Question 10 - A-Level Maths

CAIE P3 2008 June — Question 10 10 marks

Exam Board	CAIE
Module	P3 (Pure Mathematics 3)
Year	2008
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Find parameter value for geometric condition
Difficulty	Standard +0.8 This is a multi-part Further Maths vectors question requiring: (i) showing skew lines via non-intersection (testing direction vectors and solving simultaneous equations), and (ii) using the angle condition with dot product formula to derive a quadratic equation in parameter t. While it involves several standard techniques (vector equations, dot products, angle formulas), the combination and the geometric setup with the 60° angle condition requires solid understanding and careful algebraic manipulation across multiple steps, placing it moderately above average difficulty.
Spec	1.10b Vectors in 3D: i,j,k notation 1.10c Magnitude and direction: of vectors 1.10f Distance between points: using position vectors

10 The points $A$ and $B$ have position vectors, relative to the origin $O$, given by $$\overrightarrow { O A } = \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } .$$ The line $l$ has vector equation $$\mathbf { r } = ( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }$$

Show that $l$ does not intersect the line passing through $A$ and $B$.
The point $P$ lies on $l$ and is such that angle $P A B$ is equal to $60 ^ { \circ }$. Given that the position vector of $P$ is $( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }$, show that $3 t ^ { 2 } + 7 t + 2 = 0$. Hence find the only possible position vector of $P$.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(i) State a vector equation for the line through $A$ and $B$, e.g. $\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + s(\mathbf{i} - \mathbf{j})$	B1
Equate at least two pairs of components of general points on $AB$ and $l$, and solve for $s$ or for $t$	M1
Obtain correct answer for $s$ or $t$, e.g. $s = -6, 2, -2$ when $t = 3, -1, -1$ respectively	A1
Verify that all three component equations are not satisfied	A1	[4 marks]
(ii) State or imply a direction vector for $AP$ has components $(-2t, 3 + t, -1-t)$, or equivalent	B1
State or imply \(\cos 60° = \frac{\vec{AP}.\vec{AB}}{	\vec{AP}
Carry out correct processes for expanding the scalar product and expressing the product of the moduli in terms of $t$, in order to obtain an equation in $t$ in any form	M1(dep$^*$)
Obtain the given equation $3t^2 + 7t + 2 = 0$ correctly	A1
Solve the quadratic and use a root to find a position vector for $P$	M1
Obtain position vector $5\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$ from $t = -2$, having rejected the root $t = -\frac{1}{3}$ for a valid reason	A1	[6 marks]

**(i)** State a vector equation for the line through $A$ and $B$, e.g. $\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + s(\mathbf{i} - \mathbf{j})$ | B1 |

Equate at least two pairs of components of general points on $AB$ and $l$, and solve for $s$ or for $t$ | M1 |

Obtain correct answer for $s$ or $t$, e.g. $s = -6, 2, -2$ when $t = 3, -1, -1$ respectively | A1 |

Verify that all three component equations are not satisfied | A1 | [4 marks]

**(ii)** State or imply a direction vector for $AP$ has components $(-2t, 3 + t, -1-t)$, or equivalent | B1 |

State or imply $\cos 60° = \frac{\vec{AP}.\vec{AB}}{|\vec{AP}||\vec{AB}|}$ | M1$^*$ |

Carry out correct processes for expanding the scalar product and expressing the product of the moduli in terms of $t$, in order to obtain an equation in $t$ in any form | M1(dep$^*$) |

Obtain the given equation $3t^2 + 7t + 2 = 0$ correctly | A1 |

Solve the quadratic and use a root to find a position vector for $P$ | M1 |

Obtain position vector $5\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$ from $t = -2$, having rejected the root $t = -\frac{1}{3}$ for a valid reason | A1 | [6 marks]

Show LaTeX source

10 The points $A$ and $B$ have position vectors, relative to the origin $O$, given by

$$\overrightarrow { O A } = \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } .$$

The line $l$ has vector equation

$$\mathbf { r } = ( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }$$

(i) Show that $l$ does not intersect the line passing through $A$ and $B$.\\
(ii) The point $P$ lies on $l$ and is such that angle $P A B$ is equal to $60 ^ { \circ }$. Given that the position vector of $P$ is $( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }$, show that $3 t ^ { 2 } + 7 t + 2 = 0$. Hence find the only possible position vector of $P$.

\hfill \mbox{\textit{CAIE P3 2008 Q10 [10]}}

This paper (10 questions)

View full paper

Q1 4 Q2 5 Q3 6 Q4 7 Q5 7 Q6 8 Q7 9 Q8 9 Q9 10 Q10 10

Answer	Marks	Guidance
(i) State a vector equation for the line through \(A\) and \(B\), e.g. \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + s(\mathbf{i} - \mathbf{j})\)	B1
Equate at least two pairs of components of general points on \(AB\) and \(l\), and solve for \(s\) or for \(t\)	M1
Obtain correct answer for \(s\) or \(t\), e.g. \(s = -6, 2, -2\) when \(t = 3, -1, -1\) respectively	A1
Verify that all three component equations are not satisfied	A1	[4 marks]
(ii) State or imply a direction vector for \(AP\) has components \((-2t, 3 + t, -1-t)\), or equivalent	B1
State or imply \(\cos 60° = \frac{\vec{AP}.\vec{AB}}{	\vec{AP}
Carry out correct processes for expanding the scalar product and expressing the product of the moduli in terms of \(t\), in order to obtain an equation in \(t\) in any form	M1(dep\(^*\))
Obtain the given equation \(3t^2 + 7t + 2 = 0\) correctly	A1
Solve the quadratic and use a root to find a position vector for \(P\)	M1
Obtain position vector \(5\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) from \(t = -2\), having rejected the root \(t = -\frac{1}{3}\) for a valid reason	A1	[6 marks]