CHAPTER 6 : VECTORS

6. Cartesian Vectors
Vector is a quantity that has both magnitude and direction. It is typically
Vectors express in terms of i, j and k, which are unit vectors parallel to the
represented by an arrow whose direction is the same as that of the quantity
and whose length is proportional to the quantity's magnitude. x, y and z axes respectively. y
a
 
Theorems PQ = ai + bj + ck =  b  and PQ =│PQ │= a 2  b2  c 2
 
1. Parallel Vectors : a and b are parallel if and only if a = kb for k ≠ 0. c 0 x
z

2. Non-Parallel Vectors : If a and b are not parallel then

eg 1 │3i – 4j + 5k│= 32  ( 4)2  52 = 5 2
(a) ma = nb, for m = n = 0
(b) pa + qb =  a +  b, for p =  and q = 
eg 2 The points P, Q and R are the midpoints of the sides BC, CA and AB
respectively of the triangle ABC. The lines AP and BQ meet at the point G,
P
3. Vectors Addition / Subtraction
 where AG = m AP and BG = n BQ.
a–b a+b
(a) PQ + QR = PR (a) Show that AG = 1 m AB + 1 m AC and AG = (1 – n) AB + 1 n AC.
a
2 2 2
(b) PQ – PR = –QR = RQ   R 2 2
Deduce that AG = AP and BG = BQ.
–b Q b 3 3
4. Position Vectors (b) Show that CR meets AP and BQ at G, where CG = 2 CR.
3
 vectors that begin from origin, O

(Generally OA is denoted by a ) (a) AG = m AP = m ( 1 AB + 1 AC) = 1 m AB + 1 m AC _____ (1)

eg. Position vector of A = OA 1 1 1 1 2 2

Note : RQ = OQ – OR (1) R (1) A

 
B
AG = 1 n AB +
n  (1 n)
n
n  (1 n)
AQ 
(n) (m )
G
5. Ratio Theorem of Vectors
1 (m ) N (n) R
= (1 – n) AB + 1 n AC _____ (2) (1)  (1 – n)
(1)
For QN : NR = m : n ,    Since (1) = (2) :
2
P  (1 – m) Q
– 1 m = 1 – n and 1 m = 1 n
m n (1) (1)
2 2 2
PN =
mn
PR +
mn
PQ
n P  m = n = 2  AG = 2 AP and BG = 2 BQ C
) 3 3 3
(
Proof : QN : NR = m : n  n1QN = m NR (b) CR = CA + AR = –AC + 1 AB = 1 (AB – 2 AC)
2 2
n (PN – PQ) = m (PR – PN)  (m– + n) PN = m PR + n PQ CG = CA + AG = –AC + ( 1 AB + 1 AC) = 1 (AB – 2 AC) = 2 CR
3 3 3 3
m n  C, G and R are collinear points  CR meets AP and BQ at G
PN = PR + m PQ
mn m)  n
(
n
) 6-1
(
m
)
eg 3 The position vectors of the points Q and R, relative to the origin O, eg 4 The position vectors of points A, B, C and D relative to the origin are
are q and r respectively. i + j – 2k, 2i – j + k, 3i + j and 2i + 3j – 3k respectively.
Show that ABCD is a parallelogram.
(a) Find an expression for the position vector of the point N which lies
on the line QR such that QN : NR = 3 : 2.
AB = (2i – j + k) – (i + j – 2k) = i – 2j + 3k
(b) The point P with position vector p, relative to O, is such that O is
the midpoint of the line PN. Prove that 5p + 2q + 3r = 0. DC = (3i + j) – (2i + 3j – 3k) = i – 2j + 3k

(c) The point L lies on the line RP such that RL : LP = 5 : 3.

 AB = DC  ABCD is a parallelogram
Show that the points Q, O and L are collinear, and find QO : OL.
Alternative 1
(d) The point M lies on the line PQ such that MOR is a straight line.
Find PM : MQ. AD = (2i + 3j – 3k) – (i + j – 2k) = i + 2j – k

2 OQ + 3 OR BC = (3i + j) – (2i – j + k) = i + 2j – k
(a) QN : NR = 3 : 2  ON =  ON = 2 q + 3 r
23 23 5 5
 AD = BC  ABCD is a parallelogram
(b) PO = ON : –p = 2 q + 3 r  5p + 2 q + 3 r = 0
5 5
Alternative 2
(c) OL = 5 OP + 3 OR = 5 (– 2 q – 3 r) + 3 r = – 1 q = – 1 OQ
53 53 8 5 5 8 4 4 1
 Q, O and L are collinear points, with QO : OL = 4 : 1. OM = [(i + j – 2k) + (3i + j)] = 2i + j – k
2
1
(d) Let PM = m PQ and OM = –k OR = –kr : ON = [(2i – j + k) + (2i + 3j – 3k)] = 2i + j – k
2
 OM = m OQ + (1 – m)OP  (1 – m)( 2 q + 3 r) = mq + kr
5 5  OM = ON  ABCD is a parallelogram
 2 (1 – m) = m  m= 2
5 7 Alternative 3

