Edexcel Paper 2 2024 June — Question 7 5 marks

Exam BoardEdexcel
ModulePaper 2 (Paper 2)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypePoint on line satisfying distance or other condition
DifficultyStandard +0.8 This question requires finding a vector between two points (routine), then solving a distance ratio condition on a line. Part (b) requires setting up a parametric equation, applying the magnitude condition |AP| = 2|BP|, squaring to eliminate square roots, and solving a quadratic equation. The two solutions (P between A and B, and P beyond B) require careful algebraic manipulation. This goes beyond standard textbook exercises but uses familiar techniques, making it moderately challenging for A-level Further Maths.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2ce10759-9ce6-47a1-b55d-d22082f88f55-16_330_654_246_751} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of the straight line \(l\).
Line \(l\) passes through the points \(A\) and \(B\).
Relative to a fixed origin \(O\)
  • the point \(A\) has position vector \(2 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k }\)
  • the point \(B\) has position vector \(5 \mathbf { i } + 6 \mathbf { j } + 8 \mathbf { k }\)
    1. Find \(\overrightarrow { A B }\)
Given that a point \(P\) lies on \(l\) such that $$| \overrightarrow { A P } | = 2 | \overrightarrow { B P } |$$
  • find the possible position vectors of \(P\).
    • Question 7(a):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    \(\overrightarrow{AB} = 3\mathbf{i}+9\mathbf{j}+3\mathbf{k}\)B1 Allow \(\begin{pmatrix}3\\9\\3\end{pmatrix}\) but not \(\begin{pmatrix}3\mathbf{i}\\9\mathbf{j}\\3\mathbf{k}\end{pmatrix}\) and not \((3,9,3)\). If part (a) not attempted and correct \(\overrightarrow{AB}\) seen in (b), B1 awarded there.
    Question 7(b):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Correct complete strategy for at least one position of \(P\), e.g. \(\overrightarrow{OP}=\overrightarrow{OA}+2\overrightarrow{AB}=2\mathbf{i}-3\mathbf{j}+5\mathbf{k}+2(3\mathbf{i}+9\mathbf{j}+3\mathbf{k})\) or \(\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{AB}=5\mathbf{i}+6\mathbf{j}+8\mathbf{k}+(3\mathbf{i}+9\mathbf{j}+3\mathbf{k})\)M1 May be implied by at least 2 correct components
    \(8\mathbf{i}+15\mathbf{j}+11\mathbf{k}\) or \(4\mathbf{i}+3\mathbf{j}+7\mathbf{k}\)A1 Allow coordinates \((8,15,11)\) or \((4,3,7)\) or \(x=\ldots, y=\ldots, z=\ldots\)
    Correct complete strategy to find both positions of \(P\)dM1 May be implied by at least 2 correct components for both positions
    \(8\mathbf{i}+15\mathbf{j}+11\mathbf{k}\) and \(4\mathbf{i}+3\mathbf{j}+7\mathbf{k}\)A1 Both correct position vectors and no others
    Alternative 1 (vector equation of line \(l\)):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Form \(\mathbf{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB}=\begin{pmatrix}2\\-3\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\9\\3\end{pmatrix}\), use \(\\overrightarrow{AP}\ =2\
    One correct positionA1 See main scheme
    Finds both positionsdM1 See main scheme
    Both correctA1 See main scheme
    Alternative 2 (P as general point):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Set \(\overrightarrow{OP}=\begin{pmatrix}x\\y\\z\end{pmatrix}\), form \(\overrightarrow{AP}=\begin{pmatrix}x-2\\y+3\\z-5\end{pmatrix}\), \(\overrightarrow{BP}=\begin{pmatrix}x-5\\y-6\\z-8\end{pmatrix}\), use \(\\overrightarrow{AP}\ =2\
    One correct positionA1 See main scheme
    Both positions founddM1 See main scheme
    \(\begin{pmatrix}4\\3\\7\end{pmatrix}\) and \(\begin{pmatrix}8\\15\\11\end{pmatrix}\)A1 See main scheme 
    ## Question 7(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = 3\mathbf{i}+9\mathbf{j}+3\mathbf{k}$ | B1 | Allow $\begin{pmatrix}3\\9\\3\end{pmatrix}$ but **not** $\begin{pmatrix}3\mathbf{i}\\9\mathbf{j}\\3\mathbf{k}\end{pmatrix}$ and **not** $(3,9,3)$. If part (a) not attempted and correct $\overrightarrow{AB}$ seen in (b), B1 awarded there. |

