# IB DP Math AA : Topic : AHL 3.16: Geometric interpretation of v×w . HL Paper 2

## Question

The diagram shows a cube OABCDEFG.

Let O be the origin, (OA) the x-axis, (OC) the y-axis and (OD) the z-axis.

Let M, N and P be the midpoints of [FG], [DG] and [CG], respectively.

The coordinates of F are (2, 2, 2).

(a)     Find the position vectors $$\overrightarrow {{\text{OM}}}$$, $$\overrightarrow {{\text{ON}}}$$ and $$\overrightarrow {{\text{OP}}}$$ in component form.

(b)     Find $$\overrightarrow {{\text{MP}}} \times \overrightarrow {{\text{MN}}}$$.

(c)     Hence,

(i)     calculate the area of the triangle MNP;

(ii)     show that the line (AG) is perpendicular to the plane MNP;

(iii)     find the equation of the plane MNP.

(d)     Determine the coordinates of the point where the line (AG) meets the plane MNP.

## Markscheme

(a)     $$\overrightarrow {{\text{OM}}} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 2 \end{array}} \right)$$, $$\overrightarrow {{\text{ON}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 2 \end{array}} \right)$$ and $$\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ 1 \end{array}} \right)$$     A1A1A1

[3 marks]

(b)     $$\overrightarrow {{\text{MP}}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ 0 \\ { – 1} \end{array}} \right)$$ and $$\overrightarrow {{\text{MN}}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ { – 1} \\ 0 \end{array}} \right)$$     A1A1

$$\overrightarrow {{\text{MP}}} \times \overrightarrow {{\text{MN}}} = \left( {\begin{array}{*{20}{c}} i&j&k \\ { – 1}&0&{ – 1} \\ { – 1}&{ – 1}&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 1 \end{array}} \right)$$     (M1)A1

[4 marks]

(c)     (i)     area of MNP $$= \frac{1}{2}\left| {\overrightarrow {{\text{MP}}} \times \overrightarrow {{\text{MN}}} } \right|$$     M1

$$= \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 1 \end{array}} \right)} \right|$$

$$= \frac{{\sqrt 3 }}{2}$$     A1

(ii)     $$\overrightarrow {{\text{OA}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 0 \end{array}} \right)$$, $$\overrightarrow {{\text{OG}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ 2 \end{array}} \right)$$

$$\overrightarrow {{\text{AG}}} = \left( {\begin{array}{*{20}{c}} { – 2} \\ 2 \\ 2 \end{array}} \right)$$     A1

since $$\overrightarrow {{\text{AG}}} = 2(\overrightarrow {{\text{MP}}} \times \overrightarrow {{\text{MN}}} )$$ AG is perpendicular to MNP     R1

(iii)     $$r \cdot \left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 2 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 1 \end{array}} \right)$$     M1A1

$$r \cdot \left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 1 \end{array}} \right) = 3$$ (accept $$– x + y + z = 3$$)     A1

[7 marks]

(d)     $$r = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 0 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 2} \\ 2 \\ 2 \end{array}} \right)$$     A1

$$\left( {\begin{array}{*{20}{c}} {2 – 2\lambda } \\ {2\lambda } \\ {2\lambda } \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 1 \end{array}} \right) = 3$$     M1A1

$$– 2 + 2\lambda + 2\lambda + 2\lambda = 3$$

$$\lambda = \frac{5}{6}$$     A1

$$r = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 0 \end{array}} \right) + \frac{5}{6}\left( {\begin{array}{*{20}{c}} { – 2} \\ 2 \\ 2 \end{array}} \right)$$     M1

coordinates of point $$\left( {\frac{1}{3},\frac{5}{3},\frac{5}{3}} \right)$$     A1

[6 marks]

Total [20 marks]

## Examiners report

This was the most successfully answered question in part B, with many candidates achieving full marks. There were a few candidates who misread the question and treated the cube as a unit cube. The most common errors were either algebraic or arithmetic mistakes. A variety of notation forms were seen but in general were used consistently. In a few cases, candidates failed to show all the work or set it properly.

Scroll to Top