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Vector. Unit Vector: A vector of unit magnitude in the direction of a vector a

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1. Vectors & Their Representation:

Vector quantities are specified by definite magnitude and definite directions. A vector is generally represented by a directed line segment, say AB. A is called the initial point & B is called the terminal point. The magnitude of vector AB is expressed by AB.

Zero Vector:

A vector of zero magnitude is a zero vector. i.e. which has the same initial & terminal point, is called a Zero Vector. It is denoted by O. The direction of zero vector is indeterminate.

Unit Vector:

A vector of unit magnitude in the direction of a vector a is called unit

vector along a and is denoted by aˆ symbolically, ,

| a | a aˆ    . Equal Vectors:

Two vectors are said to be equal if they have the same magnitude, direction & represent the same physical quantity.

Collinear Vectors:

Two vectors are said to be collinear if their directed line segments are parallel irrespective of their directions. Collinear vectors are also called parallel vectors. If they have the same direction they are named as like vectors otherwise unlike vectors.

Symbolically, two non zero vectorsa andb are collinear if and only if,,a Kb, where K  R

Vectors a = a1iˆ + a2jˆ + a3kˆ and b = b1iˆ + b2jˆ + b3kˆ are collinear

if 1 1 b a = 2 2 b a = 3 3 b a Coplanar Vectors:

A given number of vectors are called coplanar if their line segments are all parallel to the same plane. Note that “TWO VECTORS ARE ALWAYS COPLANAR”.

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1.Example

Find values of x & y for which the vectors a = (x + 2)iˆ – (x – y)jˆ + kˆ

b= (x – 1)iˆ + (2x + y)jˆ + 2kˆ are parallel.

Solution

a and b are parallel if x 1 2 x   = y x 2 x y   = 2 1 x = – 5, y = – 20

2. Angle Between two Vectors

It is the smaller angle formed when the initial points or the terminal points of the two vectors are brought together. It should be noted that 0º    180º .

3. Addition Of Vectors:

 If two vectors a &b are represented by OA &OB , then their sum a isb a vector represented byOC , where OC is the diagonal of the parallelogram OACB.

 ab b a (commutative)  (ab) c a( bc) (associativity)  a 0  a   0a  a ( a)  0  ( a)  a

 |ab||a||b|  |ab|||a||b||

 ab = |a|2 |b|2 2|a||b|cos where  is the angle between the vectors

 A vector in the direction of the bisector of the angle between the two

vectors a&b is     a a b b

 . Hence bisector of the angle between the two

vectorsaand b is 

a  b

, where  R+. Bisector of the exterior angle between

 

a&b is 

a  b

,  R+.

4. Multiplication Of A Vector By A Scalar:

Ifa is a vector & m is a scalar, then m a is a vector parallel to a whose modulus is m times that of a. This multiplication is called SCALAR MULTIPLICATION. If a and bare vectors & m, n are scalars, then:

a m m ) a ( ) a ( m      m(na)n(ma) (mn)a a n a m a ) n m (       m(ab) ma  mb

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2. Example

I f aiˆ2jˆ3kˆ a n d b 2iˆ4jˆ5kˆ r e p re s e n t t w o a d ja c e n t s i de s of a parall elogram, fi nd uni t vectors paral l el to the diagonals of the parallelogram.

Solution.

Let ABCD be a parallelogram such that AB = a and BC = b. Then,AB + BC = AC  AC = ab = 3iˆ6jˆ2kˆ and AB + BD = AD  AD + AD = AB  BD = AD AB = ba Now, AC = 3iˆ6jˆ2kˆ  |AC| = 9364 = 7 and, BD = iˆ2jˆ8kˆ  |BD| = 1464 = 69

 Unit vector along AC = | AC | AC = 7 1

3iˆ6jˆ2kˆ

and, Unit vector along BD = | BD | BD = 69 1

iˆ2jˆ8kˆ

3. Example

ABCDE is a pentagon. Prove that the resultant of the forces AB, AE, BC, DC, ED and AC is 3AC.

Solution.