 PM = 2 PQ   1 3 1 1 2  0   2  2 1 3 1 3 
PM : MQ = 2 : 5 M=  , ,  = (2, 1, –1) N=  , ,  = (2, 1, –1)
7  2 2 2   2 2 2 

 M=N  ABCD is a parallelogram

6-2
Scalar Product of Vectors (Dot Product of Vectors) eg 2 The position vectors of points A, B and C relative to the origin are
2i – j + 3k, 4i + 3j + ak and –4i + 5j + 6k respectively. Point P
b
Definition : a  b = │a│.│b│cos θ divides the line AC in the ratio 1 : 2. Find the position vector of P.
where θ is the angle between a and b θ Hence, determine the value of a such that APB is 90o.
a
(1 + 2) OP = 2(2i – j + 3k) + (–4i + 5j + 6k)  OP = j + 4k
Properties of Scalar / Dot Product
PA = OA – OP = 2i – 2j – k PB = OB – OP = 4i + 2j + (a – 4)k
1. Commutative : ab=ba
APB is 90o : (2i – 2j – k)  [4i + 2j + (a – 4)k] = 0
 8 – 4 – (a – 4) = 0  a = 8
2. a  a = │a│2  ii=1, jj=1 , kk=1
eg 3 Prove that the diagonals of a rhombus are perpendicular to each other.
3. ab=0 if and only if ab S
PR  QS = (a + b)  (a – b) a b
= aa – ab + ba – bb
 ij=ji=ik=ki=jk=kj=0 = │a│2 – │b│2 (Since │a│= │b│) P R
=0 a
 Diagonals of a rhombus are perpendicular to each other
b
4. Distributive : a  (b + c) = a  b + a  c Q

 (ai + bj + ck)  (pi + qj + rk) = ap + bq + cr

eg 4 Given that (4a + b) is perpendicular to a, and │b│= 2│a│.
Find the angle between (a + b) and b.
5. To find angle between two vectors
(4a + b)  a = 0  4│a│2 + b  a = 0

(ai + bj + ck)  (pi + qj + rk) = a 2  b2  c 2 . p2  q 2  r 2 . cos θ (a + b)  b = a  b + │b│2 = – 4│a│2 + │2a│2 = 0

ap  bq  cr  Angle between (a + b) and b is 90o.
 θ = cos –1( )
a 2  b2  c 2 . p2  q 2  r 2
eg 5 Find the values of a if the angle between the vectors i + aj + 2k
and –3i + 4j + 5k is 45o.
eg 1 Find the angle between the vectors 3i – 4j + 5k and 2i + 3j – 6k.
(i + aj + 2k)  (–3i + 4j + 5k) = 12  a 2  22 . ( 3)2  42  52 . cos 45o
(3i – 4j + 5k)  (2i + 3j – 6k) = 32  ( 4)2  52 . 22  32  ( 6)2 . cos θ  –3 + 4a + 10 = 5  a 2 . 5 2 .
1
 4a + 7 = 5 5  a 2
36 2
 6 – 12 – 30 = 5 2 (7) cos θ  θ = cos –1( ) = 136.66o
35 2  16a2 + 56a + 49 = 25(5 + a2)  9a2 – 56a + 76 = 0
38
 (9a – 38)(a – 2) = 0  a= or 2
9

6-3
Vector Product of Vectors (Cross Product of Vectors) eg 1 Given that a = 4i + 2j – 3k and b = 2i – 2j + k.
Find │a x b│, │b x a│, (b x a)  a and (b x a)  b.
axb
Definition : 1. │a x b│ = │a│.│b│sin θ  4  2  4 
     
where θ is the angle between a and b b a x b =  2  x  2  =  10   │a x b│= (4)2  (10)2  (12)2 = 2 65
θ  3   1  12 
2. (a x b) is perpendicular to a and to b      
[ i.e. (a x b)  a = 0 and (a x b)  b = 0 ] a
 2  4  4
     
b x a =   2  x  2  =  10   │a x b│= 42  102  122 = 2 65
 1  3   12 
Properties of Vector / Cross Product   

(b x a)  a = (4i + 10j + 12k)  (4i + 2j – 3k) = 16 + 20 – 36 = 0

1. Non-Commutative : a x b = – (b x a ) , same magnitude, but opposite direction (b x a)  b = (4i + 10j + 12k)  (2i – 2j + k) = 8 – 20 + 12 = 0

eg 2 A and B are points with position vectors 2i – 3j + k and 4i + 5j – 3k

2. axa=0  ixi=0, jxj=0 , kxk=0
respectively. Find │OA x OB│, │AO x AB│ and │BO x BA│.
Hence, deduce the area of triangle OAB.
3. ixj=k , jxk=i , kxi=j
 2  4  4 
j x i = –k k x j = –i i x k = –j      
OA x OB =  3  x  5  =  10   │OA x OB│= 42  102  222 = 10 6
 1  3   22 
     
4. Distributive : a x (b + c) = a x b + a x c
 2   2   4 
     
i j k  a   p   br  cq  AO x AB =  3  x  8  =  10   │AO x AB│= (4)2  (10)2  (22)2 = 10 6
     1  4   22 
 (ai + bj + ck) x (pi + qj + rk) = a b c or  b  x  q  =   (ar  cp )       
     
p q r  c   r   aq  bp   4   2   4 
= (br – cq)i – (ar – cp)j + (aq – bp)k      
BO x BA =  5  x  8  =  10   │BO x BA│= 42  102  222 = 10 6
 3   4   22 
     

b 1
1 1  Area of triangle OAB = │OA x OB│= 5 6
5. Area of triangle = │a│.│b│sin θ = │a x b│ 2
2 2 θ
a
Note : 1. Cross product is NOT suitable in finding angle between vectors.
1 1
2. Area of triangle OAB = │OA x OB│≠ │AO x OB│.
2 2