---

## Question 7(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct complete strategy for at least one position of $P$, e.g. $\overrightarrow{OP}=\overrightarrow{OA}+2\overrightarrow{AB}=2\mathbf{i}-3\mathbf{j}+5\mathbf{k}+2(3\mathbf{i}+9\mathbf{j}+3\mathbf{k})$ or $\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{AB}=5\mathbf{i}+6\mathbf{j}+8\mathbf{k}+(3\mathbf{i}+9\mathbf{j}+3\mathbf{k})$ | M1 | May be implied by at least 2 correct components |
| $8\mathbf{i}+15\mathbf{j}+11\mathbf{k}$ **or** $4\mathbf{i}+3\mathbf{j}+7\mathbf{k}$ | A1 | Allow coordinates $(8,15,11)$ or $(4,3,7)$ or $x=\ldots, y=\ldots, z=\ldots$ |
| Correct complete strategy to find **both** positions of $P$ | dM1 | May be implied by at least 2 correct components for both positions |
| $8\mathbf{i}+15\mathbf{j}+11\mathbf{k}$ **and** $4\mathbf{i}+3\mathbf{j}+7\mathbf{k}$ | A1 | Both correct position vectors and no others |

**Alternative 1 (vector equation of line $l$):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Form $\mathbf{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB}=\begin{pmatrix}2\\-3\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\9\\3\end{pmatrix}$, use $\|\overrightarrow{AP}\|=2\|\overrightarrow{BP}\|$ to produce quadratic in $\lambda$, solve to get $\lambda=2, \dfrac{2}{3}$ | M1 | If $\mathbf{i}+3\mathbf{j}+\mathbf{k}$ used for direction, should get $\lambda=6,2$ |
| One correct position | A1 | See main scheme |
| Finds both positions | dM1 | See main scheme |
| Both correct | A1 | See main scheme |

**Alternative 2 (P as general point):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Set $\overrightarrow{OP}=\begin{pmatrix}x\\y\\z\end{pmatrix}$, form $\overrightarrow{AP}=\begin{pmatrix}x-2\\y+3\\z-5\end{pmatrix}$, $\overrightarrow{BP}=\begin{pmatrix}x-5\\y-6\\z-8\end{pmatrix}$, use $\|\overrightarrow{AP}\|=2\|\overrightarrow{BP}\|$, square and equate components: $(x-2)^2=4(x-5)^2\Rightarrow x=4,8$; $(y+3)^2=4(y-6)^2\Rightarrow y=3,15$; $(z-5)^2=4(z-8)^2\Rightarrow z=7,11$ | M1 | Sets $P$ as general point, forms $\overrightarrow{AP}$ and $\overrightarrow{BP}$, uses $\|\overrightarrow{AP}\|=2\|\overrightarrow{BP}\|$, squares and equates to find at least one position |
| One correct position | A1 | See main scheme |
| Both positions found | dM1 | See main scheme |
| $\begin{pmatrix}4\\3\\7\end{pmatrix}$ and $\begin{pmatrix}8\\15\\11\end{pmatrix}$ | A1 | See main scheme |
    7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{2ce10759-9ce6-47a1-b55d-d22082f88f55-16_330_654_246_751}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

Figure 2 shows a sketch of the straight line $l$.\\
Line $l$ passes through the points $A$ and $B$.\\
Relative to a fixed origin $O$

\begin{itemize}
  \item the point $A$ has position vector $2 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k }$
  \item the point $B$ has position vector $5 \mathbf { i } + 6 \mathbf { j } + 8 \mathbf { k }$
\begin{enumerate}[label=(\alph*)]
\item Find $\overrightarrow { A B }$
\end{itemize}

Given that a point $P$ lies on $l$ such that

$$| \overrightarrow { A P } | = 2 | \overrightarrow { B P } |$$
\item find the possible position vectors of $P$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel Paper 2 2024 Q7 [5]}}
    This paper (15 questions)
    View full paper
    Q1 5 Q2 5 Q3 4 Q4 5 Q5 3 Q6 7 Q7 5 Q8 7 Q9 7 Q10 6 Q11 5 Q12 12 Q13 9 Q14 8 Q15 12