| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a standard C4 vectors question testing perpendicularity (dot product = 0) and intersection conditions (equating position vectors). Part (i) is routine application of perpendicularity; part (ii) requires solving simultaneous equations from vector components, which is slightly more involved but still follows standard procedures with no novel insight required. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| State/imply scalar product of any two vectors \(= 0\) | M1 | |
| Scalar product of correct two vectors \(= 4 + 2a - 6\) | A1 | \((4+2a-6=0 \rightarrow\) M1A1) |
| \(a = 1\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempt to produce at least two relevant equations | M1 | e.g. \(2t = 3+2s\ldots\) |
| Solve two not containing '\(a\)' for \(s\) and \(t\) | M1 | |
| Obtain at least one of \(s = -\frac{1}{2}\), \(t=1\) | A1 | |
| Substitute in third equation & produce \(a = -2\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Method for finding magnitude of any vector | M1 | possibly involving '\(a\)' |
| Using \(\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{ | \mathbf{a} | |
| \(107, 108\) (\(107.548\)) or \(72, 73, 72.4, 72.5\) (\(72.4516\)) c.a.o. | A1 | 3 marks \(1.87, 1.88\) (\(1.87707\)) or \(1.26\) |
# Question 6(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| State/imply scalar product of any two vectors $= 0$ | M1 | |
| Scalar product of correct two vectors $= 4 + 2a - 6$ | A1 | $(4+2a-6=0 \rightarrow$ M1A1) |
| $a = 1$ | A1 | **3 marks** |
# Question 6(ii)(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to produce at least two relevant equations | M1 | e.g. $2t = 3+2s\ldots$ |
| Solve two not containing '$a$' for $s$ and $t$ | M1 | |
| Obtain at least one of $s = -\frac{1}{2}$, $t=1$ | A1 | |
| Substitute in third equation & produce $a = -2$ | A1 | **4 marks** |
# Question 6(ii)(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Method for finding magnitude of any vector | M1 | possibly involving '$a$' |
| Using $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ for the pair of direction vectors | M1 | possibly involving '$a$' |
| $107, 108$ ($107.548$) or $72, 73, 72.4, 72.5$ ($72.4516$) c.a.o. | A1 | **3 marks** $1.87, 1.88$ ($1.87707$) or $1.26$ |
**Total: 10 marks**
---6 Lines $l _ { 1 }$ and $l _ { 2 }$ have vector equations
$$\mathbf { r } = \mathbf { j } + \mathbf { k } + t ( 2 \mathbf { i } + a \mathbf { j } + \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = 3 \mathbf { i } - \mathbf { k } + s ( 2 \mathbf { i } + 2 \mathbf { j } - 6 \mathbf { k } )$$
respectively, where $t$ and $s$ are parameters and $a$ is a constant.\\
(i) Given that $l _ { 1 }$ and $l _ { 2 }$ are perpendicular, find the value of $a$.\\
(ii) Given instead that $l _ { 1 }$ and $l _ { 2 }$ intersect, find
\begin{enumerate}[label=(\alph*)]
\item the value of $a$,
\item the angle between the lines.
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2010 Q6 [10]}}