6-4
eg 3 A, B and C are points with position vectors 2i – 2j + 3k, 4i – 2j + k and eg 5 Three vectors a = pi + qj, b= –5i + j and c = 4i + 7j are such that a and b
2i + 2j – k respectively. Find a vector which is perpendicular to the plane are perpendicular and the scalar product of a and c is 78.
that contains the triangle OAB and triangle ABC respectively.
(a) Determine the values of p and q.
 2  4  4   2
        (b) Find the angle between a and c.
OA x OB =  2  x  2  =  10   perpendicular vector to OAB =  5 
 3   1  4   2 (a) (pi + qj)  (–5i + j) = 0 : –5p + q = 0 .......(1)
       
(pi + qj)  (4i + 7j) = 78 : 4p + 7q = 78 .......(2)
 2  0  8   1  p = 2 , q = 10
       
AB x AC =  0  x  4  =  8   perpendicular vector to ABC =  1
 2   4   8   1 (b) | a | = 4  100 = 104 | c | = 16  49 = 65
       
a  c = | a |.| c |cos θ : 78 = 104 . 65 cos θ  θ = 18.43o

eg 4 The points A, B and C are the vertices of the triangle ABC. The position
vectors of the points A, B and C relative to the origin are eg 6 P(1 , 3, –2), Q(2, –1, 1) and R(4, –2, 0) are three vertices of a triangle.
4i – 8j + k, 2i + j – 2k and 3i – 4j + 2k respectively.
(a) Find PQ x PR. Hence, flnd the area of the triangle PQR.
(a) Find a vector which is perpendicular to the vectors AB and AC.
(b) Calculate the area of the triangle ABC. (b) Given that PQ' = 2 PQ and PR' = 2 PR.
Determine the ratio of the area ΔPQR : ΔPQ'R'.
(a) AB = –2i + 9j – 3k & AC = –i + 4j + k
(a) PQ = (2i – j + k) – (i + 3j – 2k) = i – 4j + 3k
Vector which is perpendicular to the AB and AC
= (–2i + 9j – 3k) x (–i + 4j + k) = 21i + 5j + k PR = (4i – 2j) – (i + 3j – 2k) = 3i – 5j + 2k
Alternative PQ x PR = (i – 4j + 3k) x (3i – 5j + 2k) = (–8 + 15)i – (2 – 9)j + (–5 + 12)k
= 7i + 7j + 7k
(a) AB = (2i + j – 2k) – (4i – 8j + k) = –2i + 9j – 3k
1 1 2 7
Area of the triangle PQR = | 7i + 7j + 7k | = 7  72  72 = 3
AC = (3i – 4j + 2k) – (4i – 8j + k) = –i + 4j + k 2 2 2
(xi + yj + zk)  (–2i + 9j – 3k) = 0  –2x + 9y – 3z = 0 ____(1)
(xi + yj + zk)  (–i + 4j + k) = 0  –x + 4y + z = 0 ____(2)
(1) – (2) x 2 : y – 5z = 0 1 1
Let z = 1 : y = 5 , x = 21  Perpendicular vector is 21i + 5j + k
(b) ΔPQ'R' = | PQ' x PR' | = | 2 PQ x 2 PR |  ΔPQR : ΔPQ'R' = 1 : 4
2 2
] Note : This Alternative cannot lead to Part (b)
Alternative (b)
(b) ΔPQ'R' is an enlargement of ΔPQR with scaling factor 2

(b) Area of the triangle ABC =

1
| 21i + 5j + k | =
1
212  52  12 =
1
467  ΔPQR : ΔPQ'R' = 1 : 4
2 2 2

6-5
eg 7 The position vectors of points A, B, C and D relative to the origin are Alternative (b) – 1
i + j – 2k, 2i – j + k, 3i + j and 2i + 3j – 3k respectively. Point P divides
the line AC in the ratio 2 : 1. (b) AB = i – 2j + 3k , AD = i + 2j – k : (i + 2j – k) x (i – 2j + 3k) = 4i – 4j – 4k

(a) Show that ABCD is a parallelogram. Area = | 4i – 4j – 4k | = 42  (4)2  (4)2 = 4 3

(b) Calculate the exact area of the parallelogram ABCD.
Alternative (b) – 2
(c) Find the position vector of P and the angle APB.
(b) AB = i – 2j + 3k , AD = i + 2j – k
(a) AB = (2i – j + k) – (i + j – 2k) = i – 2j + 3k
DC = (3i + j) – (2i + 3j – 3k) = i – 2j + 3k (i + 2j – k)●(i – 2j + 3k) = 12  (2)2  32 . 12  22  (1)2 cos θ

1 4  3 6 48  48 
 AB = DC  ABCD is a parallelogram cos θ = =  sin θ =  Area = 14. 6  = 4 3
 84 
14(6) 84 84  
Alternative (a) – 1

(a) AD = (2i + 3j – 3k) – (i + j – 2k) = i + 2j – k

BC = (3i + j) – (2i – j + k) = i + 2j – k (c) AP = 2(PC)  OP – OA = 2(OC – OP)

