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Date	November 2019	Marks available	1	Reference code	19N.1.AHL.TZ0.H_11
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Comment	Question number	H_11	Adapted from	N/A

Question

Points $A$ ${\text{A}}$ (0 , 0 , 10) , $B$ ${\text{B}}$ (0 , 10 , 0) , $C$ ${\text{C}}$ (10 , 0 , 0) , $V$ ${\text{V}}$ ( $p$ $p$ , $p$ $p$ , $p$ $p$ ) form the vertices of a tetrahedron.

Consider the case where the faces $ABV$ ${\text{ABV}}$ and $ACV$ ${\text{ACV}}$ are perpendicular.

The following diagram shows the graph of $θ$ $\theta$ against $p$ $p$ . The maximum point is shown by $X$ ${\text{X}}$ .

Show that $- - \to AB \times - - \to AV = - 10 ⎛ ⎜ ⎝ \begin{matrix} 10 - 2 p p p \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AV}}} = - 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right)$ and find a similar expression for $- - \to AC \times - - \to AV$ $\overrightarrow {{\text{AC}}} \times \overrightarrow {{\text{AV}}}$ .

[3]

a.i.

Hence, show that, if the angle between the faces $ABV$ ${\text{ABV}}$ and $ACV$ ${\text{ACV}}$ is $θ$ $\theta$ , then $cos θ = \frac{p (3 p - 20)}{6 p^{2} - 40 p + 100}$ ${\text{cos}}\,\theta = \frac{{p\left( {3p - 20} \right)}}{{6{p^2} - 40p + 100}}$ .

[5]

a.ii.

Find the two possible coordinates of $V$ ${\text{V}}$ .

[3]

b.i.

Comment on the positions of $V$ ${\text{V}}$ in relation to the plane $ABC$ ${\text{ABC}}$ .

[1]

b.ii.

At $X$ ${\text{X}}$ , find the value of $p$ $p$ and the value of $θ$ $\theta$ .

[3]

c.i.

Find the equation of the horizontal asymptote of the graph.

[2]

c.ii.

Markscheme

$- - \to AV = ⎛ ⎜ ⎝ \begin{matrix} p p p - 10 \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AV}}} = \left( {\begin{array}{*{20}{c}} p \\ p \\ {p - 10} \end{array}} \right)$ A1

$- - \to AB \times - - \to AV = ⎛ ⎜ ⎝ \begin{matrix} 010 - 10 \end{matrix} ⎞ ⎟ ⎠ \times ⎛ ⎜ ⎝ \begin{matrix} p p p - 10 \end{matrix} ⎞ ⎟ ⎠ = ⎛ ⎜ ⎝ \begin{matrix} 10 (p - 10) + 10 p - 10 p - 10 p \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AV}}} = \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { - 10} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} p \\ p \\ {p - 10} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {10\left( {p - 10} \right) + 10p} \\ { - 10p} \\ { - 10p} \end{array}} \right)$ A1

$= ⎛ ⎜ ⎝ \begin{matrix} 20 p - 100 - 10 p - 10 p \end{matrix} ⎞ ⎟ ⎠ = - 10 ⎛ ⎜ ⎝ \begin{matrix} 10 - 2 p p p \end{matrix} ⎞ ⎟ ⎠$ $= \left( {\begin{array}{*{20}{c}} {20p - 100} \\ { - 10p} \\ { - 10p} \end{array}} \right) = - 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right)$ AG

$- - \to AC \times - - \to AV = ⎛ ⎜ ⎝ \begin{matrix} 100 - 10 \end{matrix} ⎞ ⎟ ⎠ \times ⎛ ⎜ ⎝ \begin{matrix} p p p - 10 \end{matrix} ⎞ ⎟ ⎠ = ⎛ ⎜ ⎝ \begin{matrix} 10 p 100 - 20 p 10 p \end{matrix} ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ = 10 ⎛ ⎜ ⎝ \begin{matrix} p 10 - 2 p p \end{matrix} ⎞ ⎟ ⎠ ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AC}}} \times \overrightarrow {{\text{AV}}} = \left( {\begin{array}{*{20}{c}} {10} \\ 0 \\ { - 10} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} p \\ p \\ {p - 10} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {10p} \\ {100 - 20p} \\ {10p} \end{array}} \right)\left( { = 10\left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right)} \right)$ A1

