \\ 0 \\ 1 \end{array}} \right)$ , $\overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ 2 \end{array}} \right)$ A1

$\overrightarrow {{\text{CB}}} \times \overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} { - 4} \\ 6 \\ { - 8} \end{array}} \right)$ A1

$\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ { - 1} \end{array}} \right)$ , $\overrightarrow {{\text{BD}}} = \left( {\begin{array}{*{20}{c}} { - 10} \\ { - 12} \\ { - 4} \end{array}} \right)$ A1

$\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{BD}}} = \left( {\begin{array}{*{20}{c}} { - 12} \\ {18} \\ { - 24} \end{array}} \right)$ A1

Note: Other vectors which might be used are $\overrightarrow {{\text{DA}}} = \left( {\begin{array}{*{20}{c}} {14} \\ {16} \\ {5} \end{array}} \right)$ , $\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} {4} \\ {4} \\ {1} \end{array}} \right)$ , $\overrightarrow {{\text{DC}}} = \left( {\begin{array}{*{20}{c}} {12} \\ {12} \\ {3} \end{array}} \right)$ .