Let R be the resultant force

 R = AB + AE + BC + DC + ED + AC  R = (AB + BC) + (AE + ED + DC) + AC = AC + AC + AC

= 3AC. Hence proved.

5. Position Vector Of A Point:

let O be a fixed origin, then the position vector of a point P is the vector OP. If a and b are position vectors of two points A and B, then,

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DISTANCE FORMULA

Distance between the two points A(a) and B(b) is AB = ab

SECTION FORMULA

If a and b are the position vectors of two points A & B then the p.v. of a point which divides AB in the ratio m: n is given by:

n m b m a n r       .

Note p.v. of mid point of AB = 2

b a .

4. Example

ABCD is a parallelogram. If L, M be the middle point of BC and CD, express AL and AM in terms of AB and AD, if AL + AM = k AC. find k

Solution.

Let the position vectors of points B and D be respectively b and a referred to A as origin of reference.

Then AC = AD + DC = AD + AB [ DC = AB] = d + b  AB = b, AD = d

i.e. position vector of C referred to A is d + b  AL = p.v. of L, the mid point of BC.

= 2 1 [p.v. of D + p.v. of C] = 2 1

b d b

     = AB + 2 1 AD AM = 2 1

ddb

= AD + 2 1 AB  AL + AM = b + 2 1 d + d+ 2 1 b = 2 3 b + 2 3 d = 2 3 (b + d) = 2 3 AC.

6. Scalar Product Of Two Vectors:

Geometrical interpretation of Scalar Product

Let a and b be vectors represented by y OA and OB respectively. Let  be the angle between OA and OB. Draw BL  OA and AM  OB.

From s OBL and OAM, we have OL = OB cos  and OM = OA cos . Here OL and OM are known as projections of b on a and a on b respectively.ely. Now, a.b = |a| | b| cos 

= |a| (OB cos  ) = |a| (OL)

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Againa.b = |a| | b| cos  = | b| (|a| cos  ) = | b| (OA cos )

= | b| (OM)

= (magnitude of b) (Projection of a on b) ...(ii)

Thus geometrically interpreted, the scalar product of two vectors is the product of modulus of either vector and the projection of the other in its direction.

1. i.i = j.j = k.k = 1; i.j = j.k = k.i = 0

 projection of | b | b . a b on a       2. ifa = a1i + a2j + a3k &b = b1i + b2j + b3k then a b. = a1b1 + a2b2 + a3b3  a  a12a22a32 ,  b  b12b22b32

3. the angle  betweena & b is given by |a| |b| b . a cos      0  4. a.b a b cos(0)     ,

note that if  is acute then a b . > 0 & if  is obtuse then a b. < 0 5. a.a a2a2,a.bb.a (commutative)  a.(bc)a.ba.c (distributive)

6. a.b 0 a b (a0 b0)  

7.

(

m a

)

.b =a.

(

m b

)

= m

(

a b .

)

(associative) where m is scalar..

Note :

(i) Maximum value ofa . b is a  b (ii) Minimum value ofa .b is – a  b

(iii) Any vectora can be written as,a =

 

a i i.  

 

a.  j j

a k k.  

.

5. Example

If a + b + c = 0, |a| = 3, |b| = 5 and |c| = 7, find the angle between a and b. Solution. We have, abc0  ab = –c 

ab

.

ab

=

 

c .

 

c  ab 2 = 2 | c |  a 2 + b 2 + 2a.b = c 2  a 2 + 2 b + 2 a b cos  = c 2  9 + 25 + 2 (C) (5) cos  = 49  cos  = 2 1   = 3  .

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6. Example

D is the mid point of the side BC of a triangle ABC, if AB2 + AC2 = k (AD2

+ BD2) find k Solution. We have AB = AD + DB  AB2 = 2 ) DB AD (  = AD2 + DB + AD 2 . DB ...(i) Also we have AC = AD + DC  AC2 = (ADDC)2 = AD2 + DC2 + AD 2 . DC ...(ii)

Adding (i) and (ii), we get

AB2 + AC2 = 2AD2 + 2BD2 + 2

AD . (DBDC) = 2(DA2 + DB2), for

DB + DC = 0

Solved Example

If a = iˆ + jˆ + kˆ and b = 2iˆ – jˆ + 3kˆ, then find (i) Component of b along a.

