OCR MEI Further Pure Core 2021 November — Question 11 9 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2021
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeShortest distance between two skew lines
DifficultyStandard +0.8 This is a substantial Further Maths question requiring multiple vector techniques: dot product, cross product, angle between planes, and shortest distance between skew lines. While each individual technique is standard, the multi-part structure, the need to recognize how part (a) connects to part (b), and the skew lines distance formula (which is less routine than other vector topics) elevate this above average difficulty. The algebraic manipulation and the specific form required for the answer add modest challenge.
Spec4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line
11
  1. Given that \(\mathbf { u } = \lambda \mathbf { i } + \mathbf { j } - 3 \mathbf { k }\) and \(\mathbf { v } = \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k }\), find the following, giving your answers in terms of \(\lambda\).
    1. u.v
    2. \(\mathbf { u } \times \mathbf { v }\)
  2. Hence determine
    1. the acute angle between the planes \(2 x + y - 3 z = 10\) and \(x + 2 y - 2 z = 10\),
    2. the shortest distance between the lines \(\frac { x - 3 } { 3 } = \frac { y } { 1 } = \frac { z - 2 } { - 3 }\) and \(\frac { x } { 1 } = \frac { y - 4 } { 2 } = \frac { z + 2 } { - 2 }\), giving your answer as a multiple of \(\sqrt { 2 }\).
Question 11:
AnswerMarks Guidance
11(a) (i)
=+8B1
[1]1.1
11(a) (ii)
uv= 1  2 =2−3 oe i, j, k form
     
AnswerMarks
−3 −2 2−1B1
B1
AnswerMarks
[2]1.1a
1.1For one of i,j,k correct
For all i,j,k correct
AnswerMarks Guidance
11(b) (i)
taking 2 ,  =
1 0
c o s  =
1 4 9
AnswerMarks
 = 27.0 or 0.472 radM1
M1
A1
AnswerMarks
[3]3.1a
1.1
1.1
AnswerMarks Guidance
11(b) (ii)
3  =
with =3, uv=4i+3j+5k
 4   3 − 0 
1 2 0
d i s t a n c e = 3 . 0− − 4− = = 2 2
5 0 5 2
AnswerMarks
5 2 ( 2 )M1
M1
A1
AnswerMarks
[3]3.1a
1.1
1.1
Question 11:
11 | (a) | (i) | 1 1 2 ( 3 ) ( 2 )  u . v =  +  + −  −
=+8 | B1
[1] | 1.1
11 | (a) | (ii) |   1   4 
uv= 1  2 =2−3 oe i, j, k form
     
−3 −2 2−1 | B1
B1
[2] | 1.1a
1.1 | For one of i,j,k correct
For all i,j,k correct
11 | (b) | (i) | angle between 2 i + j − 3 k and i + 2 j − 2 k
taking 2 ,  =
1 0
c o s  =
1 4 9
 = 27.0 or 0.472 rad | M1
M1
A1
[3] | 3.1a
1.1
1.1
11 | (b) | (ii) | direction vectors are 3 i + j − 3 k and i + 2 j − 2 k , so take
3  =
with =3, uv=4i+3j+5k
 4   3 − 0 
1 2 0
d i s t a n c e = 3 . 0− − 4− = = 2 2
5 0 5 2
5 2 ( 2 ) | M1
M1
A1
[3] | 3.1a
1.1
1.1
11
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathbf { u } = \lambda \mathbf { i } + \mathbf { j } - 3 \mathbf { k }$ and $\mathbf { v } = \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k }$, find the following, giving your answers in terms of $\lambda$.
\begin{enumerate}[label=(\roman*)]
\item u.v
\item $\mathbf { u } \times \mathbf { v }$
\end{enumerate}\item Hence determine
\begin{enumerate}[label=(\roman*)]
\item the acute angle between the planes $2 x + y - 3 z = 10$ and $x + 2 y - 2 z = 10$,
\item the shortest distance between the lines $\frac { x - 3 } { 3 } = \frac { y } { 1 } = \frac { z - 2 } { - 3 }$ and $\frac { x } { 1 } = \frac { y - 4 } { 2 } = \frac { z + 2 } { - 2 }$, giving your answer as a multiple of $\sqrt { 2 }$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q11 [9]}}
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