[3 marks]

a.i.

attempt to find a scalar product M1

$- 10 ⎛ ⎜ ⎝ \begin{matrix} 10 - 2 p p p \end{matrix} ⎞ ⎟ ⎠ ∙ 10 ⎛ ⎜ ⎝ \begin{matrix} p 10 - 2 p p \end{matrix} ⎞ ⎟ ⎠ = 100 (3 p^{2} - 20 p)$ $- 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right) \bullet 10\left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right) = 100\left( {3{p^2} - 20p} \right)$

OR $- ⎛ ⎜ ⎝ \begin{matrix} 10 - 2 p p p \end{matrix} ⎞ ⎟ ⎠ ∙ ⎛ ⎜ ⎝ \begin{matrix} p 10 - 2 p p \end{matrix} ⎞ ⎟ ⎠ = 3 p^{2} - 20 p$ $- \left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right) = 3{p^2} - 20p$ A1

attempt to find magnitude of either $- - \to AB \times - - \to AV$ $\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AV}}}$ or $- - \to AC \times - - \to AV$ $\overrightarrow {{\text{AC}}} \times \overrightarrow {{\text{AV}}}$ M1

$∣ ∣ ∣ ∣ - 10 ⎛ ⎜ ⎝ \begin{matrix} 10 - 2 p p p \end{matrix} ⎞ ⎟ ⎠ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 10 ⎛ ⎜ ⎝ \begin{matrix} p 10 - 2 p p \end{matrix} ⎞ ⎟ ⎠ ∣ ∣ ∣ ∣ = 10 \sqrt{{(10 - 2 p)}^{2} + 2 p^{2}}$ $\left| { - 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right)} \right| = \left| {10\left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right)} \right| = 10\sqrt {{{\left( {10 - 2p} \right)}^2} + 2{p^2}}$ A1

$100 (3 p^{2} - 20 p) = 100 {(\sqrt{{(10 - 2 p)}^{2} + 2 p^{2}})}^{2} cos θ$ $100\left( {3{p^2} - 20p} \right) = 100{\left( {\sqrt {{{\left( {10 - 2p} \right)}^2} + 2{p^2}} } \right)^2}{\text{cos}}\,\theta$

$cos θ = \frac{3 p^{2} - 20 p}{{(10 - 2 p)}^{2} + 2 p^{2}}$ ${\text{cos}}\,\theta = \frac{{3{p^2} - 20p}}{{{{\left( {10 - 2p} \right)}^2} + 2{p^2}}}$ A1

Note: Award A1 for any intermediate step leading to the correct answer.

$= \frac{p (3 p - 20)}{6 p^{2} - 40 p + 100}$ $= \frac{{p\left( {3p - 20} \right)}}{{6{p^2} - 40p + 100}}$ AG

Note: Do not allow FT marks from part (a)(i).

[8 marks]

a.ii.

$p (3 p - 20) = 0 \Rightarrow p = 0$ $p\left( {3p - 20} \right) = 0 \Rightarrow p = 0$ or $p = \frac{20}{3}$ $p = \frac{{20}}{3}$ M1A1

coordinates are (0, 0, 0) and $(\frac{20}{3}, \frac{20}{3}, \frac{20}{3})$ $\left( {\frac{{20}}{3}{\text{, }}\frac{{20}}{3}{\text{, }}\frac{{20}}{3}} \right)$ A1

Note: Do not allow column vectors for the final A mark.

[3 marks]

b.i.

two points are mirror images in the plane
or opposite sides of the plane
or equidistant from the plane
or the line connecting the two Vs is perpendicular to the plane R1

[1 mark]

b.ii.

geometrical consideration or attempt to solve $- 1 = \frac{p (3 p - 20)}{6 p^{2} - 40 p + 100}$ $- 1 = \frac{{p\left( {3p - 20} \right)}}{{6{p^2} - 40p + 100}}$ (M1)