(ii) Component of b perpendicular to along a.

Solution. (i) Component of b  along a is         2 | a | b . a    a Here a . b = 2 – 1 + 3 = 4 2 | a | = 3 Hence        2 | a | b . a    a = 3 4 a = 3 4 (iˆ + jˆ + kˆ) (ii) Component of b  perpendicular to along a is b –        2 | a | b . a    a. = 3 1

2iˆ7jˆ5kˆ

7. Vector Product Of Two Vectors:

1. Ifa&b are two vectors &  is the angle between them thena x b   a b sin  n, wheren is the unit vector perpendicular to botha&b such that a b, & n forms a right handed screw system.

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2. Geometricallya x b = area of the parallelogram whose two adjacent sides  are represented bya & b.

3. iˆiˆ jˆjˆkˆkˆ0 ; iˆjˆkˆ,jˆkˆ iˆ,kˆiˆ jˆ

4. Ifa = a1 +a2jˆ + a3kˆ &  b = b1 + b2jˆ + b3kˆ then 3 2 1 3 2 1 b b b a a a kˆ jˆ iˆ b a    5.a x bb x a (not commutative)

6.

(

m a

)

b =a 

(

m b

)

= m

(

a  b

)

(associative) where m is a scalar.. 7. a x b c(  )  (a x b )  (a x c ) (distributive)

8. a x b0  a& b are parallel (collinear) (a0 , b0 i.e.) aK b , where K is a scalar.

9. Unit vector perpendicular to the plane of  

   

a b is n a x b

a x b &   

 A vector of magnitude ‘r’ & perpendicular to the palne of  

    a b is r a x b

a x b & 

 If  is the angle between 

  

 

a b then a x b

a b & sin  

 If a b, & are the pv’s of 3 points A, B & C then the vector area of trianglec

AB C = 1

2       a x b  b x c  cx a . T h e p oi n t s A , B & C a r e c ol l i n e a r i f 0 a x c c x b b x a         

 Area of any quadrilateral whose diagonal vectors ared1 & d2 is given by

1

2 1 2

 

d x d

 Lagrange's Identity: for any two vectors       

        a b a x b a b a b a a a b a b b b & ; ( ) ( . ) . . . . 2 2 2 2 7. Example

Find a vector of magnitude 9, which is perpendicular to both the vectors kˆ 3 jˆ iˆ 4   and 2iˆ jˆ2kˆ. Solution.

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b a = 2 1 2 3 1 4 kˆ jˆ iˆ    = (2 – 3) iˆ – (–8 + 6) jˆ + (4 – 2) = iˆ2jˆ2kˆ  |ab| = (1)22222 = 3  Required vector = 9          | b a | b a     = 3 9 ) kˆ 2 jˆ 2 iˆ (   = 3iˆ6jˆ6kˆ 8. Example

For any vector a, if | a iˆ|2 + | a jˆ|2 + | a kˆ|2 = k |a|2 find k

Solution.

Let a = a1iˆa2jˆa3kˆ. Then iˆ

a  = (a1iˆa2jˆa3kˆ) × iˆ = a1( iˆ iˆ)+ a2( jˆ iˆ) + a3 ( kˆ iˆ) = –a2kˆa3iˆ

 | a iˆ|2 = a22 + a32 jˆ a  = (a1iˆa2jˆa3kˆ) × jˆ = a1kˆa3iˆ  | a jˆ|2 = a21 + a32  | a|2 = a 1 2 + a 2 2  | a iˆ|2 + | a jˆ|2 + | a kˆ|2 = a22 + a33 + a12 + a32 + a12 + a22 2 (d12 + a 2 2 + a 3 2) = 2 |a|2 9. Example

Let OA = a, OB = 10a + 2b and OC = b where O is origin. Let p denote the area of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides. find k if p = kq.

Solution.