1
 AD = BC  ABCD is a parallelogram
3(OP) = OA + 2(OC) or OP = [OA + 2(OC)]
3
1
Alternative (a) – 2
OP = [(i + j – 2k) + 2(3i + j)] = 7 i + j – 2 k
3 3 3

(a) OM =
1
[(i + j – 2k) + (3i + j)] = 2i + j – k Alternative (c) – 1
2 1 1 1 7 2
1 x = [1 + 2(3)] , y = [1 + 2(1)] , z = [–2 + 2(0)]  OP = i+j– k
ON = [(2i – j + k) + (2i + 3j – 3k)] = 2i + j – k 3 3 3 3 3
2
Alternative (c) – 2
 OM = ON  ABCD is a parallelogram 2 2 4
(c) AP = (AC) = [(3i + j) – (i + j – 2k)] = (i + k)
3 3 3
Alternative (a) – 3
4 7 2
 1 3 1 1 2  0   2  2 1 3 1 3  OP = OA + AP = (i + j – 2k) + (i + k) = i + j – k
(a) M =  , ,  = (2, 1, –1) N=  , ,  = (2, 1, –1) 3 3 3
 2 2 2   2 2 2 
4 1
 M=N  ABCD is a parallelogram PA = (–i – k) , PB = (–i – 6j + 5k)
3 3
(–i – k) ● (–i – 6j + 5k) = (1)2  (1)2 . (1)2  (6)2  52 cos θ
(b) AB = i – 2j + 3k , AD = i + 2j – k : (i – 2j + 3k) x (i + 2j – k) = –4i + 4j + 4k 2
 cos θ =  θ = 111.05o
31
Area = | –4i + 4j + 4k | = (4)2  42  42 = 4 3

6-6
eg 8 A tetrahedron OABC has a base OAB and a vertex C, with
OA = 2i + j + k, OB = 4i – j + 3k and OC = 2i – j – 3k.
(a) Show that OC is perpendicular to both OA and OB. Alternative 3

(b) Calculate, to the nearest 0.1o, the angle between the edge AC (b) AC = (2i – j – 3k) – (2i + j + k) = –2j – 4k
and the base OAB of the tetrahedron.
|(–2j – 4k) ● (2i – j – 3k)| = 20 . 14 sin θ
(c) Calculate the area of the base OAB and the volume of the tetrahedron.
 |0 + 2 + 12| = 20(14) sin θ  θ = 56.8o
(a) OC●OA = 4 – 1 – 3 = 0 OC●OB = 8 + 1 – 9 = 0  OAC = 56.8o (1dp)
 OC is perpendicular to both OA and OB
Alternative
 2  4 i j k
    (c) OA x OB = (2i + j + k) x (4i – j + 3k) = 4i – 2j – 6k
(a) OA x OB =  1  x  1  or 2 1 1 = 4i – 2j – 6k = 2 OC
 1  3 4 1 3
    1
Area of OAB =
1
| 4i – 2j – 6k | = 16  4  36 = 14
 OC is perpendicular to both OA and OB 2 2
1 1 14
Volume = (Area of OAB)(OC) = ( 14 )( 14 ) =
3 3 3
(b) The angle between AC and OAB is OAC.
Alternative
OC 4  1 9
 tan OAC = =  OAC = 56.8o (1dp)
OA 4  1 1 (c) OA x OB = (2i + j + k) x (4i – j + 3k) = 4i – 2j – 6k

Alternative 1 1 1
Area of OAB = |4i – 2j – 6k | = 16  4  36 = 14
2 2
(b) AC = (2i – j – 3k) – (2i + j + k) = –2j – 4k
1 1 14
Volume = (2i – j – 3k)●(4i – 2j – 6k) = (8 + 2 + 18) =
OC 4  1 9 OA 4  1 1 6 6 3
 sin OAC = = or cos OAC = =
AC 0  4  16 AC 0  4  16
 OAC = 56.8o (1dp)

Alternative 2

(b) AC = (2i – j – 3k) – (2i + j + k) = –2j – 4k

(–2j – 4k)●(2i – j – 3k) = 0  4  16 . 4  1 9 cos θ
 0 + 2 + 12 = 20(14) cos θ  θ = 33.2o
 OAC = 56.8o (1dp)

6-7
LINES eg
A
A. Vector Equation of Line R u  1  3 
    x 1 y2 z3
a 1. r =  2  + t  1  ≡ x = 1 + 3t , y = 2 – t , z = 3 + t ≡ = =
r  3  1  3 1 1
OR = OA + AR  r = a + tu , t  O    

eg 1 Given that A(1, 2, 3) and B(4, 1, 4),

0  3 
Line AB : r = (i + 2j + 3k) + t(3i – j + k) , for t  or     x y2 z3
2. r =  2  + t  1  ≡ x = 3t , y = 2 – t , z = 3 + t ≡ = =
3  1  3 1 1
r = (4i + j + 4k) + (3i – j + k) , for   or    
r = (i + 2j + 3k) + s(–3i + j – k) , for s  or

r = (4i + j + 4k) + (–3i + j – k) , for    1 3

    x 1 z3
3. r =  2 + t  0 ≡ x = 1 + 3t , y = 2 , z = 3 + t ≡ = ,y=2
Note : OR = r = xi + yj + zk and R(x, y, z)  3  1 3 1
   

eg 2 Given that P(3, 4, 5) and Q(6, 7, 8),

Line PQ : r = (3i + 4j + 5k) + t(i + j + k) , for t  or C. Line with Point – Foot of Perpendicular (N) & Shortest Distance

r = (6i + 7j + 8k) + (i + j + k) , for   eg 1 The line l has equation r = (i + 2j) + (i + j – k), where  is parameter.
If N is the foot of perpendicular of B(8, 7, 3) to the line, find the position
B. Types of Line Equation vector of N and the shortest distance of B to the line.

