AQA C4 2010 January — Question 8 11 marks

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2010
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeFoot of perpendicular from general external point to line
DifficultyStandard +0.3 This is a standard multi-part vectors question requiring routine techniques: verifying a point on a line (substitution), finding a vector between points (subtraction), using scalar multiples of vectors, and applying perpendicularity condition (dot product = 0). All parts follow textbook methods with no novel insight required, making it slightly easier than average.
Spec1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles
8 The points \(A , B\) and \(C\) have coordinates \(( 2 , - 1 , - 5 ) , ( 0,5 , - 9 )\) and \(( 9,2,3 )\) respectively. The line \(l\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 2 \\ - 1 \\ - 5 \end{array} \right] + \lambda \left[ \begin{array} { r } 1 \\ - 3 \\ 2 \end{array} \right]\).
  1. Verify that the point \(B\) lies on the line \(l\).
  2. Find the vector \(\overrightarrow { B C }\).
  3. The point \(D\) is such that \(\overrightarrow { A D } = 2 \overrightarrow { B C }\).
    1. Show that \(D\) has coordinates \(( 20 , - 7,19 )\).
    2. The point \(P\) lies on \(l\) where \(\lambda = p\). The line \(P D\) is perpendicular to \(l\). Find the value of \(p\).
Question 8(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(0 = 2 + \lambda \Rightarrow \lambda = -2\)M1
Check: \(-1 + -2\times-3 = -1+6 = 5\); \(-5-2\times2 = -5\times-4 = -9\)A1 OE
Question 8(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{BC} = \begin{bmatrix}9\\2\\3\end{bmatrix} - \begin{bmatrix}0\\5\\-9\end{bmatrix} = \begin{bmatrix}9\\-3\\12\end{bmatrix}\)M1 A1 \(\pm\left(\overrightarrow{OC} - \overrightarrow{OB}\right)\)
Question 8(c)(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD} = \overrightarrow{OA} + 2\overrightarrow{BC}\)M1
\(\overrightarrow{OD} = \begin{bmatrix}2\\-1\\-5\end{bmatrix} + \begin{bmatrix}18\\-6\\24\end{bmatrix} = \begin{bmatrix}20\\-7\\19\end{bmatrix}\)
\(D\) is \((20, -7, 19)\)A1 AG
Question 8(c)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{PD} = \overrightarrow{OD} - \overrightarrow{OP} = \begin{bmatrix}20\\-7\\19\end{bmatrix} - \begin{bmatrix}2+p\\-1-3p\\-5+2p\end{bmatrix} = \begin{bmatrix}18-p\\-6+3p\\24-2p\end{bmatrix}\)M1 A1 Find \(\overrightarrow{PD}\) in terms of \(p\); condone \(\overrightarrow{PD} = \overrightarrow{OP} - \overrightarrow{OD}\)
\(\overrightarrow{PD} \cdot \begin{bmatrix}1\\-3\\2\end{bmatrix} = 0\)B1
\((18-p)\times1 + (-6+3p)\times-3 + (24-2p)\times2 = 0\)m1
\(p = 6\)A1 CSO; OE working with \(\overrightarrow{DP}\) 
## Question 8(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 2 + \lambda \Rightarrow \lambda = -2$ | M1 | |
| Check: $-1 + -2\times-3 = -1+6 = 5$; $-5-2\times2 = -5\times-4 = -9$ | A1 | OE |

## Question 8(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{BC} = \begin{bmatrix}9\\2\\3\end{bmatrix} - \begin{bmatrix}0\\5\\-9\end{bmatrix} = \begin{bmatrix}9\\-3\\12\end{bmatrix}$ | M1 A1 | $\pm\left(\overrightarrow{OC} - \overrightarrow{OB}\right)$ |

## Question 8(c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD} = \overrightarrow{OA} + 2\overrightarrow{BC}$ | M1 | |
| $\overrightarrow{OD} = \begin{bmatrix}2\\-1\\-5\end{bmatrix} + \begin{bmatrix}18\\-6\\24\end{bmatrix} = \begin{bmatrix}20\\-7\\19\end{bmatrix}$ | | |
| $D$ is $(20, -7, 19)$ | A1 | AG |

## Question 8(c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{PD} = \overrightarrow{OD} - \overrightarrow{OP} = \begin{bmatrix}20\\-7\\19\end{bmatrix} - \begin{bmatrix}2+p\\-1-3p\\-5+2p\end{bmatrix} = \begin{bmatrix}18-p\\-6+3p\\24-2p\end{bmatrix}$ | M1 A1 | Find $\overrightarrow{PD}$ in terms of $p$; condone $\overrightarrow{PD} = \overrightarrow{OP} - \overrightarrow{OD}$ |
| $\overrightarrow{PD} \cdot \begin{bmatrix}1\\-3\\2\end{bmatrix} = 0$ | B1 | |
| $(18-p)\times1 + (-6+3p)\times-3 + (24-2p)\times2 = 0$ | m1 | |
| $p = 6$ | A1 | CSO; OE working with $\overrightarrow{DP}$ |

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8 The points $A , B$ and $C$ have coordinates $( 2 , - 1 , - 5 ) , ( 0,5 , - 9 )$ and $( 9,2,3 )$ respectively.

The line $l$ has equation $\mathbf { r } = \left[ \begin{array} { r } 2 \\ - 1 \\ - 5 \end{array} \right] + \lambda \left[ \begin{array} { r } 1 \\ - 3 \\ 2 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item Verify that the point $B$ lies on the line $l$.
\item Find the vector $\overrightarrow { B C }$.
\item The point $D$ is such that $\overrightarrow { A D } = 2 \overrightarrow { B C }$.
\begin{enumerate}[label=(\roman*)]
\item Show that $D$ has coordinates $( 20 , - 7,19 )$.
\item The point $P$ lies on $l$ where $\lambda = p$. The line $P D$ is perpendicular to $l$. Find the value of $p$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C4 2010 Q8 [11]}}
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