We have,

p = Area of the quadrilateral OABC

= 2 1 | AC OB |  = 2 1 | ) OA OC ( OB |   = 2 1 | ) a b ( ) b 2 a 10 ( |     = 2 1 | ) a b ( 2 ) b b ( 2 ) a a ( 10 b a ( 10 |       = 2 1 | ) b a ( 2 0 0 ) b a ( 10 |     

and, q = Area of the parallelogram with OA and OC as adjacent sides = |OAOC| = |a b|

 

 ...(ii)

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8. Scalar Triple Product:

 The scal ar tri pl e product of three vectors a b, & c i s defi ned as:      

a x b c. a b c sin cos where  is the angle betweena & b &  is the angle

between a x b& c. It is also written as[  a b c]and spelled as box product.

 Scalar triple product geometrically represents the volume of the parallelopiped whose three coterminous edges are represented by

      a b, &c i e V. . [a b c]

 In a scalar triple product the position of dot & cross can be interchanged i.e. a. (b x c ) ( a x b c  ). OR [a b c  ] [b c a  ] [  c a b]

 a. ( b x c) a c x b i e  .( ) . . [a b c  ]  [a c b  ]

 If a= a1i+a2j+a3k;b = b 1i+b2j+b3k &c = c1i+c2j+c3k then[a b c  ]

a a a b b b c c c  1 2 3 1 2 3 1 2 3 . In general, ifaa l1 a m2 a n3;b b l1 b m2  b n3 &   c c l1 c m2 c n3 then

a b c  

  

a a a b b b c c c l m n  1 2 3 1 2 3 1 2 3

; where,m &n are non coplanar vectors.

 Ifa b c  , , are coplanar[a b c  ]0.

 Scalar product of three vectors, two of which are equal or parallel is 0 i.e.[  a b c  0] ,

 If  a b c, , are non  coplanar then[a b c  0  ] for right handed system & [  a b c  0] for left handed system.

 [i j k] = 1 [K a b c  ]K a b c[  ] [ (a b c d  ) ][a c d  ][b c d  ]

 The volume of the tetrahedron OABC with O as origin & the pv’s of A, B

and C beinga b , &c respectively is given by V 1 a b c

6 [ ]

  

 The positon vector of the centroid of a tetrahedron if the pv’s of its

vertices area b c  , , &d are given by 1

4 [ ]

   

a  b cd .

Note that this is also the point of concurrency of the lines joining the vertices to the centroids of the opposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it is equidistant from the vertices and the four faces of the tetrahedron.

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10. Example

Find the volume of a parallelopiped whose sides are given by 3iˆ7jˆ5kˆ, kˆ 3 jˆ 7 iˆ 5    and 7iˆ5jˆ3kˆ Solution.

Let a3iˆ7jˆ5kˆ, b5iˆ7jˆ3kˆ and c7iˆ5jˆ3kˆ.

We know that the volume of a parallelopiped whose three adjacent edges are a,b,c is [a,b,c]    . Now, [abc] = 3 5 7 3 7 5 5 7 3      = –3 (–21 – 15) – 7 (15 + 21) + 5 (25 – 49) = 108 – 252 – 120 = –264

So, required volume of the parallelopiped = [a,b,c] = | – 264 | = 264 cubic units

9. Vector Triple Product:

Leta b c  , , be any three vectors, then the expressiona x b x c( ) is a vector & is called a vector triple product.

Geometrical Interpretation of a x b x c ( )

Consider the expressiona x b x c( ) which itself is a vector, since it is a

cross product of two vectorsa& ( b x c). Nowa x b x c ( ) is a vector perpendicular

to the plane containinga& ( b x c) butb x c  is a vector perpendicular to the

plane containingb & c, thereforea x b x c ( ) is a vector which lies in the

plane ofb& c and perpendicular toa. Hence we can expressa x b x c ( ) in

terms ofb&c i.e.a x b x c ( ) =xb yc where x & y are scalars.  a x b x c ( )= ( . )  a c b( . )a b c    (a x b x c)  =( . )  a c b( . )b c a  

 (a x b x c)   a x b x c ( ), in general

11. Example

For any vector a, P.T. iˆ(aiˆ)jˆ(ajˆ)kˆ(ak) = 2a

Solution.

