| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| 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 line equations, checking perpendicularity via dot product, finding intersection by equating parameters, and calculating distance. All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| (i) For (either point) + t(diff between posn vectors) | M1 | "\(\vec{r} =\)" not necessary for the M mark |
| \(\vec{r} = \) (either point) \(+ t(\pm 2j - 3k)\) or \(\vec{i} + 2j + 3k)\) | A1 | 2 marks |
| (ii) r = \((i + 2j - k) + s(2i - 3j - k)\) or \((i + 2j - k)\) | B1 | |
| Eval scalar product of \(i + 2j - k\) & their vect in (i) | M1 | |
| Show as \((1x1) + (2x-2) + (-4)(-1 \times -3\) or \(3)\) \(= 0\) | A1 | and state perpendicular |
| (iii) For at least two equations with diff parameters | M1 | This is just one example of numbers involved: e.g. \(5 + t = s, -2t = 2s, -9 - 3t = -s\) |
| Obtain \(t = -2\) or \(s = 3\) (possibly -3 or 2 or -2) | A1 | Check if \(t = 2,1\) or \(-1\) |
| Subst. into eqn \(AB\) or \(OT\) and produce \(3i + 6j - 3k\) | A1 | 3 marks |
| (iv) Indicate that \(\overrightarrow{OC}\) is to be found | M1 | where C is their point of intersection |
| \(\sqrt{54}\) i.t. \(\sqrt{a^2 + b^2 + c^2}\) from \(ai + bj + ck\) in (iii) | √A1 | 2 marks |
(i) For (either point) + t(diff between posn vectors) | M1 | "$\vec{r} =$" not necessary for the M mark
$\vec{r} = $ (either point) $+ t(\pm 2j - 3k)$ or $\vec{i} + 2j + 3k)$ | A1 | 2 marks | ... but it is essential for the A mark; Accept any parameter, including $t$
(ii) r = $(i + 2j - k) + s(2i - 3j - k)$ or $(i + 2j - k)$ | B1 |
Eval scalar product of $i + 2j - k$ & their vect in (i) | M1 |
Show as $(1x1) + (2x-2) + (-4)(-1 \times -3$ or $3)$ $= 0$ | A1 | and state perpendicular | A1 | 4 marks
(iii) For at least two equations with diff parameters | M1 | This is just one example of numbers involved: e.g. $5 + t = s, -2t = 2s, -9 - 3t = -s$
Obtain $t = -2$ or $s = 3$ (possibly -3 or 2 or -2) | A1 | Check if $t = 2,1$ or $-1$
Subst. into eqn $AB$ or $OT$ and produce $3i + 6j - 3k$ | A1 | 3 marks
(iv) Indicate that $\overrightarrow{OC}$ is to be found | M1 | where C is their point of intersection
$\sqrt{54}$ i.t. $\sqrt{a^2 + b^2 + c^2}$ from $ai + bj + ck$ in (iii) | √A1 | 2 marks
In the above question, accept any vectorial notation. $t$ and $s$ may be interchanged, and values stated above need to be treated with caution. In (iii), if the point of intersection is correct, it is more than likely that the whole part is correct – but check.The position vectors of the points $P$ and $Q$ with respect to an origin $O$ are $5\mathbf{i} + 2\mathbf{j} - 9\mathbf{k}$ and $4\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}$ respectively.
\begin{enumerate}[label=(\roman*)]
\item Find a vector equation for the line $PQ$. [2]
\end{enumerate}
The position vector of the point $T$ is $\mathbf{i} + 2\mathbf{j} - \mathbf{k}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Write down a vector equation for the line $OT$ and show that $OT$ is perpendicular to $PQ$. [4]
\end{enumerate}
It is given that $OT$ intersects $PQ$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the position vector of the point of intersection of $OT$ and $PQ$. [3]
\item Hence find the perpendicular distance from $O$ to $PQ$, giving your answer in an exact form. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2007 Q10 [11]}}