OCR MEI C4 — Question 5 17 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeLine-plane intersection and related angle/perpendicularity
DifficultyStandard +0.3 This is a standard C4 vectors question involving routine calculations: finding vector lengths and angles, verifying a normal vector, finding plane equations, and calculating angles between planes. All techniques are textbook exercises requiring no novel insight, though the multi-part structure and coordinate geometry in 3D places it slightly above average difficulty.
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 plane
A tent has vertices ABCDEF with coordinates as shown in Fig. 7. Lengths are in metres. The \(Oxy\) plane is horizontal. \includegraphics{figure_3}
  1. Find the length of the ridge of the tent DE, and the angle this makes with the horizontal. [4]
  2. Show that the vector \(\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}\) is normal to the plane through A, D and E. Hence find the equation of this plane. Given that B lies in this plane, find \(a\). [7]
  3. Verify that the equation of the plane BCD is \(x + z = 8\). Hence find the acute angle between the planes ABDE and BCD. [6]
Question 5:
AnswerMarks Guidance
5(i) DE = [(5)2 + 02 + 12] = 26
26
1
5
cos  = 5/26 oe
AnswerMarks
  = 11.3M1 A1
M1
A1
AnswerMarks
[4]oe
oe using scalar products eg –5i + k with i oe
or better (or 168.7). Allow radians.
AnswerMarks Guidance
5(ii) 1 5 
 
   
AE 4 , ED 0
   
   
3 1
1 1 
   
4 . 4 116150
   
   
3 5 
5  1 
   
0 . 4 5050
   
   
1 5 
 i  4j + 5k is normal to AED
x 1  0  1 
 eqn of AED is        
y . 4  4 . 4
       
       
z 5  0  5 
 x  4y + 5z = 16
B lies in plane if 8  4(a) + 5.0 = 16
AnswerMarks
 a = 2B1
B1
B1
M1
A1
M1
A1
AnswerMarks
[7]two relevant direction vectors (or 6i + 4j + 2k oe)
scalar product with a direction vector in the plane (including evaluation and = 0)
(OR M1 forms vector cross product with at least two correct terms in solution)
scalar product with second direction vector, with evaluation.
(following OR above, A1 all correct ie a multiple of i – 4j + 5k)
(NB finding only one direction vector and its scalar product is B1 only.)
for x – 4y + 5z = c oe
M1A0 for i – 4j + 5k = 16
allow M1 for subst in their plane or if found from say scalar product of normal
with vector EB can also get M1A1
For first five marks above
SC1, if states, ‘if i – 4j + 5k is normal then of form x – 4y + 5z = c’ and substitutes
one coordinate gets M1A1, then substitutes other two coordinates A2 (not
A1,A1).Then states so1  is normal can get B1 provided that there is a clear
 
4
 
 5  
argument ie M1A1A2B1. Without a clear argument this is B0.
SC2, if finds two relevant vectors, B1 and then finds equation of the plane from
vector form, r = a + μb + λc gets B1. Eliminating parameters B1cao.
If then states ‘so 1  is normal’ can get B1 (4/5).
 
4
 
 5  
Question 5:
5 | (i) | DE = [(5)2 + 02 + 12] = 26
26
1
5
cos  = 5/26 oe
  = 11.3 | M1 A1
M1
A1
[4] | oe
oe using scalar products eg –5i + k with i oe
or better (or 168.7). Allow radians.
5 | (ii) | 1 5 
 
   
AE 4 , ED 0
   
   
3 1
1 1 
   
4 . 4 116150
   
   
3 5 
5  1 
   
0 . 4 5050
   
   
1 5 
 i  4j + 5k is normal to AED
x 1  0  1 
 eqn of AED is        
y . 4  4 . 4
       
       
z 5  0  5 
 x  4y + 5z = 16
B lies in plane if 8  4(a) + 5.0 = 16
 a = 2 | B1
B1
B1
M1
A1
M1
A1
[7] | two relevant direction vectors (or 6i + 4j + 2k oe)
scalar product with a direction vector in the plane (including evaluation and = 0)
(OR M1 forms vector cross product with at least two correct terms in solution)
scalar product with second direction vector, with evaluation.
(following OR above, A1 all correct ie a multiple of i – 4j + 5k)
(NB finding only one direction vector and its scalar product is B1 only.)
for x – 4y + 5z = c oe
M1A0 for i – 4j + 5k = 16
allow M1 for subst in their plane or if found from say scalar product of normal
with vector EB can also get M1A1
For first five marks above
SC1, if states, ‘if i – 4j + 5k is normal then of form x – 4y + 5z = c’ and substitutes
one coordinate gets M1A1, then substitutes other two coordinates A2 (not
A1,A1).Then states so1  is normal can get B1 provided that there is a clear
 
4
 
 5  
argument ie M1A1A2B1. Without a clear argument this is B0.
SC2, if finds two relevant vectors, B1 and then finds equation of the plane from
vector form, r = a + μb + λc gets B1. Eliminating parameters B1cao.
If then states ‘so 1  is normal’ can get B1 (4/5).
 
4
 
 5  
A tent has vertices ABCDEF with coordinates as shown in Fig. 7. Lengths are in metres. The $Oxy$ plane is horizontal.

\includegraphics{figure_3}

\begin{enumerate}[label=(\roman*)]
\item Find the length of the ridge of the tent DE, and the angle this makes with the horizontal. [4]

\item Show that the vector $\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ is normal to the plane through A, D and E.

Hence find the equation of this plane. Given that B lies in this plane, find $a$. [7]

\item Verify that the equation of the plane BCD is $x + z = 8$.

Hence find the acute angle between the planes ABDE and BCD. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4  Q5 [17]}}
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Q1 18 Q2 4 Q3 7 Q4 18 Q5 17