Let aa1iˆa2jˆa3kˆ. Then, iˆ(aiˆ)jˆ(aˆjˆ)kˆ(akˆ) = {(iˆ.iˆ)a(iˆ.a)iˆ} + {(jˆ. jˆ)a(jˆ.a)jˆ} + {(kˆ.kˆ)a(kˆ.a)kˆ} = {(a(iˆ.a)iˆ}{a(jˆ.a)jˆ} + {a(kˆ.a)kˆ}

= 3a{(iˆ.a)iˆ(jˆ.a)jˆ(kˆ.a)kˆ = 3a(a1iˆa2jˆa3kˆ)

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12. Example

Let a = iˆ + 2jˆ – 3kˆ, b = iˆ + 2jˆ – 2kˆ and c = 2iˆ – jˆ + kˆ. Find the value(s) of , if any, such that

ab

 

 bc

×

ca

= 0. Find the vector product when  = 0. Solution.

 

ab  bc

×

ca

=

a b c

b ×

ca

=

a b c

   

a.b c b.c a

which vanishes if (i)

 

a.b c =

 

b.c a (ii)

a b c

= 0

(i)

 

a.b c =

 

b.c a leads to the equation 2 3 + 10  + 12 = 0, 2 + 6 = 0 and 6 – 6 = 0, which do not have a common solution.

(ii)

a b c

  = 0  1 2 2 2 1 3 2       = 0  3 = 2   = 3 2 when  = 0,

a b c

= – 10, a.b = 6, b.c 

= 0 and the vector product is – 60

2 iˆ kˆ

.

13. Example

If A B = a, A . a = 1 and AB = b, then find A and B.

Solution. Given A B a ...(i)  a.

AB

= a.a  a.A a    . B = a.a  1 + a.B = |a|2  a.B = |a|2 – 1 Given ABb  a

AB

= a × b 

a.B

A  –

a.A

B ab 

|a|2 1

A B = ab [using equation (B)] solving equation (1) and (5), simultaneously, we get

A = 2 | a | a b a       and B =

2

2 | a | 1 | a | a a b       

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14. Example

Solve for r , the simultaneous equations rbcb, r.a 0 provided a is not perpnedicular to b.

Solution

) c r

( × b = 0 r c and b are collinearrckb  r = ckb ...(i) a . r   = 0  (ckb) . a = 0  k = – b . a c . a    

putting in (i) we get

b . a c . a c r         b

10. Reciprocal System Of Vectors:

If a b c  , , &   a' , b' , c' are two sets of non coplanar vectors such that      

a. a' = b. b' = c. c' = 1 then the two systems are called Reciprocal System of vectors. Note:

c b a b x a c c b a a x c b c b a c x b = a                  11. Linear Combinations:

Given a finite set of vectorsa b c  , , ,... then the vector rxa  yb  zc  ...is called a linear combination of  a b c, , ,... for any x, y, z... R. We have the following results:

(a) If a b, are non zero, noncollinear vectors then xayb x a'y b' xx y' ; y' (b) FundamentalTheorem: Let a b, be non zero, non collinear vectors. Then

any vectorr coplanar with a b, can be expressed uniquely as a linear combination of a b,

i.e. There exist some uniquly x, y  R such thatxayb r. (c) If  a b c, , are nonzero, noncoplanar vectors then:

xa yb zc x a y b z c   ' ' ' xx' , yy' ,zz'

(d) Fundamental Theorem In Space: Leta b c  , , be nonzero, noncoplanar vectors in space. Then any vectorr, can be uniquly expressed as a linear com bi nat i on of  a b c, , i .e. There ex i st som e un i que x,y  R su ch thatxa  yb zc  r.

(e) Ifx x 1, 2,...xn are n non zero vectors, & k1, k2,...kn are n scalars & if the linear combination k x1 1 k x2 2 ...k xnn 0k1 0,k2 0...kn 0 then we say that vectorsx x 1, 2,...xn are LINEARLY INDEPENDENT VECTORS.

