CAIE FP1 2013 November — Question 8 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with plane
DifficultyStandard +0.3 This is a standard Further Maths vectors question requiring routine techniques: finding a plane equation via cross product of two vectors in the plane, finding line-plane intersection by substituting parametric equations, and calculating angle between line and plane using dot product formula. All steps are algorithmic with no novel insight required, making it slightly easier than average for FM content.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point

8 The points \(A , B , C\) have position vectors $$4 \mathbf { i } + 5 \mathbf { j } + 6 \mathbf { k } , \quad 5 \mathbf { i } + 7 \mathbf { j } + 8 \mathbf { k } , \quad 2 \mathbf { i } + 6 \mathbf { j } + 4 \mathbf { k }$$ respectively, relative to the origin \(O\). Find a cartesian equation of the plane \(A B C\). The point \(D\) has position vector \(6 \mathbf { i } + 3 \mathbf { j } + 6 \mathbf { k }\). Find the coordinates of \(E\), the point of intersection of the line \(O D\) with the plane \(A B C\). Find the acute angle between the line \(E D\) and the plane \(A B C\).

Question 8:
Part (finding normal and Cartesian equation):
AnswerMarks Guidance
\(\overrightarrow{AB} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\), \(\overrightarrow{AC} = -2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\)
\(\overrightarrow{AB} \times \overrightarrow{AC} = -6\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\)M1A1
\(6x + 2y - 5z = \text{constant} = 24 + 10 - 30 = 4\)M1A1 (4)
Part (finding where OD meets plane):
AnswerMarks Guidance
Equation of \(OD\): \(\mathbf{r} = t(6\mathbf{i} + 3\mathbf{j} + 6\mathbf{k})\)B1
\(\Rightarrow 36t + 6t - 30t = 12t = 4 \Rightarrow t = \frac{1}{3}\)M1A1
\(E\) is the point \((2,1,2)\)A1 (4)
Part (angle between planes):
AnswerMarks Guidance
Using \((6\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}) \cdot 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\)
\(\Rightarrow 12 + 2 - 10 = \sqrt{36+4+25}\sqrt{4+1+4}\sin\theta\)M1A1
\(\Rightarrow \theta = 9.5°\) (0.166 rad)A1 (3)
# Question 8:

## Part (finding normal and Cartesian equation):
| $\overrightarrow{AB} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$, $\overrightarrow{AC} = -2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ | | |
|---|---|---|
| $\overrightarrow{AB} \times \overrightarrow{AC} = -6\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}$ | M1A1 | |
| $6x + 2y - 5z = \text{constant} = 24 + 10 - 30 = 4$ | M1A1 | **(4)** |

## Part (finding where OD meets plane):
| Equation of $OD$: $\mathbf{r} = t(6\mathbf{i} + 3\mathbf{j} + 6\mathbf{k})$ | B1 | |
|---|---|---|
| $\Rightarrow 36t + 6t - 30t = 12t = 4 \Rightarrow t = \frac{1}{3}$ | M1A1 | |
| $E$ is the point $(2,1,2)$ | A1 | **(4)** |

## Part (angle between planes):
| Using $(6\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}) \cdot 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$ | | |
|---|---|---|
| $\Rightarrow 12 + 2 - 10 = \sqrt{36+4+25}\sqrt{4+1+4}\sin\theta$ | M1A1 | |
| $\Rightarrow \theta = 9.5°$ (0.166 rad) | A1 | **(3)** |

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8 The points $A , B , C$ have position vectors

$$4 \mathbf { i } + 5 \mathbf { j } + 6 \mathbf { k } , \quad 5 \mathbf { i } + 7 \mathbf { j } + 8 \mathbf { k } , \quad 2 \mathbf { i } + 6 \mathbf { j } + 4 \mathbf { k }$$

respectively, relative to the origin $O$. Find a cartesian equation of the plane $A B C$.

The point $D$ has position vector $6 \mathbf { i } + 3 \mathbf { j } + 6 \mathbf { k }$. Find the coordinates of $E$, the point of intersection of the line $O D$ with the plane $A B C$.

Find the acute angle between the line $E D$ and the plane $A B C$.

\hfill \mbox{\textit{CAIE FP1 2013 Q8 [11]}}