[MAA 3.17-3.

19] PLANES
SOLUTIONS
Compiled by: Christos Nikolaidis

O. Practice questions

2 1 2

1. (a) r   1     0     1 
   
5  3  
    4

 1   2   3 
     
(b) b c  01   2 
 3  4  1 
     
 3   2
(c) n   2  (the normal vector) a   1  (the position vector of the point)
 
1 5
   
(d) 3 x  2 y  z  d
We plug in a to find d  6  2  5  1
The Cartesian equation of the plane is 3 x  2 y  z  1

 1  1  1  1   12 



  


  





      
2. (a) AB   0  , AC   2  , AB  AC   0    2    9 
6  3 6  3 2
         
(b) 12 x  9 y  2 z  d
We plug in any point to find d . For example pick B(1, 1, 1),
12 x  9 y  2 z  23







(c) AB  AC  12 2 9 2  22  229

229
(d)
2

 1
 
3. (a) n = 1 , hence equation of L through A(2, 5, –1) is given
 1
 
x  2 y  5 z 1
by   .
1 1 1
(b) A general point on L is (2 + λ, 5 + λ, –1 + λ).
At intersection of line L and the plane
5
(2 + λ) + (5 + λ) + (–1 + λ) –1 = 0 3λ = –5 λ = –
3
 1 10 8 
 point of intersection  , , 
3 3 3

1
 (c) Let A(x, y, z) be the reflection of A.
10  4 5 13 
EITHER At A′λ = –  A′ =   , , 
3  3 3 5
OR Since point of intersection of L and the plane is midpoint of AA′
 4 5 13 
 A′ =   , , 
 3 3 5

4.

0
(d) the y-axis has direction vector  1  sinθ 
7
 θ  51.06
 
0 81
 
(e)

(f)

(g)

2
A. Exam style questions (SHORT)

1  2   4  3   7 
5. (a)
       
u × v =  2    1   6  2    4  = 7i + 4j – 5k.
 3   2   1  4   5 
       
   2 
 
(b) w =  2   
 3  2  
 
The line of intersection of the planes is parallel to u × v. Now,
w . (u × v) = 7 + 14 + 8 – 4 –15 – 10 = 0 for all , .
Therefore, w is perpendicular to the line of intersection of the given planes.
OR
The line of intersection of the planes is perpendicular to u and to v,
so it will be perpendicular to the plane containing u and v, that is,
to all vectors of the form u + v = w.

1
 
6. The normal vector to the plane is  3  .
 2
 
 4  1
   
 1 3 = 3
 2 2
   
EITHER  is the angle between the line and the plane.
sin   3  = 10.1(= 0.176)
14 21
OR  is the angle between the line and the normal to the plane.
3  3 
cos θ       = 79.9 (= 1.394 ...)
14 21  7 6 
The required angle is 90 – 10.1 (= 0.176)

7.

3
8.

 0
 
9. z-axis has direction vector  0 
1
 
Let  equal the angle between the line and the plane.
0  3 
   
 0 .  2
1  4 
    4
sin θ =
2
1 3 2 42 2 29

θ = 48°.

10. The point (2, 4, 7) lies on the plane.

 1 
 
The vector joining (2, 4, 7) and (1,  1, 4) is  5 
 3 
 

 1   2   12  4
 5  ×  1  =  3  . We can consider the parallel vector n = 1
       
 3   3  9   3 
       
Cartesian equation of plane is 4x + y  3z = 9 (or equivalent).

11. Let d1 and d2 be the direction vectors of the two lines. Then the normal to the plane is

1  3   7 
 
d1 × d2 =  2  ×
 3  =  2  (or equivalent)
   
1 5  3
     
Then equation of the plane is for the form –7x – 2y + 3z = c
Using the point (1, 1, 2) which is in the plane gives the equation of the plane
–7x – 2y + 3z = –3
OR
1  1   3 
     
r =  1   λ – 2      3  (or equivalent)
 2  1   5 
     

4
12.