 1  1   8   7   1 N l
1. Vector Equation : r = (xAi + yAj + zAk) + (u1i + u2j + u3k) , for  
         
BN = ON – OB =  2  +   1  –  7  =  5  +   1 
●
 0  1   3   3   1 
         
 xA   u1   7   1   1 
   
      ●B
or r =  y A  +   u2  , for   BN  l :  5  +   1     1  = 0  (–7 – 5 + 3) + (1 + 1 + 1) = 0   = 3
z  u   1    1 
 A  3  3 
     
 1  1  4  7   1   4 
           
2. Parametric Equation : x = xA + u1 , y = yA + u2 , z = zA + u3  ON =  2  + 3  1  =  5  and BN =  5  + 3  1  =  2 
 0  1   3   3   1   6 
           

x  xA y  yA z  zA  Shortest distance = (4)2  (2)2  (6)2 = 2 14

3. Cartesian Equation : = =
u1 u2 u3

6-8
D. Intersecting Lines – Angle between Lines & Intersection Point eg 1 Find the equation of the plane that contaiin the points
A(1, 2, 3), B(5, –2, –5) and C(–1, 4, –3).
Note : Angle between lines may be acute or obtuse.
AB = (5i – 2j – 5k) – (i + 2j + 3k) = 4i – 4j – 8k = 4(i – j – 2k)
eg 1 The lines l1 and l2 have equations r = (i + 2j + 3k) + t(2i + j – k) and AC = (–i + 4j – 3k) – (i + 2j + 3k) = –2i + 2j – 6k = –2(i – j + 3k)
r = (i + j – 2k) + (3i + 2j + k) respectively, where t and  are parameters.
Find the intersection point and angle between the lines. BC = (–i + 4j – 3k) – (5i – 2j – 5k) = –6i + 6j + 2k = 2(–3i + 3j + k)

 1  2   1 3  1  2 t  1  3 ___(1) Plane ABC : r = (i + 2j + 3k) + t(i – j – 2k) + (i – j + 3k)

        r = (5i – 2j – 5k) + t(i – j – 2k) + (i – j + 3k)
For intersection,  2  + t  1  =  1  +   2   2  t  1  2 ___( 2)
3  1   2  1  r = (–i + 4j – 3k) + t(i – j – 2k) + (i – j + 3k)
        3  t  2   ___( 3)
By (2) & (3) :  = 2 , t = 3 r = (i + 2j + 3k) + t(i – j – 2k) + (–3i + 3j + k)
From (1) :  = 2  t = 3 (consistent) r = (5i – 2j – 5k) + t(i – j – 2k) + (–3i + 3j + k)
 r = (i + j – 2k) + 2(3i + 2j + k) = 7i + 5j  Int. point = (7, 5, 0) r = (–i + 4j – 3k) + t(i – j – 2k) + (–3i + 3j + k)

(2i + j – k) ● (3i + 2j + k) = 22  12  ( 1)2 . 32  22  12 cos θ r = (i + 2j + 3k) + t(i – j + 3k) + (–3i + 3j + k)

7 r = (5i – 2j – 5k) + t(i – j + 3k) + (–3i + 3j + k)
6+2–1= 6 . 14 cos θ  θ = cos 1( ) = 40.20o r = (–i + 4j – 3k) + t(i – j + 3k) + (–3i + 3j + k)
6(14)
 Angle between lines is 40.20o  1  1  5   1  1  3   5   1  1  3   10   1
                       
 1 x  1 =  5  =  5  1  1 x  3  =  5  = 5  1  1 x  3  =  10  =  10  1
 2   3   0   0  2   1   0   0  3   1  0   0
                       
PLANES
Plane ABC : r ● (i + j) = (i + 2j + 3k) ● (i + j) = 3
A. Vector Equation of Plane n
(a) In terms of two non parallel r ● (i + j) = (5i – 2j – 5k) ● (i + j) = 3
vectors on the plane R r ● (i + j) = (–i + 4j – 3k) ● (i + j) = 3
●
OR = OA + AR v
A● u eg 2 Find the equation of the plane that contaiin the points
r = a + t u + v, for t, 
A(1, –2, 3) and the line r = (3i + 4j + 5k) + t(i + 3j – 2k).

