CAIE P1 2018 November — Question 6 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeVector geometry in 3D shapes
DifficultyStandard +0.3 This is a straightforward 3D vector problem requiring students to establish coordinates, form two vectors, apply the scalar product formula, and find an angle. While it involves multiple steps and 3D visualization, it uses standard techniques (dot product for angles) with no conceptual surprises, making it slightly easier than average for A-level.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors

6 \includegraphics[max width=\textwidth, alt={}, center]{d5d94eb8-7f41-4dff-b503-8be4f20e21b7-08_743_897_260_623} The diagram shows a solid figure \(O A B C D E F G\) with a horizontal rectangular base \(O A B C\) in which \(O A = 8\) units and \(A B = 6\) units. The rectangle \(D E F G\) lies in a horizontal plane and is such that \(D\) is 7 units vertically above \(O\) and \(D E\) is parallel to \(O A\). The sides \(D E\) and \(D G\) have lengths 4 units and 2 units respectively. Unit vectors \(\mathbf { i } , \mathbf { j }\) and \(\mathbf { k }\) are parallel to \(O A , O C\) and \(O D\) respectively. Use a scalar product to find angle \(O B F\), giving your answer in the form \(\cos ^ { - 1 } \left( \frac { a } { b } \right)\), where \(a\) and \(b\) are integers.

Question 6:
AnswerMarks Guidance
\((\mathbf{BO}) = -8\mathbf{i} - 6\mathbf{j}\)B1 OR \((\mathbf{OB}) = 8\mathbf{i} + 6\mathbf{j}\)
\((\mathbf{BF}) = -6\mathbf{j} - 8\mathbf{i} + 7\mathbf{k} + 4\mathbf{i} + 2\mathbf{j} = -4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\)B1 OR \((\mathbf{FB}) = 4\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}\)
\((\mathbf{BF.BO}) = (-4)(-8) + (-4)(-6)\)M1 OR \((\mathbf{FB.OB})\). Expect 56. Accept one reversed but award final A0
\(\mathbf{BF} \times
Angle \(OBF = \cos^{-1}\left(\frac{\text{their }56}{\text{their }90}\right) = \cos^{-1}\left(\frac{56}{90}\right)\) or \(\cos^{-1}\left(\frac{28}{45}\right)\)DM1A1 All M marks dependent on use of \((\pm)\mathbf{BO}\) and \((\pm)\mathbf{BF}\)
## Question 6:

| $(\mathbf{BO}) = -8\mathbf{i} - 6\mathbf{j}$ | B1 | OR $(\mathbf{OB}) = 8\mathbf{i} + 6\mathbf{j}$ |
|---|---|---|
| $(\mathbf{BF}) = -6\mathbf{j} - 8\mathbf{i} + 7\mathbf{k} + 4\mathbf{i} + 2\mathbf{j} = -4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}$ | B1 | OR $(\mathbf{FB}) = 4\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$ |
| $(\mathbf{BF.BO}) = (-4)(-8) + (-4)(-6)$ | M1 | OR $(\mathbf{FB.OB})$. Expect 56. Accept one reversed but award final A0 |
| $|\mathbf{BF}| \times |\mathbf{BO}| = \sqrt{4^2+4^2+7^2} \times \sqrt{8^2+6^2}$ | M1 | Expect 90. At least one magnitude methodically correct |
| Angle $OBF = \cos^{-1}\left(\frac{\text{their }56}{\text{their }90}\right) = \cos^{-1}\left(\frac{56}{90}\right)$ or $\cos^{-1}\left(\frac{28}{45}\right)$ | DM1A1 | All M marks dependent on use of $(\pm)\mathbf{BO}$ and $(\pm)\mathbf{BF}$ |

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\includegraphics[max width=\textwidth, alt={}, center]{d5d94eb8-7f41-4dff-b503-8be4f20e21b7-08_743_897_260_623}

The diagram shows a solid figure $O A B C D E F G$ with a horizontal rectangular base $O A B C$ in which $O A = 8$ units and $A B = 6$ units. The rectangle $D E F G$ lies in a horizontal plane and is such that $D$ is 7 units vertically above $O$ and $D E$ is parallel to $O A$. The sides $D E$ and $D G$ have lengths 4 units and 2 units respectively. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O C$ and $O D$ respectively. Use a scalar product to find angle $O B F$, giving your answer in the form $\cos ^ { - 1 } \left( \frac { a } { b } \right)$, where $a$ and $b$ are integers.\\

\hfill \mbox{\textit{CAIE P1 2018 Q6 [6]}}