13. The direction vector, i + 2j + lk, for the line, is perpendicular to 6i – 2j + k,

the normal of the plane.
Therefore, (i + 2j +lk) . (6i – 2j + k) = 0
Therefore, 6 – 4 + l = 0 so l = –2
OR
x = t + 1, y = 2t – 1, z = lt + 3
6t + 6 – 4t + 2 + lt + 3
2t + lt = 0, so l = –2

14. Substituting gives,

2(2 + 4) + 3(– – 2) – (3 + 2) = 2
 4 + 8 – 3 – 6 – 3 – 2 = 2
 –2 = 2   = –1
Intersection is (2, –1, –1)

x  3 y 1 z 1
15. Let   = , then x = 2 – 3, y = – + 1, z = 2 + 1
2 1 2
Substituting into P gives;
4 – 6 – 3 + 3 – 2 – 1 = –5  = 1
Therefore x = –1, y = 0, z = 3
Therefore the point of intersection is (–1, 0, 3)

16. x = 1 + µ,
y = –µ,
z = 1 + 2µ
2(1 + µ) + µ + 1 + 2µ + 2 = 0. µ = –1
P is(0, 1, –1)

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17.

0 3
   
18 EITHER Gaussian elimination gives r =   1  λ11
  1 1
   
OR
 y  z  2
Let x = 0    3z = –3, z = –1, y = –1  (0, –1, –1)
  y  2 z  1
  4 x  y  2
Let z = 0    –x = –3, x = 3, y = 10  (3, 10, 0)
 3 x  y  1
 0 3
   
The equation of the line of intersection is r =   1  λ11 (or equivalent)
  1 1
   

19. METHOD 1 Gaussian elimination

R1: 1 2 –3 2
R2: 2 3 –5 3
0 0 0 0

R1: 1 2 –3 2
R2 – 2R1: 0 –1 1 –1
0 0 0 0

Let z = t, then y = t + 1 and x = t.

Therefore the line of intersection is x = t, y = t + 1, z = t (or equivalent).
METHOD 2
Let z = 0 x + 2y = 2
2x + 3y = 3
 x = 0, y = 1
The direction vector of the line of intersection is n1×n2 == – i – j – k
 0  1
   
Therefore, the line of intersection is r =  1   t 1 (or equivalent)
 0  1
   
 x  1 2
     
20. Equation of line is  y    1     1 
 z  9  – 1
     
Coordinates of foot satisfy
2(1 + 2) + (1 + ) – (9 – ) = 6  6 = 12   = 2
Foot of perpendicular is (5, 3, 7)

6
21.

 2 2 4
22. (a)
     
OP  OA  OB =  1  +  1 =  0   P = (4, 0, –3)
 2   1  3 
     
 2   1   3
     
OQ  OA  OC =  1  +  2  =  3   Q = (3, 3, 0)
 2   2   0 
     
 2   1   3
     
OR  OB  OC =  1 +  2  =  1   R = (3, 1, 1)
 1   2   1 
     
 4  1  5 
     
OS  OP  OC =  0  +  2  =  2   S = (5, 2, –1)
 3   2   1
     
 2 2  3 
(b)
   
OA  OB =  1  ×  1 =
 2 
 
 2   1  4 
     
 2   2
   
An equation of the plane is 3x + 2y + 4z = 0 OR r =   1      1
  2   1
   
 3   1 
(c)
   
V =   2  i  2   = –3 – 4 – 8= 0
 4   2 
   

7
INTERSECTION OF THREE PLANES

23.
[MAA 1.10] Geometric relationship of the three planes
Exercise (vector equation of the line of intersection if applicable)

(i) The intersection point of the three planes is (1, 1,1)

(ii) There is no point of intersection. The three planes form a prism
1 0 1
(iii) The three planes intersect in the straight line: r   3     2 
   
0 1
   

2 The intersection point of the three planes is (1,-1,2)

3 The intersection point of the three planes is (-1,2,3)

4 The intersection point of the three planes is (1.2, 0.6, 1.6)

  1 / 12   1/ 6
The three planes intersect in the straight line    
5 r   1/ 6    2 / 3 
 0   1 
   

 4  3
The three planes intersect in the straight line    
6 r   1    2
 0 1
   

If k  3 : There is no point of intersection. The three planes form a prism

7 1  7 / 5 
If k  3 : The three planes intersect in the line
r   1     11/ 5 
0  1 
   

If k  5 : There is no point of intersection. The three planes form a prism

8  3  3 
If k  5 : The three planes intersect in the line    
r    1     3 
 0  1 
   

If k  3, 1 : There is a unique point of intersection.

