| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard C4 vectors question testing intersection of lines and angle between direction vectors. Part (a) requires equating components and solving simultaneous equations (routine but multi-step), while part (b) uses the standard dot product formula. Both parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(4x = -7 - 3t\) ... (1) | ||
| \(7 - 3x = 1\) ... (2) | ||
| \(-4 + s = 8 + 2t\) ... (3) | ||
| (2) \(\Rightarrow x = 2\), sub. (1) \(\Rightarrow t = -5\) | B1 M1 | |
| check (3): \(-4 + 2 = 8 - 10\), true \(\therefore\) intersect | A1 | |
| intersect at \((7\hat{i} - 4\hat{k}) + 2(4\hat{i} - 3\hat{j} + \hat{k}) = (8\hat{i} - \hat{j} - 2\hat{k})\) | A1 | |
| (b) \(= \cos^{-1}\left(\frac{ | 4 \times (-3) + (-3) \times 0 + 1 \times 2 | }{\sqrt{16 + 9 + 1} \times \sqrt{9 + 0 + 4}}\right)\) |
| \(= \cos^{-1}\left(\frac{-10}{\sqrt{26 \times 13}}\right) = 57.0°\) (1dp) | M1 A1 | (9 marks total) |
**(a)** $4x = -7 - 3t$ ... (1) | |
$7 - 3x = 1$ ... (2) | |
$-4 + s = 8 + 2t$ ... (3) | |
(2) $\Rightarrow x = 2$, sub. (1) $\Rightarrow t = -5$ | B1 M1 |
check (3): $-4 + 2 = 8 - 10$, true $\therefore$ intersect | A1 |
intersect at $(7\hat{i} - 4\hat{k}) + 2(4\hat{i} - 3\hat{j} + \hat{k}) = (8\hat{i} - \hat{j} - 2\hat{k})$ | A1 |
**(b)** $= \cos^{-1}\left(\frac{|4 \times (-3) + (-3) \times 0 + 1 \times 2|}{\sqrt{16 + 9 + 1} \times \sqrt{9 + 0 + 4}}\right)$ | M1 A1 |
$= \cos^{-1}\left(\frac{-10}{\sqrt{26 \times 13}}\right) = 57.0°$ (1dp) | M1 A1 | (9 marks total) |Relative to a fixed origin, two lines have the equations
$$\mathbf{r} = (7\mathbf{i} - 4\mathbf{k}) + s(4\mathbf{i} - 3\mathbf{j} + \mathbf{k}),$$
and
$$\mathbf{r} = (-7\mathbf{i} + \mathbf{j} + 8\mathbf{k}) + t(-3\mathbf{i} + 2\mathbf{k}),$$
where $s$ and $t$ are scalar parameters.
\begin{enumerate}[label=(\alph*)]
\item Show that the two lines intersect and find the position vector of the point where they meet. [5]
\item Find, in degrees to 1 decimal place, the acute angle between the lines. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [9]}}