Question 2 - A-Level Maths

CAIE FP1 2011 November — Question 2 5 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2011
Session	November
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Volume of tetrahedron using scalar triple product
Difficulty	Standard +0.3 This is a straightforward application of the cross product formula and scalar triple product for volume. Students must compute a×b using the determinant method (routine), find its magnitude for the triangle area (standard), then use the given volume formula with the scalar triple product. While it involves multiple steps, each is a direct application of learned techniques with no novel insight required, making it slightly easier than average.
Spec	4.04g Vector product: a x b perpendicular vector

2 The position vectors of points $A , B , C$, relative to the origin $O$, are $\mathbf { a } , \mathbf { b } , \mathbf { c }$, where $$\mathbf { a } = 3 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } , \quad \mathbf { b } = 4 \mathbf { i } - 3 \mathbf { j } + 2 \mathbf { k } , \quad \mathbf { c } = 3 \mathbf { i } - \mathbf { j } - \mathbf { k }$$ Find $\mathbf { a } \times \mathbf { b }$ and deduce the area of the triangle $O A B$. Hence find the volume of the tetrahedron $O A B C$, given that the volume of a tetrahedron is $\frac { 1 } { 3 } \times$ area of base × perpendicular height.

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Question 2:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 4 & -3 & 2 \end{vmatrix} = \mathbf{i} - 10\mathbf{j} - 17\mathbf{k}$	M1 A1	Finds $\mathbf{a} \times \mathbf{b}$
Area of base $= \frac{1}{2}\sqrt{1^2 + (-10)^2 + (-17)^2} = \frac{1}{2}\sqrt{390}$ $(= 9.87)$	A1	Finds area of base; Part mark: 3
Height $= \frac{(3\mathbf{i}-\mathbf{j}-\mathbf{k})\cdot(\mathbf{i}-10\mathbf{j}-17\mathbf{k})}{\sqrt{1^2+(-10)^2+(-17)^2}} = \frac{30}{\sqrt{390}}$ $(= 1.519)$	M1	Attempts to find height
Volume $= \frac{1}{3} \times \frac{1}{2}\sqrt{390} \times \frac{30}{\sqrt{390}} = 5$	A1	Finds volume; Part mark: 2; Total: 5

# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{a} \times \mathbf{b} = \begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 4 & -3 & 2 \end{vmatrix} = \mathbf{i} - 10\mathbf{j} - 17\mathbf{k}$ | M1 A1 | Finds $\mathbf{a} \times \mathbf{b}$ |
| Area of base $= \frac{1}{2}\sqrt{1^2 + (-10)^2 + (-17)^2} = \frac{1}{2}\sqrt{390}$ $(= 9.87)$ | A1 | Finds area of base; Part mark: 3 |
| Height $= \frac{(3\mathbf{i}-\mathbf{j}-\mathbf{k})\cdot(\mathbf{i}-10\mathbf{j}-17\mathbf{k})}{\sqrt{1^2+(-10)^2+(-17)^2}} = \frac{30}{\sqrt{390}}$ $(= 1.519)$ | M1 | Attempts to find height |
| Volume $= \frac{1}{3} \times \frac{1}{2}\sqrt{390} \times \frac{30}{\sqrt{390}} = 5$ | A1 | Finds volume; Part mark: 2; Total: 5 |

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2 The position vectors of points $A , B , C$, relative to the origin $O$, are $\mathbf { a } , \mathbf { b } , \mathbf { c }$, where

$$\mathbf { a } = 3 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } , \quad \mathbf { b } = 4 \mathbf { i } - 3 \mathbf { j } + 2 \mathbf { k } , \quad \mathbf { c } = 3 \mathbf { i } - \mathbf { j } - \mathbf { k }$$

Find $\mathbf { a } \times \mathbf { b }$ and deduce the area of the triangle $O A B$.

Hence find the volume of the tetrahedron $O A B C$, given that the volume of a tetrahedron is $\frac { 1 } { 3 } \times$ area of base × perpendicular height.

\hfill \mbox{\textit{CAIE FP1 2011 Q2 [5]}}

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Q1 5 Q2 5 Q3 7 Q4 7 Q5 7 Q6 8 Q7 11 Q8 11 Q9 12 Q10 13 Q11 EITHER Q11 OR