MEI Core 4
Vectors
Chapter Assessment
1. Find:
1  3 
(i) 2     
 4   1
 2   2
(ii)   .  
 3   4 
[2]
[2]
 2   1
(iii)  1  .  3 
0  2 
  
[2]
2. (i) Work out the angle between the vectors:
2
 3
(a)
  and  
 1
 2 
(b)
2
1
  and  
1
 2 
4
3
 
 
[3]
[3]
 2
 1
 1
 2
 
 
 
 
(ii) Show that the lines r   1   λ  2  and r   3   μ  5  are perpendicular to
0
4
2
 3
 
 
 
 
each other.
[3]
3. The points A, B and C have coordinates (2, -1), (3, 2) and (-4, 0) respectively.
(i) Write down the vectors AB , AC and BC .
[3]
(ii) Find BC .
[2]
(iii) Find ABC .
[3]
4. (i) Find the vector equation of the line joining A(2, -1, 0) to B(3, -2, -5).
(ii) Verify that A and B both lie on the plane 2 x  3 y  z  7 .
(iii) Write down the vector equation of the line passing through A which is
perpendicular to the plane.
[3]
[2]
[2]
5. The points A, B and C have coordinates (2, -1, 3), (4, -2, 0) and (1, -5, 0)
respectively.
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MEI C4 Vectors Assessment solutions
1
1


(i) Work out AB.  1 and BC.  1 .
1
1
 
 
(ii) Find the Cartesian equation of the plane ABC.
[4]
[3]
2
 4 
6. Work out the coordinates of the point where the lines r     λ   and
 1
 3
5
2
[6]
r     μ   intersect.
 3 
 1
7. The position vectors of A and B are as follows:
a  i  2 j  3k
A:
b  2i  4 j  k
B:
(i) Find the vector equation of the line AB.
(ii) The line AB meets the plane 6 x  y  3z  13  0 at the point C.
Find the position vector of C.
[3]
[4]
Total 50 marks
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MEI C4 Vectors Assessment solutions
Solutions to Chapter Assessment
1.
 1  3  2  3   5 
(i) 2              
 4   1   8   1   7 
 2  2
(ii)   .    2  2  ( 3)  4  4  ( 12)  8
 3   4 
 2   1 
   
(iii)  1  .  3   2 ( 1)  1  3  0  2  2  3  1
0  2 
   
2. (i) Using the formula: cos  
[2]
[2]
[2]
a.b
a b
2
(a) Let a     a  2 2  ( 1)2  4  1  5
 1 
 3
and b     b  32  ( 2)2  9  4  13
 2 
2 3
also a.b    .    2  3  ( 1) ( 2)  6  2  8
 1   2 
8
 0.992...    7.13 to 3 s.f.
So cos  
5 13
2
 
(b) Let a   1   a  2 2  1 2  42  4  1  16  21
 4
 
 1 
 
and b   2   b  1 2  ( 2)2  3 2  1  4  9  14
 3 
 
[3]
2  1 
  
also a.b   1  .  2   2  1  1 ( 2)  4  3  2  ( 2)  12  12
 4  3 
  
12
 0.699...    45.6 to 3 s.f.
So cos  
[3]
14 21
(ii) The lines are perpendicular if the angle between the two direction vectors is
90°. This means the dot (or scalar) product of the two direction vectors is 0.
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MEI C4 Vectors Assessment solutions
 1   2 
 

So  2  .  5   ( 1)  2  2 ( 5 )  4  3  2  ( 10)  12  0
4 3 
 

as required.
[3]
3  2   1
3. (i) AB  OB  OA         
 2   1   3 
 4   2   6 
AC  OC  OA         
 0   1   1 
 4   3   7 
BC  OC  OB        

 0   2   2 
(ii) BC  ( 7)2  ( 2)2  49  4  53
[3]
[2]
(iii) ABC is the angle between BC and BA
 1 
BA   AB   
 3 
Using the formula: cos  
BC.BA
BC BA
 7   1 
BC.BA  
 .    ( 7) ( 1)  ( 2) ( 3)  7  6  13
 2   3 
BC  53 (from part (ii))
BA  ( 1)2  ( 3)2  1  9  10
13
 0.564...    55.6 to 3 s.f.
So cos  
53 10
[3]
4. (i) The vector equation of a line is r  OA   AB
 3  2  1 

   

and AB  OB  OA   2    1    1 
 5   0   5 

   

2
 1 
 


So the equation of the line is r   1     1 
 0 
 5 
 


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MEI C4 Vectors Assessment solutions
(ii) Substitute the coordinates into the equation of the plane:
A(2, -1, 0): 2  2  3 ( 1)  0  4  3  7
B(3, -2, -5): 2  3  3 ( 2)  ( 5 )  6  6  5  7 as required.
[2]
(iii) A vector normal to the plane is found by the coefficients of x, y and z in the
 2 
 
equation of the plane ie.  3 
 1 
 
2
2 
 
 
So the equation is: r   1     3  .
0
 1 
 
 
[2]
 4 2 2 
     
5. (i) AB  OB  OA   2    1    1 
 0   3   3 
     
 1  2  1 
    
AB.  1    1  .  1   2  1  3  0
 1   3   1 
    
 1   4   3 

    
BC  OC  OB   5    2    3 
 0   0   0 

    
 1   3   1 
    
BC.  1    3  .  1   3  3  0  0
 1   0  1 
    
[4]
 1 
 
(ii) The vector  1  is perpendicular to two lines on the plane and so it must be
 1 
 
perpendicular to the plane.
The equation of the plane is r.n  a.n
 1 2 1
    
r.  1    1  .  1   2  1  3  6
 1 3 1
    
So the Cartesian equation of the plane is x  y  z  6
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MEI C4 Vectors Assessment solutions
2
 4   5 
2
6. At intersection,             
 1 
 3   3 
 1 
2  4  5  2 
Reading across:
x:
1  3  3  
y:
2  6  6  2 
Multiply  by 2:
2  4  5  2 
+
2  6  6  2 
+



2  1     21
x  2 
 4 
Substitute    21 into r           :
 y   1 
 3
x  2 
 4   4 
r        21    
1
 y   1 
 3   2 2 
So the point of intersection is at (4, 2 21 ) .
2  1   1 
     
7. (i) Direction vector is b  a   4    2    6 
 1   3   4 
     
 1 
 1 
 
 
The equation of the line is r   2     6  .
 3 
 4 
 
 
[6]
[3]
x  1 
 1 
   
 
(ii) The line AB is: r   y    2     6 
z   3 
 4 
   
 
So reading across: x  1  
y  2  6
z  3  4
Substituting into the equation of the plane 6 x  y  3z  13  0 :
6(1   )  ( 2  6 )  3(3  4 )  13  0
12  12   0
   1
x  1 
 1   0 
   
   
Substitute   1 into the line AB: r   y    2   1  6    8 
z   3 
 4   7 
   
   
So the position vector of C is: c  8 j  7k
[4]
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