| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a straightforward vectors question requiring dot product for perpendicularity (giving a trigonometric equation to solve) and triangle area formula. The trigonometric equation 6sin(α)cos(α) + 8sin(α)cos(α) = 3 simplifies nicely using double angle formulas, and the area calculation is routine application of ½|OA||OB|. Slightly above average due to the trigonometric manipulation required, but still a standard multi-part question with well-signposted steps. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3\sin\alpha\times 2\cos\alpha + 2\cos\alpha\times 4\sin\alpha + -1\times 3\) | M1 | allow one sign error or one coefficient error |
| \(6\sin\alpha\cos\alpha + 8\sin\alpha\cos\alpha - 3 = 0\) soi | A1 | |
| substitution of \(\sin\alpha\cos\alpha = \frac{1}{2}\sin 2\alpha\) | M1 | NB \(7\sin 2\alpha = 3\) |
| \(\alpha =\) awrt \(12.7°\) | A1 | awrt \(0.221\) |
| \(\alpha =\) awrt \(77.3°\) | A1 | awrt \(1.35\); if A0A0, SC1 for \(13°\) and \(77°\) or \(0.22\) and \(1.4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Their \(\alpha = 12.7°\) substituted in \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\); or in \( | \overrightarrow{OA} | \) and \( |
| \(\sqrt{(3\sin\alpha)^2 + (2\cos\alpha)^2 + (-1)^2}\) or \(\sqrt{(2\cos\alpha)^2 + (4\sin\alpha)^2 + 3^2}\) | M1* | Allow omission of brackets, one slip in arithmetic and one sign error |
| \(\frac{1}{2}\sqrt{9\sin^2\alpha + 4\cos^2\alpha + 1}\sqrt{4\cos^2\alpha + 16\sin^2\alpha + 9}\) | M1*dep | May be implied by numerical value for lengths; allow one sign or coefficient error; \(\alpha\) may be unspecified or any acute angle for these method marks |
| awrt 4.22 | A1 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\sin\alpha\times 2\cos\alpha + 2\cos\alpha\times 4\sin\alpha + -1\times 3$ | M1 | allow one sign error or one coefficient error |
| $6\sin\alpha\cos\alpha + 8\sin\alpha\cos\alpha - 3 = 0$ soi | A1 | |
| substitution of $\sin\alpha\cos\alpha = \frac{1}{2}\sin 2\alpha$ | M1 | NB $7\sin 2\alpha = 3$ |
| $\alpha =$ awrt $12.7°$ | A1 | awrt $0.221$ |
| $\alpha =$ awrt $77.3°$ | A1 | awrt $1.35$; if A0A0, SC1 for $13°$ and $77°$ or $0.22$ and $1.4$ |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Their $\alpha = 12.7°$ substituted in $\overrightarrow{OA}$ and $\overrightarrow{OB}$; or in $|\overrightarrow{OA}|$ and $|\overrightarrow{OB}|$ | M1 | |
| $\sqrt{(3\sin\alpha)^2 + (2\cos\alpha)^2 + (-1)^2}$ or $\sqrt{(2\cos\alpha)^2 + (4\sin\alpha)^2 + 3^2}$ | M1* | Allow omission of brackets, one slip in arithmetic and one sign error |
| $\frac{1}{2}\sqrt{9\sin^2\alpha + 4\cos^2\alpha + 1}\sqrt{4\cos^2\alpha + 16\sin^2\alpha + 9}$ | M1*dep | May be implied by numerical value for lengths; allow one sign or coefficient error; $\alpha$ may be unspecified or any acute angle for these method marks |
| awrt 4.22 | A1 | |
---8 The points $A$ and $B$ have position vectors relative to the origin $O$ given by
$$\overrightarrow { O A } = \left( \begin{array} { c }
3 \sin \alpha \\
2 \cos \alpha \\
- 1
\end{array} \right) \text { and } \overrightarrow { O B } = \left( \begin{array} { c }
2 \cos \alpha \\
4 \sin \alpha \\
3
\end{array} \right)$$
where $0 ^ { \circ } < \alpha < 90 ^ { \circ }$. It is given that $\overrightarrow { O A }$ and $\overrightarrow { O B }$ are perpendicular.\\
(i) Calculate the two possible values of $\alpha$.\\
(ii) Calculate the area of triangle $O A B$ for the smaller value of $\alpha$ from part (i).
\hfill \mbox{\textit{OCR C4 2016 Q8 [9]}}