| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of triangle from given side vectors or coordinates |
| Difficulty | Standard +0.8 This is a substantial multi-part vector question requiring: (a) solving simultaneous equations to find intersection, (b) verifying perpendicularity via dot product, and (c) finding triangle area using cross product. Part (c) requires finding specific points from coordinates, then computing |PQ × PR|/2, which involves multiple vector operations and simplification. More demanding than standard C4 questions but uses established techniques. |
| Spec | 4.04a 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 |
|---|---|---|
| \(l_1\): \((1+\lambda, 3+2\lambda, 5-\lambda)\); \(l_2\): \((-2+2\mu, 3+\mu, -4+4\mu)\) | M1 | Writing parametric coordinates |
| Equating: \(1+\lambda = -2+2\mu\), \(3+2\lambda = 3+\mu\), \(5-\lambda = -4+4\mu\) | M1 | Setting equal |
| From equation 2: \(2\lambda = \mu\) | A1 | Correct relation |
| Substituting into equation 1: \(1+\lambda = -2+4\lambda \Rightarrow \lambda=1, \mu=2\) | M1 | Solving |
| Check equation 3: \(5-1=4\), \(-4+8=4\) ✓ | A1 | Verification |
| \(Q = (2, 5, 4)\) | A1 | Correct coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Direction vectors: \(l_1\): \((1,2,-1)\); \(l_2\): \((2,1,4)\) | M1 | Identifying direction vectors |
| \((1)(2)+(2)(1)+(-1)(4) = 2+2-4=0\) | A1 | Dot product \(=0\), perpendicular shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=3\) on \(l_1\): \(1+\lambda=3 \Rightarrow \lambda=2\), \(P=(3,7,3)\) | B1 | Finding P |
| \(x=4\) on \(l_2\): \(-2+2\mu=4 \Rightarrow \mu=3\), \(R=(4,6,8)\) | B1 | Finding R |
| \(\overrightarrow{PQ} = (-1,-2,1)\), \(\overrightarrow{PR} = (1,-1,5)\) | M1 | Finding vectors from P |
| \(\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&-2&1\\1&-1&5\end{vmatrix}\) | M1 | Cross product method |
| \(= (-10+1)\mathbf{i} - (-5-1)\mathbf{j} + (1+2)\mathbf{k} = (-9, 6, 3)\) | A1 | Correct cross product |
| \( | \overrightarrow{PQ}\times\overrightarrow{PR} | = \sqrt{81+36+9} = \sqrt{126} = 3\sqrt{14}\) |
| Area \(= \frac{1}{2} | {\overrightarrow{PQ}\times\overrightarrow{PR}} | = \frac{3\sqrt{14}}{2}\) |
# Question 9:
## Part (a)
| $l_1$: $(1+\lambda, 3+2\lambda, 5-\lambda)$; $l_2$: $(-2+2\mu, 3+\mu, -4+4\mu)$ | M1 | Writing parametric coordinates |
| Equating: $1+\lambda = -2+2\mu$, $3+2\lambda = 3+\mu$, $5-\lambda = -4+4\mu$ | M1 | Setting equal |
| From equation 2: $2\lambda = \mu$ | A1 | Correct relation |
| Substituting into equation 1: $1+\lambda = -2+4\lambda \Rightarrow \lambda=1, \mu=2$ | M1 | Solving |
| Check equation 3: $5-1=4$, $-4+8=4$ ✓ | A1 | Verification |
| $Q = (2, 5, 4)$ | A1 | Correct coordinates |
## Part (b)
| Direction vectors: $l_1$: $(1,2,-1)$; $l_2$: $(2,1,4)$ | M1 | Identifying direction vectors |
| $(1)(2)+(2)(1)+(-1)(4) = 2+2-4=0$ | A1 | Dot product $=0$, perpendicular shown |
## Part (c)
| $x=3$ on $l_1$: $1+\lambda=3 \Rightarrow \lambda=2$, $P=(3,7,3)$ | B1 | Finding P |
| $x=4$ on $l_2$: $-2+2\mu=4 \Rightarrow \mu=3$, $R=(4,6,8)$ | B1 | Finding R |
| $\overrightarrow{PQ} = (-1,-2,1)$, $\overrightarrow{PR} = (1,-1,5)$ | M1 | Finding vectors from P |
| $\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&-2&1\\1&-1&5\end{vmatrix}$ | M1 | Cross product method |
| $= (-10+1)\mathbf{i} - (-5-1)\mathbf{j} + (1+2)\mathbf{k} = (-9, 6, 3)$ | A1 | Correct cross product |
| $|\overrightarrow{PQ}\times\overrightarrow{PR}| = \sqrt{81+36+9} = \sqrt{126} = 3\sqrt{14}$ | | |
| Area $= \frac{1}{2}|{\overrightarrow{PQ}\times\overrightarrow{PR}}| = \frac{3\sqrt{14}}{2}$ | A1 | Correct exact area |9. The equations of the lines $l _ { 1 }$ and $l _ { 2 }$ are given by
$$\begin{array} { l l }
l _ { 1 } : & \mathbf { r } = \mathbf { i } + 3 \mathbf { j } + 5 \mathbf { k } + \lambda ( \mathbf { i } + 2 \mathbf { j } - \mathbf { k } ) , \\
l _ { 2 } : & \mathbf { r } = - 2 \mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k } + \mu ( 2 \mathbf { i } + \mathbf { j } + 4 \mathbf { k } ) ,
\end{array}$$
where $\lambda$ and $\mu$ are parameters.
\begin{enumerate}[label=(\alph*)]
\item Show that $l _ { 1 }$ and $l _ { 2 }$ intersect and find the coordinates of $Q$, their point of intersection.
\item Show that $l _ { 1 }$ is perpendicular to $l _ { 2 }$.
The point $P$ with $x$-coordinate 3 lies on the line $l _ { 1 }$ and the point $R$ with $x$-coordinate 4 lies on the line $l _ { 2 }$.
\item Find, in its simplest form, the exact area of the triangle $P Q R$.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q9 [14]}}