(b) In terms of the normal to the plane (3i + 4j + 5k) – (i – 2j + 3k) = 2i + 6j + 2k = 2(i + 3j + k)
(i + 3j + k) x (i + 3j – 2k) = –9i + 3j = –3(3i – j)
AR  n : AR ● n = 0  (OR – OA) ● n = 0 Plane : r ● (3i – j) = (3i + 4j + 5k) ● (3i – j) = 9 – 4 + 0 = 5

(r – a ) ● n = 0  r●n = a●n = d r ● (3i – j) = (i – 2j + 3k) ● (3i – j) =3+2+0=5

r ● (–3i + j) = (i – 2j + 3k) ● (–3i + j) = –3 –2 + 0 = –5
Note : 1. (u x v) is parallel to n
2. Generally equation of plane is express in terms of its normal. r ● (–3i + j) = (3i + 4j + 5k) ● (–3i + j) = –9 + 4 + 0 = –5

6-9
B. Types of Plane Equation D. Plane & Line – Intersection Point & Angle between Plane and Line

a  u1   n1 
 n1   a   n1       
      Given line r =  b  +   u2  and plane r ●  n2  = d
1. Vector Equation : r ●  n2  =  b  ●  n2  = d   u  n 
n      c  3  3
 3  c   n3 
 a   u1    n1 
     
1. If line lies in plane, then  b  +   u2   ●  n2  = d, for all valuies of .
a  u1   v1 
       c   u   n 
or r =  b  +   u2  +   v2  , for ,      3   3 
  u  v 
c  3  3
 u1   n1 
   
2. If line is parallel to plane, then  u2  ●  n2  = 0.
u  n 
2. Cartesian Equation : n1.x + n2.y + n3.z = d  3  3

 a   u1    n1 
eg 1 r ● (3i + 4j + 5k) = 6  3x + 4y + 5 z = 6      
3. To find intersection point, solve  b  +   u2   ●  n2  = d, for a value of .
r ● (2i – 3j – 5k) = 8  2x – 3y – 5 z = 8
 c   u   n 
   3   3 

C. Plane with Point – Foot of Perpendicular (N) & Shortest Distance 4. To find angle between line and plane, find θ such that

If P(x1, y1, z1) lies on the plane r ● (n1i + n2j + n3k) = d, then  u1   n1 
    2 2 2 2 2 2
(x1i + y1j + z1k) ●(n1i + n2j + n3k) = d  u2  ●  n2  = u1  u2  u3 . n1  n2  n3 . cos θ
u  n 
Note : 1. If (x1i + y1j + z1k) ●(n1i + n2j + n3k) ≠ d, then P does not lies on the plane.  3  3

Case 1 : If 0o < θ < 90o, then angle between is (90o – θ)

eg 1 Given the point A(1, 5, 8) and a plane x + 2y + 3z = 7. If the foot of Case 2 : If 90o < θ < 180o, then angle between is (θ – 90o)
perpendicular of A to the plane is N, find the position vector of N.
Hence evaluate the shortest distance A from the plane.

ON = r = (i + 5j + 8k) + (i + 2j + 3k) Plane : r ● (i + 2j + 3k) = 7

A●
 [(i + 5j + 8k) + (i + 2j + 3k)] ● (i + 2j + 3k) = 7
 (1 + 10 + 24) + (1 + 4 + 9) = 7   = –2 n
N●
 ON = (i + 5j + 8k) – 2(i + 2j + 3k) = –i + j + 2k
 Shortest distance = AN = | – 2(i + 2j + 3k) | = 2 12  22  32 = 2 14

6-10
  1 3 5  1 0  1  1  1  2
                 
eg 1 The line r =  2  +   p  is perpendicular to the plane r =  0  + s  1 + t  1  , eg 2 The lines l1 and l2 have equations r =  2  + s  1  and r =  3  + t  3 
 3  1  1  q  2  0  1   1  1 
                 
where p and q are constants and , s, t  . respectively, where s and t are parameters.

(a) Determine the values of p and q. (a) Show that the lines l1 and l2 intersect.
(b) Find the position vector of the point of intersection of the line and the plane. (b) Find the Cartesian equation of the plane which contains the lines l1 and l2.
 3  0  3   1  1  1   1  2
                1 + s = 1 + 2t
(a)  p  ●  1  = 0 and  p  ●  1 = 0 (a)  2  + s  1  =  3  + t  3 
 1   2  1  q  0  1   1   1  2 + s = 3 + 3t  : t = –1 , s = –2
                –s = 1 – t
0 + p + 2 = 0  p = –2 2 – p + q = 0  q = –5
From (3rd) :
Alternative t = –1  s = –2 or s = –2  t = –1 or t = –1 & s = –2  LHS = RHS
 1 0  2  q  i j k  lines l1 and l2 intersect
     
(a)  1 x  1  =  2  or 1 1 q = (–2 – q)i – 2j + k
 q  2  1  Alternative 1
      0 1 2
(a) 1 + s = 1 + 2t & 2 + s = 3 + 3t
 2  q  3
    By any 2 eq.s : t = –1 , s = –2
  2  = m  p   p = –2 , q = –5
 1   1  1  1   1  1  2   1
               
l1 : r =  2  + (–2)  1  =  0  and l2 : r =  3  + (–1)  3  =  0 
 0  1   2   1  1   2 
           
 1   3   3   5   3 
           lines l1 and l2 intersect
(b)  2  +   p   ●  2  =  0  ●  2 
 3   1   1   1  1 
          Alternative 2
(3 – 4 + 3) + (9 – 2p + 1) = 15 + 0 + 1  =1 x 1 y2 z x 1 y3 z 1
(a) = = and = =
 1  3  4 1 1 1 2 3 1
     
 r =  2 +  2   position vector of the point of intersection =  0  By any 2 eq.s : x = –1 , y = 0 (or with z = 2)
 3  1  4 rd
      Sub. to find the 3 variable : z = 2 from both eqs.  lines l1 and l2 intersect

