OCR C4 — Question 7 12 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeShow lines intersect and find intersection point
DifficultyStandard +0.3 This is a standard multi-part 3D vectors question covering routine techniques: finding a line equation from two points, showing lines intersect by solving simultaneous equations, and using perpendicularity conditions. All parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level.
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 vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting
7. The line \(l _ { 1 }\) passes through the points \(A\) and \(B\) with position vectors ( \(3 \mathbf { i } + 6 \mathbf { j } - 8 \mathbf { k }\) ) and ( \(8 \mathbf { j } - 6 \mathbf { k }\) ) respectively, relative to a fixed origin.
  1. Find a vector equation for \(l _ { 1 }\). The line \(l _ { 2 }\) has vector equation $$\mathbf { r } = ( - 2 \mathbf { i } + 10 \mathbf { j } + 6 \mathbf { k } ) + \mu ( 7 \mathbf { i } - 4 \mathbf { j } + 6 \mathbf { k } ) ,$$ where \(\mu\) is a scalar parameter.
  2. Show that lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect.
  3. Find the coordinates of the point where \(l _ { 1 }\) and \(l _ { 2 }\) intersect. The point \(C\) lies on \(l _ { 2 }\) and is such that \(A C\) is perpendicular to \(A B\).
  4. Find the position vector of \(C\).
Question 7:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\overrightarrow{AB} = (8\mathbf{j} - 6\mathbf{k}) - (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) = (-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\)M1
\(\therefore\ \mathbf{r} = (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) + \lambda(-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\)A1
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3 - 3\lambda = -2 + 7\mu \quad (1)\)
\(6 + 2\lambda = 10 - 4\mu \quad (2)\)
\(-8 + 2\lambda = 6 + 6\mu \quad (3)\)B1
\((3)-(2):\ -14 = -4 + 10\mu,\ \mu = -1,\ \lambda = 4\)M1 A1
check \((1)\): \(3 - 12 = -2 - 7\), true \(\therefore\) intersectB1
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{r} = (-2\mathbf{i} + 10\mathbf{j} + 6\mathbf{k}) - (7\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}) \quad \therefore (-9,\ 14,\ 0)\)B1
Part (iv):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\overrightarrow{OC} = [(-2+7\mu)\mathbf{i} + (10-4\mu)\mathbf{j} + (6+6\mu)\mathbf{k}]\)
\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(-5+7\mu)\mathbf{i} + (4-4\mu)\mathbf{j} + (14+6\mu)\mathbf{k}]\)M1 A1
\(\therefore\ [(-5+7\mu)\mathbf{i} + (4-4\mu)\mathbf{j} + (14+6\mu)\mathbf{k}]\cdot(-3\mathbf{i}+2\mathbf{j}+2\mathbf{k}) = 0\)M1
\(15 - 21\mu + 8 - 8\mu + 28 + 12\mu = 0\)
\(\mu = 3\ \therefore\ \overrightarrow{OC} = (19\mathbf{i} - 2\mathbf{j} + 24\mathbf{k})\)M1 A1 (12) 
# Question 7:

## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = (8\mathbf{j} - 6\mathbf{k}) - (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) = (-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ | M1 | |
| $\therefore\ \mathbf{r} = (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) + \lambda(-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ | A1 | |

## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3 - 3\lambda = -2 + 7\mu \quad (1)$ | | |
| $6 + 2\lambda = 10 - 4\mu \quad (2)$ | | |
| $-8 + 2\lambda = 6 + 6\mu \quad (3)$ | B1 | |
| $(3)-(2):\ -14 = -4 + 10\mu,\ \mu = -1,\ \lambda = 4$ | M1 A1 | |
| check $(1)$: $3 - 12 = -2 - 7$, true $\therefore$ intersect | B1 | |

## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = (-2\mathbf{i} + 10\mathbf{j} + 6\mathbf{k}) - (7\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}) \quad \therefore (-9,\ 14,\ 0)$ | B1 | |

## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{OC} = [(-2+7\mu)\mathbf{i} + (10-4\mu)\mathbf{j} + (6+6\mu)\mathbf{k}]$ | | |
| $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(-5+7\mu)\mathbf{i} + (4-4\mu)\mathbf{j} + (14+6\mu)\mathbf{k}]$ | M1 A1 | |
| $\therefore\ [(-5+7\mu)\mathbf{i} + (4-4\mu)\mathbf{j} + (14+6\mu)\mathbf{k}]\cdot(-3\mathbf{i}+2\mathbf{j}+2\mathbf{k}) = 0$ | M1 | |
| $15 - 21\mu + 8 - 8\mu + 28 + 12\mu = 0$ | | |
| $\mu = 3\ \therefore\ \overrightarrow{OC} = (19\mathbf{i} - 2\mathbf{j} + 24\mathbf{k})$ | M1 A1 | **(12)** |

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7. The line $l _ { 1 }$ passes through the points $A$ and $B$ with position vectors ( $3 \mathbf { i } + 6 \mathbf { j } - 8 \mathbf { k }$ ) and ( $8 \mathbf { j } - 6 \mathbf { k }$ ) respectively, relative to a fixed origin.\\
(i) Find a vector equation for $l _ { 1 }$.

The line $l _ { 2 }$ has vector equation

$$\mathbf { r } = ( - 2 \mathbf { i } + 10 \mathbf { j } + 6 \mathbf { k } ) + \mu ( 7 \mathbf { i } - 4 \mathbf { j } + 6 \mathbf { k } ) ,$$

where $\mu$ is a scalar parameter.\\
(ii) Show that lines $l _ { 1 }$ and $l _ { 2 }$ intersect.\\
(iii) Find the coordinates of the point where $l _ { 1 }$ and $l _ { 2 }$ intersect.

The point $C$ lies on $l _ { 2 }$ and is such that $A C$ is perpendicular to $A B$.\\
(iv) Find the position vector of $C$.\\

\hfill \mbox{\textit{OCR C4  Q7 [12]}}
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