OCR Further Pure Core 2 2022 June — Question 1 6 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with line
DifficultyModerate -0.3 Part (a) is a straightforward cross product calculation requiring routine vector manipulation. Part (b) involves equating components and solving simultaneous equations to find intersection—standard Further Maths technique with no novel insight required. The multi-part structure and algebraic manipulation place it slightly below average difficulty for Further Maths content.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04g Vector product: a x b perpendicular vector

1
  1. Find a vector which is perpendicular to both \(3 \mathbf { i } - 5 \mathbf { j } - \mathbf { k }\) and \(\mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k }\). The equations of two lines are \(\mathbf { r } = 2 \mathbf { i } + 3 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + \mathbf { k } )\) and \(\mathbf { r } = \mathbf { i } + 11 \mathbf { j } - 4 \mathbf { k } + \mu ( - \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k } )\).
  2. Show that the lines intersect, stating the point of intersection.

Question 1:
AnswerMarks Guidance
1(a) (3i – 5j – k)(i + 3j – 4k) = 23i + 11j +14k
[1]1.1 or any non-zero multiple .
ISW
AnswerMarks
(b)x: 2 + λ = 1 – μ
y: 3 – 2λ = 11 + 3μ
or z: 3 + λ = –4 – 2μ
eg 4 + 2λ = 2 – 2μ => 7 = 13 + μ
μ = –6, λ = 5
eg z: LHS = 3 + 5 = 8
RHS = –4 – 2(–6) = –4 – (–12) = 8
AnswerMarks
PoI is (7, –7, 8)M1
M1
A1
B1FT
B1
AnswerMarks
[5]1.1
1.1
1.1
1.1
AnswerMarks
1.1Any correct numeric equation
Using any 2 equations to
eliminate either λ or μ.
Check in unused (or all)
equation(s)
FT their values of λ and μ
provided that LHS = RHS.
AnswerMarks
Condone as (position) vectorMust indicate which equations are
used to find λ and μ.
Calculations must be correct and
answer must be evaluated.
LHS = 8, RHS = 8 is insufficient;
some substitution must be evident.
Question 1:
1 | (a) | (3i – 5j – k)(i + 3j – 4k) = 23i + 11j +14k | B1
[1] | 1.1 | or any non-zero multiple .
ISW
(b) | x: 2 + λ = 1 – μ
y: 3 – 2λ = 11 + 3μ
or z: 3 + λ = –4 – 2μ
eg 4 + 2λ = 2 – 2μ => 7 = 13 + μ
μ = –6, λ = 5
eg z: LHS = 3 + 5 = 8
RHS = –4 – 2(–6) = –4 – (–12) = 8
PoI is (7, –7, 8) | M1
M1
A1
B1FT
B1
[5] | 1.1
1.1
1.1
1.1
1.1 | Any correct numeric equation
Using any 2 equations to
eliminate either λ or μ.
Check in unused (or all)
equation(s)
FT their values of λ and μ
provided that LHS = RHS.
Condone as (position) vector | Must indicate which equations are
used to find λ and μ.
Calculations must be correct and
answer must be evaluated.
LHS = 8, RHS = 8 is insufficient;
some substitution must be evident.
1
\begin{enumerate}[label=(\alph*)]
\item Find a vector which is perpendicular to both $3 \mathbf { i } - 5 \mathbf { j } - \mathbf { k }$ and $\mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k }$.

The equations of two lines are $\mathbf { r } = 2 \mathbf { i } + 3 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + \mathbf { k } )$ and $\mathbf { r } = \mathbf { i } + 11 \mathbf { j } - 4 \mathbf { k } + \mu ( - \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k } )$.
\item Show that the lines intersect, stating the point of intersection.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2022 Q1 [6]}}