3
1
If k  : There is no point of intersection. The three planes form a prism
9 3
 6   7 
If k  3 : The three planes intersect in the line
r   3     2 
 0   1 
   

If a  1 : There is a unique point of intersection.

10
If a  1 : There is no point of intersection. The three planes form a prism

If k  1 : There is no point of intersection. The three planes form a prism

11  11 / 3    7 / 3
If k  1 : The three planes intersect in the line    
r    4 / 3    2 / 3 
 0   1 
   

8
B. Exam style questions (LONG)

 x  3   1   3 
       
24. (a)  y     4    2     4  3x – 4y + z = 6
 z   1  11  1 
       
(b) (i) 1 + 3 × 2 – 11 = –4 P lies in π2

 3   1  1 1 1

         
(ii)   4   3    4  r =  2    4 
 1    1 13  11 13 
         
1  3  1  3 
       
(c)  3     4  = – 10  3   11   4   26
  1  1    1  1 
       
 10
cosθ = (= –0.5913)
11 26
θ = 2.2035 radians (or θ = 126.3°)
The angle between the planes is π – 2.2035 = 0.938 radians (or 180° – 126.3° = 53.7°)
25.

(b)

9
26. (a) The plane has equation 6 x  5 y  4 z  15
Since P(p, –p, p) lies on the plane: 6 p  5 p  4 p  15  p  3
 P(3, –3, 3)

Q lies in y-axis: Q(0, λ, 0): 5  15    3

 1   1   5 
   
n   3  ×  2  =  2 
 1   1   1 
     
So the plane has the form 5 x  2 y  z  d
Substituting the point we obtain d  6

27. (a) By GDC OR by Gauss elimination, back substitution gives x = 1, y = –1, z = 2.

 1   2   11 
(b)
     
v = 3 × 1 = 7 [OR 11i – 7j – 5k.]
     
 2   3   5 
     
 1   2   m  2n 
(c)
     
u = m  3  + n  1  =  3m  n 
 2   3   2m  3n 
     
Therefore, v . u = 11(m + 2n) – 7(3m + n) – 5(–2m + 3n)
= 11m + 22n – 21m – 7n + 10m – 15n
= 0, for all m and n.
That is, v is perpendicular to u for all values of m and n.
OR
v is perpendicular to both a and b [from part (b)].
Therefore, v . a = v . b = 0,
so v . u = m(v . a) + n(v . b) = 0,
and hence v is perpendicular to u for all values of m and n.

10
 3
(d)
 
The normal to the plane,  1 , and v are both perpendicular to the required line, l.
1
 
Therefore, the direction of l is given by
3  11  3   12  6
 
v ×  1 =
 7  ×
 1 =  26  or the parallel vector  13 
       
1  5  1   10   5 
         
1 6
   
Thus, an equation for l is r =  1    13 
2  5 
   

28. (a) Since the coordinates of the points P, Q and R are (4, 1, –1), (3, 3, 5)
and (1, 0, 2c), respectively
 2   3 
   
QR   3  , PR   1 
 2c  5   2c  1
   
QR  PR  QR  PR = 0  6 + 3 + (2c – 5)(2c +1) = 0
 4c2 – 8c + 4 = 0 (c – 1)2 = 0 c = 1
 3   3   3
     
(b) PS   0  PR   1  PS  PR  0
 
3 3  3
     
(c) The parametric equation of a line l which passes through the
 3   3 
   
point (3, 3, 5) and is parallel to the vector PR is given by r =  3  t  1 
5  3 
   

(d) Let P1 and P2 be points on the line l corresponding to t = 0 and t = 1, respectively.

