| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Area of triangle using cross product |
| Difficulty | Standard +0.3 This is a straightforward application of the cross product formula to find area of a triangle. Part (a) requires routine computation of a 3×3 determinant with one parameter, and part (b) involves using the standard formula Area = ½|u×v|, leading to a quadratic equation. While it requires multiple steps, all techniques are standard FP1 material with no novel insight needed, making it slightly easier than average. |
| Spec | 4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&4&-3\\a&-6&2\end{vmatrix} = \ldots\mathbf{i}-\ldots\mathbf{j}+\ldots\mathbf{k}\) | M1 | Evidence of correct method for vector product; implied by two out of three correct components; attempting \(\mathbf{v}\times\mathbf{u}\) scores M0 |
| \(= -10\mathbf{i}-(10+3a)\mathbf{j}-(30+4a)\mathbf{k}\) | A1 | Correct answer; must be a vector not coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Area}=\frac{1}{2}\ | 2\mathbf{u}\times\mathbf{v}\ | =15 \Rightarrow \ |
| \(\Rightarrow 10^2+(10+3a)^2+(30+4a)^2=15^2 \Rightarrow 25a^2+300a+875=0 \Rightarrow a^2+12a+35=0\) | M1 | Applies modulus correctly and expands correctly to reach a quadratic in \(a\) |
| \((a+5)(a+7)=0 \Rightarrow a=\ldots\) | ddM1 | Solves quadratic by any suitable means; depends on both previous M marks |
| \(a=-7\) or \(-5\) | A1 cso | Both correct values; do not condone sign errors in vector product |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&4&-3\\a&-6&2\end{vmatrix} = \ldots\mathbf{i}-\ldots\mathbf{j}+\ldots\mathbf{k}$ | M1 | Evidence of correct method for vector product; implied by two out of three correct components; attempting $\mathbf{v}\times\mathbf{u}$ scores M0 |
| $= -10\mathbf{i}-(10+3a)\mathbf{j}-(30+4a)\mathbf{k}$ | A1 | Correct answer; must be a vector not coordinates |
# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Area}=\frac{1}{2}\|2\mathbf{u}\times\mathbf{v}\|=15 \Rightarrow \|-10\mathbf{i}-(10+3a)\mathbf{j}-(30+4a)\mathbf{k}\|=15$ | M1 | Correct method for area using vector product; "1/2" must be seen or implied |
| $\Rightarrow 10^2+(10+3a)^2+(30+4a)^2=15^2 \Rightarrow 25a^2+300a+875=0 \Rightarrow a^2+12a+35=0$ | M1 | Applies modulus correctly and expands correctly to reach a quadratic in $a$ |
| $(a+5)(a+7)=0 \Rightarrow a=\ldots$ | ddM1 | Solves quadratic by any suitable means; depends on both previous M marks |
| $a=-7$ or $-5$ | A1 cso | Both correct values; do not condone sign errors in vector product |
---\begin{enumerate}
\item Vectors $\mathbf { u }$ and $\mathbf { v }$ are given by
\end{enumerate}
$$\mathbf { u } = 5 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k } \quad \text { and } \quad \mathbf { v } = a \mathbf { i } - 6 \mathbf { j } + 2 \mathbf { k }$$
where $a$ is a constant.\\
(a) Determine, in terms of $a$, the vector product $\mathbf { u } \times \mathbf { v }$
Given that
\begin{itemize}
\item $\overrightarrow { A B } = 2 \mathbf { u }$
\item $\overrightarrow { A C } = \mathbf { v }$
\item the area of triangle $A B C$ is 15\\
(b) determine the possible values of $a$.
\end{itemize}
\hfill \mbox{\textit{Edexcel FP1 AS 2024 Q3 [6]}}