Edexcel C4 — Question 4 9 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeLine intersection: unknown constant then intersect
DifficultyStandard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: finding a direction vector from two points, writing a vector equation, substituting a point to find constants, and using the scalar product formula for angles. All parts are routine C4 material with no novel problem-solving required, making it slightly easier than average.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors

4. The line \(l _ { 1 }\) passes through the points \(P\) and \(Q\) with position vectors ( \(- \mathbf { i } - 8 \mathbf { j } + 3 \mathbf { k }\) ) and ( \(2 \mathbf { i } - 9 \mathbf { j } + \mathbf { k }\) ) respectively, relative to a fixed origin.
  1. Find a vector equation for \(l _ { 1 }\). The line \(l _ { 2 }\) has the equation $$\mathbf { r } = ( 6 \mathbf { i } + a \mathbf { j } + b \mathbf { k } ) + \mu ( \mathbf { i } + 4 \mathbf { j } - \mathbf { k } )$$ and also passes through the point \(Q\).
  2. Find the values of the constants \(a\) and \(b\).
  3. Find, in degrees to 1 decimal place, the acute angle between lines \(l _ { 1 }\) and \(l _ { 2 }\).
    4. continued

AnswerMarks Guidance
(a) \(\overrightarrow{PQ} = (2\mathbf{i} - 9\mathbf{j} + \mathbf{k}) - (-\mathbf{i} - 8\mathbf{j} + 3\mathbf{k}) = (3\mathbf{i} - \mathbf{j} - 2\mathbf{k})\)M1
\(\therefore \mathbf{r} = (-\mathbf{i} - 8\mathbf{j} + 3\mathbf{k}) + \lambda(3\mathbf{i} - \mathbf{j} - 2\mathbf{k})\)A1
(b) \(6 + \mu = 2 \Rightarrow \mu = -4\)M1
\(a + 4\mu = -9 \Rightarrow a = 7\)A1
\(b - \mu = 1 \Rightarrow b = -3\)A1
(c) \(= \cos^{-1} \left\vert \frac{3 \times 1 + (-1) \times 4 + (-2) \times (-1)}{\sqrt{9+1+4} \times \sqrt{1+16+1}} \right\vert = \cos^{-1} \frac{1}{\sqrt{14}\sqrt{18}} = 86.4°\) (1dp)M1 A1 (9)
**(a)** $\overrightarrow{PQ} = (2\mathbf{i} - 9\mathbf{j} + \mathbf{k}) - (-\mathbf{i} - 8\mathbf{j} + 3\mathbf{k}) = (3\mathbf{i} - \mathbf{j} - 2\mathbf{k})$ | M1 |

$\therefore \mathbf{r} = (-\mathbf{i} - 8\mathbf{j} + 3\mathbf{k}) + \lambda(3\mathbf{i} - \mathbf{j} - 2\mathbf{k})$ | A1 |

**(b)** $6 + \mu = 2 \Rightarrow \mu = -4$ | M1 |

$a + 4\mu = -9 \Rightarrow a = 7$ | A1 |

$b - \mu = 1 \Rightarrow b = -3$ | A1 |

**(c)** $= \cos^{-1} \left\vert \frac{3 \times 1 + (-1) \times 4 + (-2) \times (-1)}{\sqrt{9+1+4} \times \sqrt{1+16+1}} \right\vert = \cos^{-1} \frac{1}{\sqrt{14}\sqrt{18}} = 86.4°$ (1dp) | M1 A1 | (9)
4. The line $l _ { 1 }$ passes through the points $P$ and $Q$ with position vectors ( $- \mathbf { i } - 8 \mathbf { j } + 3 \mathbf { k }$ ) and ( $2 \mathbf { i } - 9 \mathbf { j } + \mathbf { k }$ ) respectively, relative to a fixed origin.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for $l _ { 1 }$.

The line $l _ { 2 }$ has the equation

$$\mathbf { r } = ( 6 \mathbf { i } + a \mathbf { j } + b \mathbf { k } ) + \mu ( \mathbf { i } + 4 \mathbf { j } - \mathbf { k } )$$

and also passes through the point $Q$.
\item Find the values of the constants $a$ and $b$.
\item Find, in degrees to 1 decimal place, the acute angle between lines $l _ { 1 }$ and $l _ { 2 }$.\\

4. continued
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4  Q4 [9]}}