OCR C4 2012 January — Question 2 7 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2012
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeVector equation of a line
DifficultyModerate -0.3 This is a straightforward two-part question testing standard vector line techniques: (i) finding direction vector from two points and writing the line equation, and (ii) using the scalar product formula to find angle between direction vectors. Both are routine C4 procedures with no problem-solving insight required, making it slightly easier than average.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles

2
  1. Find, in the form \(\mathbf { r } = \mathbf { a } + t \mathbf { b }\), an equation of the line \(l\) through the points ( \(4,2,7\) ) and ( \(5 , - 4 , - 1\) ).
  2. Find the acute angle between the line \(l\) and a line in the direction of the vector \(\left( \begin{array} { l } 1 \\ 2 \\ 3 \end{array} \right)\).

Question 2(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{a} = \begin{pmatrix}4\\2\\7\end{pmatrix}\) or \(\begin{pmatrix}5\\-4\\-1\end{pmatrix}\)B1 Accept any notation
\(\mathbf{b}\) = Difference between the two pointsM1
Provided final answer is of form \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\); \(\begin{pmatrix}1\\-6\\-8\end{pmatrix}\) or \(\begin{pmatrix}-1\\6\\8\end{pmatrix}\)A1
[3]
Question 2(ii):
AnswerMarks Guidance
AnswerMarks Guidance
Method for magnitude of any vectorM1 Accept e.g. \(\sqrt{1^2 - 6^2 - 8^2}\)
Method for scalar product of any 2 vectorsM1
Using \(\cos\theta = \frac{\mathbf{c.d}}{\mathbf{c}
\(21.4\) or better \((21.444513)\); \(0.374\) or better \((0.374277)\)A1
[4]
## Question 2(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{a} = \begin{pmatrix}4\\2\\7\end{pmatrix}$ or $\begin{pmatrix}5\\-4\\-1\end{pmatrix}$ | B1 | Accept any notation |
| $\mathbf{b}$ = Difference between the two points | M1 | |
| Provided final answer is of form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$; $\begin{pmatrix}1\\-6\\-8\end{pmatrix}$ or $\begin{pmatrix}-1\\6\\8\end{pmatrix}$ | A1 | |
| **[3]** | | |

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## Question 2(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Method for magnitude of any vector | M1 | Accept e.g. $\sqrt{1^2 - 6^2 - 8^2}$ |
| Method for scalar product of any 2 vectors | M1 | |
| Using $\cos\theta = \frac{\mathbf{c.d}}{|\mathbf{c}||\mathbf{d}|}$ for their $\mathbf{b}$ and $\begin{pmatrix}1\\2\\3\end{pmatrix}$ | M1 | |
| $21.4$ or better $(21.444513)$; $0.374$ or better $(0.374277)$ | A1 | |
| **[4]** | | |

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2 (i) Find, in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$, an equation of the line $l$ through the points ( $4,2,7$ ) and ( $5 , - 4 , - 1$ ).\\
(ii) Find the acute angle between the line $l$ and a line in the direction of the vector $\left( \begin{array} { l } 1 \\ 2 \\ 3 \end{array} \right)$.

\hfill \mbox{\textit{OCR C4 2012 Q2 [7]}}