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(f) Ifx x 1, 2,...xn are not LINEARLY INDEPENDENT then they are said to be LINEARLY DEPENDENT vectors. i.e. ifk x1 1  k x22 ... k xnn 0 & if there exists at least one kr  0 thenx x 1, 2,...xn are said to be

L

INEARLY

D

EPENDENT.

 iˆ,jˆ,kˆ are Linearly Independent set of vectors. For K1iˆ+ K2jˆ + K3kˆ= 0 K1= K2= K3 = 0

 Two vectorsa & b are linearly dependent a is parallel to 

b i.e.a x b  0    linear dependence ofa & b. Conversely ifa x b  0  thena & b are linearly independent.

 If three vectors  a b c, , are linearly dependent, then they are coplanar i.e.[ , , ]a b c  0   , conversely, if[ , , ]a b c  0   , then the vectors are linearly

independent.

15. Example

Given A that the points a – 2b + 3c, 2a + 3b – 4c, – 7b + 10c, A, B, C have position vector prove that vectors AB and AC are linearly dependent.

Solution.

Let A, B, C be the given points and O be the point of reference then OA = a – 2b + 3c, OB = 2a + 3b – 4c and OC= – 7b + 10c

Now AB = p.v. of B – p.v. of A = OB – OA = (a + 5b – 7c) = – AB

 AC = AB where  = – 1. Hence AB and AC are linearly dependent

16. Example

Prove that the vectors 5a + 6b  + 7c, 7a – 8b  + 9c and 3a + 20b  + 5c are linearly dependent a,b,c being linearly independent vectors.

Solution.

We know that if these vectors are linearly dependent , then we can express one of them as a linear combination of the other two.

Now let us assume that the given vector are coplanar, then we can write 5a + 6b + 7c = ( 7a – 8b + 9c) + m (3a + 20b + 5c)

where , m are scalars

Comparing the coefficients of a,b and c on both sides of the equation

5 = 7 + 3 ...(i)

6 = – 8 + 20 m ...(ii)

7 = 9 + 5m ...(iii)

From (i) and (iii) we get

4 = 8   = 2 1

= m which evidently satisfies (ii) equation too. Hence the given vectors are linearly dependent .

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Not e: Test Of Collinearity:

Three points A,B,C with position vectors  a b c, , respectively are collinear,, if & only if there exist scalars x, y, z not all zero simultaneously such

that;xayb zc   0, where x + y + z = 0.

Note: Test Of Coplanarity:

Four points A, B, C, D with position vectors    a b c d, , , respectively are coplanar if and only if there exist scalars x, y, z, w not all zero simultaneously such thatxa + yb + zc + wd = 0    where, x + y + z + w = 0.

17. Example

Show that the vectors 2ab3c, ab2c and ab3c are non-coplanar vectors.

Solution.

Let, the given vectors be coplanar.

Then one of the given vectors is expressible in terms of the other two. Let 2ab3c = x

ab2c

+ y

ab3c

, for some scalars x and y..

2ab3c = (x + y) a (x + y) b + (–2x – 3y) c  2 = x + y, –1 = x + y and 3 = 2x – 3y.

Solving, first and third of these equations, we get x = 9 and y = – 7.

Clearly, these values do not satisfy the thrid equation. Hence, the given vectors are not coplanar.

18. Example

Prove that four points 2a 3b c     , a2b3c , 3a4b2c and a6b6c are coplanar. Solution.

Let the given four points be P, Q, R and S respectively. These points are coplanar if the vectors PQ, PR and PS are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of other two. So, let

PQ = x PR + y PS

a5b4c = x

abc

+ y

a9b7c

a5b4c = (x – y) a + (x – 9y) b + (–x + 7y) c

 x – y = –1, x – 9y = –5, –x + 7y = 4

[Equating coeff. of a,b,c on both sides]

Solving the first of these three equations, we get x = – 2 1 , y = 2 1 . These values also satisfy the third equation. Hence, the given four points are coplanar.

References

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