 1  2   2   2   1  2
           
(b)  1  x  3  =  1   Plane : r   1  =  2    1   2x – y + z = 2 – 2 + 0
 1   1   1   1   0  1
           
 2x – y + z = 0 or –2x + y – z = 0

6-11
eg 3 A plane passes through the points A(–1, –1, –4), B(0, 4, 0) and C(1, 3, –2). eg 4 A parallelepiped for which OABC, DEFG, ABFE and G F
 4  2 OCGD are rectangles is shown in the diagram. D
   
The line r =  1  +   1 ,   , meets the plane at point P. The unit vectors i and j are parallel to OA and OC E
 11  3 C
    respectively, and the unit vector k is perpendicular to B
the plane OABC, where O is the origin. The vectors O A
(a) Find the Cartesian equation of the plane.
(b) Determine the coordinates of point P. OA, OB and OD are 4i, 4i + 3j and i + 5k respectively.
13 35
(c) Calculate the acute angle between the line and the plane. (a) Show that cos BEG = .
175
(b) Calculate the area of the triangle AEG.
(a) AB = i + 5j + 4k & AC = 2i + 4j + 2k
(c) Find the equation of the plane AEG.
(i + 5j + 4k) x (2i + 4j + 2k) = –6i + 6j – 6k
Normal vector to plane is –i + j – k or i – j + k or –6i + 6j – 6k or 6i – 6j + 6k
(a) EB = EA + AB EG = EF + FG
Plane Eq : r.(–i + j – k) = 4j.(–i + j – k)  –x + y – z = 4 or x – y + z = –4
= EA + (OB – OA) = AB + FG
 1  4   4   4  4  4
           
= – 0 +  3 –  0 =  3 –  0 –  0
(b) [(4i + j + 11k) + (2i – j + 3k)]  (–i + j – k) = 4 5  0  0  0  0  0
(–4 + 1 – 11) + (–2 – 1 – 3) = 4   = –3            
= –i + 3j – 5k = –4i + 3j
OP = (4i + j + 11k) + (–3)(2i – j + 3k) = –2i + 4j + 2k  P(–2, 4, 2)
EB = 1 9  25 = 35 EG = 16  9  0 = 5
Alternative
(–i + 3j – 5k)  (–4i + 3j) = 35 (5).cos BEG
(b) –(4 + 2) + (1 – ) – (11 + 3) = 4   = –3
490 35 13 35
OP = (4i + j + 11k) + (–3)(2i – j + 3k) = –2i + 4j + 2k  P(–2, 4, 2)  cos BEG = . =
5 35 35 175

(c) (2i – j + 3k)  (–i + j – k) = 14 . 3 cos θ (b) EA = –i – 5k EG = –4i + 3j

2  1  3 6
cos θ = =–  θ = 157.8o  1   4   0  (15)  i j k
42 42      
   
0 x 3 =   (0  20 )  or 1 0 5 = [0 –(–15)]i – [0 – 20]j + [–3 – 0]k
 Acute angle between line and plane = 67.8o  5   0   30  4 3 0
     
Alternative = 15i + 20j – 3k
1
(c) (2i – j + 3k)  (i – j + k) = 14 . 3 cos θ Area of the triangle AEG = | EA x EG | = 225  400  9 = 1 634
2 2
2  1 3 6
cos θ = =  θ = 22.2o
42 42
 Acute angle between line and plane = 67.8o Normal to plane is parallel to15i + 20j – 3k :  n = 15i + 20j – 3k
r  (15i + 20j – 3k) = (4i)  (15i + 20j – 3k)  r  (15i + 20j – 3k) = 60

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 y2
eg 5 Determine whether line x – 1 = = z – 3 lies in the plane
3
2x – y + z = 3 or not. Alternative 6
1
Line : r = (i + 2j + 3k) + (i + 3j + k) Plane : r ● (2i – j + k) = 3 2x – y + z = 3 : x = (3 + y – z)
2
[(i + 2j + 3k) + (i + 3j + k)] ● (2i – j + k) = (2 – 2 + 3) + (2 – 3 + 1) = 3 1 y2
 Line lies in the plane x–1=z–3 : (3 + y – z) – 1 = z – 3  =z–3
2 3
Alternative 1  Line lies in the plane
2x – y + z = 3 : y = 2x + z – 3
Alternative 7
y2
= z – 3 : (2x + z – 3) – 2 = 3(z – 3)  x–1=z–3 Two points on Line : A(1, 2, 3) , B(2, 5, 4)
3
A : 2(1) – (2) + (3) = 3 B : 2(2) – (5) + (4) = 3
 Line lies in the plane
 A, B lie in plane  Line lies in the plane
Alternative 2
Alternative 8
2x – y + z = 3 : y = 2x + z – 3
y2 x = t + 1 , y = 3t + 2 , z = t + 3
x–1= : 3(x – 1) = (2x + z – 3) – 2  x–1=z–3 2x – y + z = 2(t + 1) – (3t + 2) + (t + 3) = 3
3
 Line lies in the plane  Line lies in the plane
Alternative 3 Alternative 9
2x – y + z = 3 : z = 3 + y – 2x Line : r = (i + 2j + 3k) + (i + 3j + k) Plane : r ● (2i – j + k) = 3
y2 (i + 2j + 3k) ● (2i – j + k) = 2 – 2 + 3 = 3  Point lies in the plane
x – 1 = z – 3 : x – 1 = (3 + y – 2x) – 2  x–1= (i + 3j + k) ● (2i – j + k) = 2 – 3 + 1 = 0
3
 Line lies in the plane  Line perpendicular to normal of the plane
Alternative 4
 Line lies in the plane
2x – y + z = 3 : z = 3 + y – 2x
y2 y2 y2
=z–3: = (3 + y – 2x) – 2  x–1=
3 3 3
 Line lies in the plane

