OCR MEI Further Pure Core 2022 June — Question 13 17 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2022
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeAcute angle between line and plane
DifficultyStandard +0.8 This is a substantial multi-part Further Maths question requiring multiple vector techniques (direction vectors, cross products, angle formulas, distance calculations) across 5 sub-parts. While each individual technique is standard for FM students, the length, coordination of methods, and part (c)(iii) requiring synthesis of previous results to find distance from a point to a line of intersection elevates this above routine exercises. However, it remains methodical application rather than requiring novel insight.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point4.04g Vector product: a x b perpendicular vector4.04j Shortest distance: between a point and a plane
13 The points A and B have coordinates \(( 4,0 , - 1 )\) and \(( 10,4 , - 3 )\) respectively. The planes \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\) have equations \(x - 2 y = 5\) and \(2 x + 3 y - z = - 4\) respectively.
  1. Find the acute angle between the line AB and the plane \(\Pi _ { 1 }\).
  2. Show that the line AB meets \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\) at the same point, whose coordinates should be specified.
    1. Find \(( \mathbf { i } - 2 \mathbf { j } ) \times ( 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } )\).
    2. Hence find the acute angle between the planes \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\).
    3. Find the shortest distance between the point A and the line of intersection of the planes \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\).
Question 13:
AnswerMarks Guidance
13(a) A⃗⃗⃗⃗B⃗ = 6𝒊+4𝒋−2𝒌
angle between line and normal is  where𝑐𝑜𝑠𝜃 =
(3𝒊+2𝒋−𝒌).(𝒊−2𝒋) −1
=
√14√5 √14√5
AnswerMarks
𝜃 = 96.9°, so angle with plane is 6.9B1
M1
A1
A1
AnswerMarks
[4]1.1
3.1a
1.1
AnswerMarks
1.1Any multiple soi
7 or better 0.1198 rad
AnswerMarks Guidance
13(b) eqn of AB is [𝒓 =]4𝒊−𝒌+𝜆(3𝒊+2𝒋−𝒌)
Substituting 𝑥 = 4+3𝜆, 𝑦 = 2𝜆, 𝑧 = −1−𝜆:
4+3𝜆−4𝜆 = 5 ⇒ 𝜆 = −1
meets  at (1, −2, 0) (say C)
1
substituting into 
2
2×1+3×(−2)−0 = −4, so C lies on 
AnswerMarks
2B1ft
M1
A1
M1
A1
AnswerMarks
[5]2.1
1.1
2.2a
3.1a
AnswerMarks
2.2aSoi Note may use another position vector as long as it is correct
or solving with 
2
AnswerMarks
Alternative methodsoi
eqn of AB is [𝒓 =]4𝒊−𝒌+𝜆(3𝒊+2𝒋−𝒌)B1
Substituting 𝑥 = 4+3𝜆, 𝑦 = 2𝜆, 𝑧 = −1−𝜆:M1
4+3𝜆−4𝜆 = 5 ⇒ 𝜆 = −1
AnswerMarks
Substituting 𝑥 = 4+3𝜆, 𝑦 = 2𝜆, 𝑧 = −1−𝜆:M1
2(4+3𝜆)+3(2𝜆)−(−1−𝜆) = −4 ⇒ 𝜆 = −1
AnswerMarks
Finding equal 𝜆 for both planesA1
(1,-2,0)A1
[5]
AnswerMarks Guidance
13(c) (i)
[1]1.1
13(c) (ii)
√5×√14sin𝜃 = √54
AnswerMarks
⇒ 𝜃 = 61.4°B1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1soi
1.07 rad
soi
AnswerMarks Guidance
13(c) (iii)
3 2 15
( 2 )×(1) = (−23)
−1 7 −1
√152+(−23)2+(−1)2
𝑑 =
√22+12+72
AnswerMarks
= 3.74B1ft
M1
M1
A1cao
AnswerMarks
[4]3.1a
1.1
1.1
AnswerMarks
1.1Soi by using in both numerator and denominator in correct
method for d
Question 13:
13 | (a) | A⃗⃗⃗⃗B⃗ = 6𝒊+4𝒋−2𝒌
angle between line and normal is  where𝑐𝑜𝑠𝜃 =
(3𝒊+2𝒋−𝒌).(𝒊−2𝒋) −1
=
√14√5 √14√5
𝜃 = 96.9°, so angle with plane is 6.9 | B1
M1
A1
A1
[4] | 1.1
3.1a
1.1
1.1 | Any multiple soi
7 or better 0.1198 rad
13 | (b) | eqn of AB is [𝒓 =]4𝒊−𝒌+𝜆(3𝒊+2𝒋−𝒌)
Substituting 𝑥 = 4+3𝜆, 𝑦 = 2𝜆, 𝑧 = −1−𝜆:
4+3𝜆−4𝜆 = 5 ⇒ 𝜆 = −1
meets  at (1, −2, 0) (say C)
1
substituting into 
2
2×1+3×(−2)−0 = −4, so C lies on 
2 | B1ft
M1
A1
M1
A1
[5] | 2.1
1.1
2.2a
3.1a
2.2a | Soi Note may use another position vector as long as it is correct
or solving with 
2
Alternative method | soi
eqn of AB is [𝒓 =]4𝒊−𝒌+𝜆(3𝒊+2𝒋−𝒌) | B1
Substituting 𝑥 = 4+3𝜆, 𝑦 = 2𝜆, 𝑧 = −1−𝜆: | M1
4+3𝜆−4𝜆 = 5 ⇒ 𝜆 = −1
Substituting 𝑥 = 4+3𝜆, 𝑦 = 2𝜆, 𝑧 = −1−𝜆: | M1
2(4+3𝜆)+3(2𝜆)−(−1−𝜆) = −4 ⇒ 𝜆 = −1
Finding equal 𝜆 for both planes | A1
(1,-2,0) | A1
[5]
13 | (c) | (i) | (𝒊−2𝒋)×(2𝒊+3𝒋−𝒌) = 2𝒊+𝒋+7𝒌 | B1
[1] | 1.1
13 | (c) | (ii) | √54
√5×√14sin𝜃 = √54
⇒ 𝜃 = 61.4° | B1
M1
A1
[3] | 3.1a
1.1
1.1 | soi
1.07 rad
soi
13 | (c) | (iii) | 2𝒊+𝒋+7𝒌 is direction vector of line of intersection
3 2 15
( 2 )×(1) = (−23)
−1 7 −1
√152+(−23)2+(−1)2
𝑑 =
√22+12+72
= 3.74 | B1ft
M1
M1
A1cao
[4] | 3.1a
1.1
1.1
1.1 | Soi by using in both numerator and denominator in correct
method for d
13 The points A and B have coordinates $( 4,0 , - 1 )$ and $( 10,4 , - 3 )$ respectively. The planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ have equations $x - 2 y = 5$ and $2 x + 3 y - z = - 4$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the acute angle between the line AB and the plane $\Pi _ { 1 }$.
\item Show that the line AB meets $\Pi _ { 1 }$ and $\Pi _ { 2 }$ at the same point, whose coordinates should be specified.
\item \begin{enumerate}[label=(\roman*)]
\item Find $( \mathbf { i } - 2 \mathbf { j } ) \times ( 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } )$.
\item Hence find the acute angle between the planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
\item Find the shortest distance between the point A and the line of intersection of the planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q13 [17]}}
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