| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from origin to line |
| Difficulty | Standard +0.3 This is a standard C4 vectors question testing routine techniques: finding a point from a parameter, calculating angles using dot product, finding perpendicular from origin to a line using the perpendicularity condition, and computing magnitude. All steps are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\) is \(\begin{pmatrix}3\\1\\1\end{pmatrix} + \begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}4\\0\\3\end{pmatrix}\) | B1 | |
| direction vector of \(\ell\) is \(\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and of \(\overrightarrow{OP}\) is their \(P\) | \(\sqrt{}\)B1 | |
| Use \(\cos\theta = \frac{\mathbf{a.b}}{ | \mathbf{a} | |
| \(\theta = 35.3°\) or better \((0.615...\) rad\()\) | A1 |
| Answer | Marks |
|---|---|
| Use \(\begin{pmatrix}1\\-1\\2\end{pmatrix} \cdot \begin{pmatrix}3+t\\1-t\\1+2t\end{pmatrix} = 0\) | M1 |
| \(1(3+t) - 1(1-t) + 2(1+2t) = 0\) | A1 |
| \(t = -\frac{2}{3}\) | A1 |
| Subst. into \(\begin{pmatrix}3+t\\1-t\\1+2t\end{pmatrix}\) to produce \(\begin{pmatrix}\frac{7}{3}\\\frac{5}{3}\\-\frac{1}{3}\end{pmatrix}\) ISW | A1 |
| Answer | Marks |
|---|---|
| Use \(\sqrt{x^2+y^2+z^2}\) where \(\begin{pmatrix}x\\y\\z\end{pmatrix}\) is part (ii) answer | M1 |
| Obtain \(\sqrt{\frac{75}{9}}\) AEF, \(2.89\) or better \((2.8867513...)\) | A1 |
# Question 9:
## Part (i):
| $P$ is $\begin{pmatrix}3\\1\\1\end{pmatrix} + \begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}4\\0\\3\end{pmatrix}$ | B1 | |
|---|---|---|
| direction vector of $\ell$ is $\begin{pmatrix}1\\-1\\2\end{pmatrix}$ and of $\overrightarrow{OP}$ is their $P$ | $\sqrt{}$B1 | |
| Use $\cos\theta = \frac{\mathbf{a.b}}{|\mathbf{a}||\mathbf{b}|}$ for $\begin{pmatrix}1\\-1\\2\end{pmatrix}$ and their OP | M1 | |
| $\theta = 35.3°$ or better $(0.615...$ rad$)$ | A1 | |
## Part (ii):
| Use $\begin{pmatrix}1\\-1\\2\end{pmatrix} \cdot \begin{pmatrix}3+t\\1-t\\1+2t\end{pmatrix} = 0$ | M1 | |
|---|---|---|
| $1(3+t) - 1(1-t) + 2(1+2t) = 0$ | A1 | |
| $t = -\frac{2}{3}$ | A1 | |
| Subst. into $\begin{pmatrix}3+t\\1-t\\1+2t\end{pmatrix}$ to produce $\begin{pmatrix}\frac{7}{3}\\\frac{5}{3}\\-\frac{1}{3}\end{pmatrix}$ ISW | A1 | |
## Part (iii):
| Use $\sqrt{x^2+y^2+z^2}$ where $\begin{pmatrix}x\\y\\z\end{pmatrix}$ is part (ii) answer | M1 | |
|---|---|---|
| Obtain $\sqrt{\frac{75}{9}}$ AEF, $2.89$ or better $(2.8867513...)$ | A1 | |
---9 The equation of a straight line $l$ is $\mathbf { r } = \left( \begin{array} { l } 3 \\ 1 \\ 1 \end{array} \right) + t \left( \begin{array} { r } 1 \\ - 1 \\ 2 \end{array} \right) . O$ is the origin.\\
(i) The point $P$ on $l$ is given by $t = 1$. Calculate the acute angle between $O P$ and $l$.\\
(ii) Find the position vector of the point $Q$ on $l$ such that $O Q$ is perpendicular to $l$.\\
(iii) Find the length of $O Q$.
\hfill \mbox{\textit{OCR C4 2010 Q9 [10]}}