Alternative 5
1
2x – y + z = 3 : x = (3 + y – z)
2
y2 1 y2 y2
x–1= : (3 + y – z) – 1 =  =z–3
3 2 3 3
 Line lies in the plane

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E. Plane & Plane – Angle between Planes & Intersection Line eg 2 The planes, p1 and p2, have equations
2x – y + z = 0 and 3x – 4y + 2z + 6 = 0 respectively.
The planes, p1 and p2, have equations r ● (n1.i+ n2.j+ n3.k) = d1 and (a) Calculate the angle between both planes.
r ● (u1.i+ u2.j+ u3.k) = d2 respectively. (b) Show that the vector 2i – j – 5k is parallel to both planes.
(c) Find the vector equation of the line of intersection of both planes.
1. If the planes are parallel, then (d) Find the Cartesian equation of a plane which passes through point (7, 3, 2)
(n1.i+ n2.j+ n3.k) = (u1.i+ u2.j+ u3.k), for a value of . and perpendicular to p1 and p2.
(a) n1 = 2i – j + k , n2 = 3i – 4j + 2k
2. To find angle between planes, find θ such that
(n1.i+ n2.j+ n3.k) ● (u1.i+ u2.j+ u3.k) (2i – j + k) ● (3i – 4j + 2k) = 22  ( 1)2  12 . 32  ( 4)2  22 cos θ
12 12
= u12  u22  u32 . n12  n22  n32 . cos θ cos θ = or  θ = 24.53o
6 29 174
Note : Angle between planes may be acute or obtuse.
Alternative (a)
3. To find intersection line, solve their Cartesian equations
(a) n1 = 2i – j + k, n2 = 3i – 4j + 2k : (2i – j + k) x (3i – 4j + 2k) = 2i – j – 5k
n1.x + n2.y + n3.z = d1 and u1.x + u2.y + u3.z = d2
simultaneously by reducing them to 2 variables. Then |(2i – j + k) x (3i – 4j + 2k)| = 2  (1)2  12 . 32  (4)2  22 sin θ
2

(a) find the coordinates of two point on the intersection line. 30 5 30

With these two points, the equation of the line can be found, OR sin θ = or or  θ = 24.53o
6 29 29 174
(b) find the coordinates of a point on the intersection line, the the
vector which is perpendicular to both the normals
(by cross product of the normals) (b) (2i – j + k) x (3i – 4j + 2k) = (–2 + 4)i – (4 – 3)j + (–8 + 3)k = 2i – j – 5k
With this point and vector, the equation of the line can be found.
 Vector 2i – j – 5k is parallel to both planes.
eg 1 Show that the point A(2, 0, 0) lies on both planes Alternative (b)
2x – y + 4z = 4 and x – 3y – 2z = 2. (2i – j – 5k) ● (2i – j + k) = 4 + 1 – 5 = 0
Hence, find the vector equation of the line of intersection of both planes. (2i – j – 5k) ● (3i – 4j + 2k) = 6 + 4 – 10 = 0

2x – y + 4z = 2(2) – 0 + 4(0) = 4 x – 3y – 2z = 2 – 3(0) – 2(0) = 2

 Vector 2i – j – 5k is perpendicular to both normals to planes.
 Point A lies on both planes  Vector 2i – j – 5k is parallel to both planes.
 2  1  14 
      (c) 2x – y + z = 0 and 3x – 4y + 2z + 6 = 0
 1  x  3  =  8   Vector equation of the line, r = 2i + (14i + 8j – 5k)
 4  2   5  3x – 4y + 2(y – 2x) + 6 = 0  x + 2y = 6
      Let x = 0 : y = 3 , z = 3  Line of intersection : r = (3j + 3k) + (2i – j – 5k)
Alternative Alternative (c)
2x – y + 4z = 4 ......(1) x – 3y – 2z = 2 .......(2) (c) Let x = 0 : – y + z = 0 and – 4y + 2z + 6 = 0  y=3,z=3
(1) + 2(2) : 4x – 7y = 8 ......(3)
Line of intersection : r = (3j + 3k) + (2i – j – 5k)
A(2, 0, 0) : 4(2) – 7(0) = 8  Point A lies on both planes
Let y = 8 : x = 16, z = –1  B(16, 8, –5)  AB = 14i + 8j – 5k
(d) (xi + yj + zk) ● (2i – j – 5k) = (7i + 3j + 2k) ● (2i – j – 5k)
Vector equation of the line, r = 2i + (14i + 8j – 5k)
2x – y – 5z = 14 – 3 – 10  2x – y – 5z = 1

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