OCR MEI Further Pure Core 2020 November — Question 15 17 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2020
SessionNovember
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePerpendicular distance point to line
DifficultyStandard +0.3 This is a multi-part question covering standard Further Maths techniques: finding when planes meet at a point (determinant condition), solving simultaneous equations with matrices, converting line forms, and computing perpendicular distances. While lengthy (5 parts), each component uses routine methods without requiring novel insight—slightly easier than average for Further Maths content.
Spec4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations4.04a Line equations: 2D and 3D, cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines
15
  1. Show that the three planes with equations $$\begin{aligned} x + \lambda y + 3 z & = - 12 \\ 2 x + y + 5 z & = - 11 \\ x - 2 y + 2 z & = - 9 \end{aligned}$$ where \(\lambda\) is a constant, meet at a unique point except for one value of \(\lambda\) which is to be determined.
  2. In the case \(\lambda = - 2\), use matrices to find the point of intersection P of the planes, showing your method clearly. The line \(l\) has equation \(\frac { x - 1 } { 2 } = \frac { y - 1 } { - 1 } = \frac { z + 2 } { - 2 }\).
  3. Find a vector equation of \(l\).
  4. Find the shortest distance between the point P and \(l\).
    1. Show that \(l\) is parallel to the plane \(x - 2 y + 2 z = - 9\).
    2. Find the distance between \(l\) and the plane \(x - 2 y + 2 z = - 9\).
Question 15:
AnswerMarks Guidance
15(a) ​​ ​ ​
= λ−3
det = 0 when λ = 3
​​
So unique point provided λ ≠ 3
AnswerMarks
​​M1
A1
B1ft
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1calculating determinant
ft their λor full attempt to solve
finding x, y or z in terms of λ
​​ ​​ ​​ ​
PPMMTT
Y420/01 Mark SchemeNovember 2020
15 (b)
M1 1.1 matrix of coefficients or or attempt to use row ops
−1
M shown
M1 2.4
A1 1.1
⇒ x = 1, y = 2, z = −3 [3]
​​ ​​ ​​
15 (c)
M1 1.1
A1 1.1 oe
[2]
15 (d)
B1ft 1.1 ft their P or using first principles
​
M1 1.1
M1 1.1
A1 1.1 or 1.37 or better
[4]
15
AnswerMarks Guidance
15(b) ⇒ x = 1, y = 2, z = −3
​​ ​​ ​​M1
M1
A1
AnswerMarks
[3]1.1
2.4
AnswerMarks
1.1matrix of coefficients or
−1
AnswerMarks Guidance
M shownor attempt to use row ops
15(c) M1
A1
AnswerMarks Guidance
[2]1.1
1.1oe
15(d) B1ft
​
M1
M1
A1
AnswerMarks
[4]1.1
1.1
1.1
AnswerMarks
1.1ft their P
or 1.37 or betteror using first principles
PPMMTT
Y420/01 Mark SchemeNovember 2020
15 (e) (i)
B1 2.1 soi
M1 2.1
⇒ line l is parallel to the plane A1 2.2a
​​
[3]
15 (e) (ii) distance between (1, 1, −2) and x – 2y + 2z = −9 or using first principles
​​ ​​ ​​
M1 3.1a ft position vector given in 15c
A1 1.1 or 1.33 or better
= [2]
16 (a) (i)
B1 3.3
[1]
16 (a) (ii)
B1 1.1 or by integration by separating
variables
when t = 0, P = A(1 − 1) = 0
​​ ​ ​ ​ ​ B1 3.4
as ​ t ​ → ∞ , e− ​ ​ kt ​ → 0 so ​ P ​ → ​ A B1 3.4
[3]
16
AnswerMarks Guidance
15(e) (i)
​​B1
M1
A1
AnswerMarks
[3]2.1
2.1
AnswerMarks Guidance
2.2asoi
15(e) (ii)
​​ ​​ ​​
AnswerMarks
=M1
A1
AnswerMarks
[2]3.1a
1.1ft position vector given in 15c
or 1.33 or betteror using first principles 
Question 15:
15 | (a) | ​​ ​ ​
= λ−3
det = 0 when λ = 3
​​
So unique point provided λ ≠ 3
​​ | M1
A1
B1ft
[3] | 3.1a
1.1
1.1 | calculating determinant
ft their λ | or full attempt to solve
finding x, y or z in terms of λ
​​ ​​ ​​ ​
PPMMTT
Y420/01 Mark SchemeNovember 2020
15 (b)
M1 1.1 matrix of coefficients or or attempt to use row ops
−1
M shown
M1 2.4
A1 1.1
⇒ x = 1, y = 2, z = −3 [3]
​​ ​​ ​​
15 (c)
M1 1.1
A1 1.1 oe
[2]
15 (d)
B1ft 1.1 ft their P or using first principles
​
M1 1.1
M1 1.1
A1 1.1 or 1.37 or better
[4]
15
15 | (b) | ⇒ x = 1, y = 2, z = −3
​​ ​​ ​​ | M1
M1
A1
[3] | 1.1
2.4
1.1 | matrix of coefficients or
−1
M shown | or attempt to use row ops
15 | (c) | M1
A1
[2] | 1.1
1.1 | oe
15 | (d) | B1ft
​
M1
M1
A1
[4] | 1.1
1.1
1.1
1.1 | ft their P
or 1.37 or better | or using first principles
PPMMTT
Y420/01 Mark SchemeNovember 2020
15 (e) (i)
B1 2.1 soi
M1 2.1
⇒ line l is parallel to the plane A1 2.2a
​​
[3]
15 (e) (ii) distance between (1, 1, −2) and x – 2y + 2z = −9 or using first principles
​​ ​​ ​​
M1 3.1a ft position vector given in 15c
A1 1.1 or 1.33 or better
= [2]
16 (a) (i)
B1 3.3
[1]
16 (a) (ii)
B1 1.1 or by integration by separating
variables
when t = 0, P = A(1 − 1) = 0
​​ ​ ​ ​ ​ B1 3.4
as ​ t ​ → ∞ , e− ​ ​ kt ​ → 0 so ​ P ​ → ​ A B1 3.4
[3]
16
15 | (e) | (i) | ⇒ line l is parallel to the plane
​​ | B1
M1
A1
[3] | 2.1
2.1
2.2a | soi
15 | (e) | (ii) | distance between (1, 1, −2) and x – 2y + 2z = −9
​​ ​​ ​​
= | M1
A1
[2] | 3.1a
1.1 | ft position vector given in 15c
or 1.33 or better | or using first principles
15
\begin{enumerate}[label=(\alph*)]
\item Show that the three planes with equations

$$\begin{aligned}
x + \lambda y + 3 z & = - 12 \\
2 x + y + 5 z & = - 11 \\
x - 2 y + 2 z & = - 9
\end{aligned}$$

where $\lambda$ is a constant, meet at a unique point except for one value of $\lambda$ which is to be determined.
\item In the case $\lambda = - 2$, use matrices to find the point of intersection P of the planes, showing your method clearly.

The line $l$ has equation $\frac { x - 1 } { 2 } = \frac { y - 1 } { - 1 } = \frac { z + 2 } { - 2 }$.
\item Find a vector equation of $l$.
\item Find the shortest distance between the point P and $l$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $l$ is parallel to the plane $x - 2 y + 2 z = - 9$.
\item Find the distance between $l$ and the plane $x - 2 y + 2 z = - 9$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q15 [17]}}
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