| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.3 This is a straightforward application of perpendicularity conditions using dot products. Part (i) requires setting up two simultaneous equations from dot product = 0, solving for b and c, then verifying unit vector property. Part (ii) is a standard angle calculation using the scalar product formula. All techniques are routine for C4 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt scalar prod \(\{\mathbf{u}\cdot(4\mathbf{i}+\mathbf{k})\) or \(\mathbf{u}\cdot(4\mathbf{i}+3\mathbf{j}+2\mathbf{k})\}=0\) | M1 | where \(\mathbf{u}\) is the given vector |
| Obtain \(\frac{12}{13}+c=0\) or \(\frac{12}{13}+3b+2c=0\) | A1 | |
| \(c=-\frac{12}{13}\) | A1 | |
| \(b=\frac{4}{13}\) | A1 | cao No ft |
| Evaluate \(\left(\frac{3}{13}\right)^2+(\text{their }b)^2+(\text{their }c)^2\) | M1 | Ignore non-mention of \(\sqrt{}\) |
| Obtain \(\frac{9}{169}+\frac{144}{169}+\frac{16}{169}=1\) AG | A1 6 | Ignore non-mention of \(\sqrt{}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\cos\theta=\frac{\mathbf{x}\cdot\mathbf{y}}{ | \mathbf{x} | |
| Correct method for finding scalar product | M1 | |
| \(36°\) (35.837653…); Accept 0.625 (rad) | A1 3 | From \(\frac{18}{\sqrt{17}\sqrt{29}}\) |
# Question 7:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt scalar prod $\{\mathbf{u}\cdot(4\mathbf{i}+\mathbf{k})$ or $\mathbf{u}\cdot(4\mathbf{i}+3\mathbf{j}+2\mathbf{k})\}=0$ | M1 | where $\mathbf{u}$ is the given vector |
| Obtain $\frac{12}{13}+c=0$ or $\frac{12}{13}+3b+2c=0$ | A1 | |
| $c=-\frac{12}{13}$ | A1 | |
| $b=\frac{4}{13}$ | A1 | cao No ft |
| Evaluate $\left(\frac{3}{13}\right)^2+(\text{their }b)^2+(\text{their }c)^2$ | M1 | Ignore non-mention of $\sqrt{}$ |
| Obtain $\frac{9}{169}+\frac{144}{169}+\frac{16}{169}=1$ AG | A1 **6** | Ignore non-mention of $\sqrt{}$ |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\cos\theta=\frac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}||\mathbf{y}|}$ | M1 | |
| Correct method for finding scalar product | M1 | |
| $36°$ (35.837653…); Accept 0.625 (rad) | A1 **3** | From $\frac{18}{\sqrt{17}\sqrt{29}}$ |
---(i) The vector $\mathbf { u } = \frac { 3 } { 13 } \mathbf { i } + b \mathbf { j } + c \mathbf { k }$ is perpendicular to the vector $4 \mathbf { i } + \mathbf { k }$ and to the vector $4 \mathbf { i } + 3 \mathbf { j } + 2 \mathbf { k }$. Find the values of $b$ and $c$, and show that $\mathbf { u }$ is a unit vector.\\
(ii) Calculate, to the nearest degree, the angle between the vectors $4 \mathbf { i } + \mathbf { k }$ and $4 \mathbf { i } + 3 \mathbf { j } + 2 \mathbf { k }$.
\hfill \mbox{\textit{OCR C4 2009 Q7 [9]}}