P1 (3, 3, 5) and P2 (0, 2, 8)
 2  1
   
SP1   2  and SP2   1 
 3 6
   
 9 

SP1  SP2  15

 
 4 
 
Using the point S(1,1,2) we obtain plane
9x – 15y + 4z = 2
  1  9 
  
 2  .   15 
PQ  n  6   4 

15
(e) Shortest distance   = (=0.836 3 s.f.)
n 322 322
ΟR we find the foot and hence the distance,

11
  1  1
   
29. (a) AB   3  , BC   1 
1 0
   
 1   1   1
(b)
     
AB BC   3    1    1 
 1  0  2 
     
1 1 6
(c) Area of ∆ABC = │ AB × BC │= 11 4 =
2 2 2
(d) The equation of the plane is of the form –x + y + 2z = d
Since the plane contains A, then –1 + 2 + 2 = d  d = 3.
Hence, an equation of the plane is –x + y + 2z = 3.
(e) Vector n above is parallel to the required line.
Therefore, x = 2 – t
y=–1+t
z = – 6 + 2t
(f) –x + y + 2z = 3
–2 + t – 1 + t – 12 + 4t = 3  –15 + 6t = 3  6t = 18  t = 3
Point of intersection E(–1, 2, 0)
(g) Distance = 3 2  3 2  6 2  54
 1
1 1  
(h) Unit vector in the direction of n is e = n= 1 Note: –e is also acceptable.
n 6  
2
(i) D(2, –1, –6)., E (–1, 2, 0) is the midpoint of DF  F(–4, 5, 6)

0 1
   
30. (a) (i) AB   1 and AC   0 
3 5
   
 0   1   5 
     
AB × AC   1   0    3  = –5i + 3j + k
 3  5  1 
     
1 35
(ii) Area = ×–5i + 3j + k=
2 2
(b) (i) The equation of the plane  is – 5x + 3y + z = c
where c = –5 + 9 + 1 = 5, that is, – 5x + 3y + z = 5
x–5 y2
(ii) Equations of L are  =z–1
–5 3
(c) L meets  where
–5(5 – 5) + 3(3 – 2) +  + 1 = 5  = 1
Point of intersection is (0, 1, 2).
(d) Perpendicular distance is 52  32  12 = 35

12
31. (a) M (3 – 2, , 9 – 2)
 4  3   3 – 2 – 4   3 – 6 
x – 4 y z 3        
(b) (i)   or r   0     1  (ii) PM       
3 1 –2  – 3  – 2   9 – 2  3 12 – 2  
       
 3 – 6   3 
   
(c) (i)      1   0  9 – 18 +  – 24 + 4 = 0   = 3
12 – 2    – 2 
   
3
 
(ii) PM   3  PM  3 2  3 2  6 2 = 54  3 6
 6
 
 3   3   12   2
       4 
(d) n = PM   3  ×  1    24  parallel to  
 6  – 2   1
     6   
2x – 4y + z = d
Substituting the point: d = 5
2x – 4y + z = 5
(e) l1is on π1 from part (d).
Testing l1 on π2 gives (3 – 2) –5() – (9 – 2) = –11.
Therefore l1 is also on π2 and is therefore the line of intersection.

32. (a) L1 : x = 2 +  ; y = 2 + 3 ; z = 3 + 
L2 : x = 2 + µ ; y = 3 + 4µ ; z = 4 + 2µ
At the point of intersection
2+=2+µ (1)
2 + 3 = 3 + 4µ (2)
3 +  = 4 + 2µ (3)
From (1),  = µ. A1
Substituting in (2), 2 + 3 = 3 + 4  = µ = –1.
These values satisfy (3). Therefore the lines intersect.
So P is (l, –1, 2).
1  1  2 
(b)
     
normal vector is given by 3 × 4 = 1
     
1  2  1 
     
The Cartesian equation of  is 2x – y + z = 2 + 1 + 2, i.e. 2x – y + z = 5
(c) The midpoint M of [PQ] is (2, 3/2, 5/2).
The direction of MS is the same as the normal to , ie 2i – j + k
The coordinates of a general point R on MS are therefore R  2  2 , 3   , 5   
 2 2 
It follows that PR = (1 + 2)i +  5    j +  1    k
2  2 
At S, length of PR is 3, ie
(1 + 2)2 + (5/2 – )2 + (1/2 + )2 = 9
1 + 4 + 42 + 25 / 4 – 5 + 2 + 1 / 4 +  + 2 = 9
6 1
62 =  =
4 2
Substituting these values,
the possible positions of S are (3, 1, 3) and (1, 2, 2)

13
33. (a) (i) (1) 2  2    2  s  t
(2) 1    3  2s  t
(3) 1  8  9   1  s  t
subtracting (3) from (1): 1  10  10   1    
(ii) On the line of intersection     an equation of the line is
 2   2  1   2  1
         
r =  1    1     3 =  1     2 
1  8    9 1  1
         
(b) The plane 3 contains, e.g. the point (2, 0, –1).
x  3   2   3 
       
The equation of the plane is  y    – 2    0    – 2   5.
 z   1   – 1  1 
       
The cartesian equation of the plane is 3x  2 y  z  5 .
 2  1 
   
(c) Intersection between line r   1  +   2  and 3 .
1  1 
   
3 x  2 y  z  5  3(2   )  2(1  2 )  1    5
This equation is satisfied by any real value of 
 2  1
   
 the 3 planes intersect at the line r =  1      2  .
1  1
   
1
34. (a)
 
Direction vector of l1:  4 
 3 
 

14
35.

OR
By GDC: m = –1 or m = 7  m = –1
 1  2   2 
(b)
     
n =  1   1  =  3 
 1  1   1 
     
Substituting the point A(2, -1, 0)
2 x  3 y  z  7
1






(c) Area ABC = BA  BC
2

 2   2   3 
36. (a)
     
(i)  2  ×  1  =  6 
 3   0   6 
     
1
 
(ii) a parallel normal vector is n =  2 
 2
 
The equation of plane has the form x  2 y  2 z  d
Substituting (3,1,5)  d  11
The equation of the plane is x  2 y  2 z  11

15
 (c) EITHER by Gaussian elimination OR finding direction of the line and a point

(e) (i)

37.

 1   1   6 
(b)
     
a × b =  1 ×  2  =  3 
 1  4  3 
     

(c) (i)

16
 (e)

38. (a) x = 3 + 2m x = 1 + 4n
y=2–m y=4–n
z = 7 + 2m z=2+n

(b) 3 + 2m = 1 + 4n  2m – 4n = – 2 (i)
2 – m = 4 – n  m – n = –2 (ii)
7 + 2m = 2 + n  2m – n = –5 (iii)
(iii) – (ii)  m = –3  n = –1
Substitute in (i), – 6 + 4 = – 2. Hence lines intersect.
Point of intersection A is (–3, 5, 1)
 2   4  1
(c)
 1 ×  1 =  6 
     
 2   1   2
     
1  3  1 1
       
r •  6   2   6  r •  6  = 29  x + 6y + 2z = 29
 2 7  2  2
       

(d) x = –8 + 3λ
y = – 3 + 8λ
z = 2λ
Substitute in equation of plane.
–8 + 3λ – 18 + 48λ + 4λ = 29  55λ = 55  λ = 1
Coordinates of B are (–5, 5, 2)

  4
  1
   
 3
(e) Coordinates of C are   4, 5,  r =  5    6
 2  3   
   2
 
 2 

17
   1   1   0   3  1   2 
39. (a) AB = ba =  2    1    1  AC = ca =  0    1     1
 3  2  1  1   2    1
           
 0
 
 
AB  AC =  2 
 2 
 
Area of triangle ABC =
1
2
 
8  2 sq. units

(b) AB = 2
1 1
2= AB h  2  h , h is the shortest distance  h = 2
2 2
 0 
 
(c)  has form r •  2   d
  2
 
1  1 
   
Since (1, 1, 2) is on the plane d =  1  •  2   2  4   2
 2   2
   
Hence 2y  2z = 2 (or y  z = 1)
(d) The equation of OD is
 0    0 
    
r =  2 ,  or r  λ  1  
  2    1 
    
1
This meets  where 2 + 2 = 1 = 
4
 1 1
Coordinates of D are  0 ,  , 
 2 2
2 2

 1 1 1
(e) OD  0        
 2 2 2

18