Vectors Work

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VECTORS

c 15 2 10 11 ˆ i J k c= − − + c r b q a p c b b a×+ ×= + +  a b c c a b c c a b 1 u 2 u ω 2 1 u u w w+ × = 2 / 1 ). (u₁×u₂ w ≤ 0 .CD = OE a a v u & w u w w+ × v 2 / 1 ). (u×v w≤ v u ⊥ ) → → b x a 0 ) ).( ( ) ).( ( )+ × × + × × = ×d b c a d c a b d c          c c b c a c c b b b a b c a b a a a                   . . . . . . . . . 0 .b ≠ a r c r− b 0 .a = r k j iˆ+2ˆ+ ˆ c b a= + ) (b c a× × c b a,, a b b b a c a a b a b c c b a_×)2₌(_×).(_×)₊(_×).(_×) kˆ kˆ kˆ c and b a,  c b O,, c b:  R R r t R r n R r t R r t  n t  n R ) ( ) .( 1 2 n t n t r r r    × × − + k J i 2 1 2 1 2 − − k j i r_.(₃ˆ− ˆ+ 11 k j iˆ+cos(β−α)ˆ+cos(γ −α)ˆ k j iˆ+ ˆ+cos(γ −β)ˆ k a j iˆ+cos(β−γ)ˆ+ ˆ a c c 15 ˆ 2 ˆ 10 ˆ 11 ˆ i J k c= − + + 0 . . . . . . = c b b b a b c a b a a a c b a 0 . . . . . . 3 3 2 2 1 1 3 2 1 3 2 1 3 2 1 = × = b a k b a j b a i c c c b b b a a a c b c a c b b b a b a b a a a b a r= +λ     + − + + − θ θ θ θ 1 2cos 1 , cos 2 1 cos 2 , cos 2 1 1     + + − + 1 2cos 1 , 2 cos 2 1 cos 2 , cos 2 1 1 θ 3 1 , 3 1 3 1 3 1 = − − − = i J k AD AD A(1, 0, 2) B(–2, 1, 3) D _{C(2, -1, 1)} q and p  1 < +q p  b a OD b a OC b OB a OA= , = , =2+3, = −2 OA OB OA DB ) ).( (BD×AC OD×OC 2 b a× 2 b a× O α k J i− + =2 β γ α γ β α γ β×)= p+q +r c b a, , b c b a×(×2)=  a b c 3 2 0 ) ( ) ( )+ × + × = ×b b c c a a β γ 2 ) ( ) ).( ( b a b a b c l × × × = 2 ) ( ) ).( ( a b a b a c × × × 66 1 = = = b c a ] ) [( ] ) [(x q p q x r q p× − × + − × 0 ] ) [ − × = × x p r r  k r q p = = = 0 ] ) [( ] ) [( ] ) [( − × + × − × + × − × = × x q p q x r q r x p r p

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→ → → + + b c a 2 → → → − −b c a → → → + +b c a 2 3 → → → + + b c a 3 5 5 → → → c b a ,, 2 2 2 2 2 2 2 2 2 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( c z b z a z c y b y a y c x b x a x − − − − − − − − − → b j iˆ 3ˆ 4 + → b → c 0 4 3 2 1+ + + = → → → → n n n n k j iˆ 2ˆ ˆ 2 + + k j i ˆ 3 2 ˆ 3 2 ˆ 3 5 ₊ ₊ k j b i k j i aˆ+ ˆ+ ˆ,ˆ+ ˆ+ ˆ k c j iˆ+ ˆ+ ˆ 1 # # #→ → → c b a 1 1 1 1 1 1 1 ₌ − + − + −a b c k j i a=3ˆ+6ˆ−2ˆ → k j i b =4ˆ− ˆ+3ˆ → → a → b → a → b a c t b r= +  a c c c b a b _       − − +( ₂). b b a c b a×(×)+(.) b c c a c e.)=  ( c b& aˆ b aˆ ) ˆ ˆ ( 3 a×b a b a bˆ−(ˆ.)ˆ AB CD PQ AB ] [abc c b× ] [abc a c× ] [abc b a× 2 2 2 ₍ ₎ , ' ₍ ₎ ) ( ' , ) ( ) ( b a b a c b a a b a b b a b a b × × = × × × = × × × ] ' ' ' [ ' ' ] ' ' ' [ ' ' ] ' ' ' [ ' ' c b a b a c c b a a c b c b a c b× ₌ × ₌ ×     ₊ ₊ _× + = a kb a b k b a a k r ₂ 1 ₂ .     c c 15 2 10 11 ˆ i J k c= − − + c r b q a p c b b a×+ ×= + +  a b c c a b c c a b 1 u 2 u ω 2 1 u u w w+ × =

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BASIC CONCEPTS OF VECTORS:

In mathematics, physics, and engineering, a Vector (sometimes called a geometric or spatial vector) is a geometric quantity having magnitude (or length) and direction expressed numerically as ordered list of

co-ordinates [x, y, z].

A vector is an object that is an input to, or output from vector functions as per vector algebra. A Vector is a directed line segment, or arrow, connecting an initial point with a terminal point. Is a vector with initial point A and terminal point B. Technically, the [x, y, z] components of vector are equal to the vector difference minus .

DEFINITION:

A scalar is a quantity, which has only magnitude but does not have a direction. For example time, mass, temperature, distance and specific

gravity etc. are scalars.

A Vector is a quantity which has magnitude, direction and follow the

law of parallelogram

(addition of two vectors). For example displacement, force, acceleration are vectors.

(a) There are different ways of denoting a vector: a



or a or a are different ways. We use for our convenience a, b, c

  

etc. to denote vectors, and a, b, c to denote their magnitude.

(b) A vector may be represented by a line segment OA and arrow

gives direction of this

vector. Length of the line segment gives the magnitude of the vector. A O H e r e i s t h e i n i t i a l p o i n t a n d i s t h e t e r m i n a l p o i n t o f O AO A CLASSIFICATION OF VECTORS: (i) Equal vectors

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Two vectors are said to be equal if and only if they have equal magnitudes and same direction.

A B

C D A B = C D

A s w e l l a s d i r e c t i o n i s s a m e

(ii) Zero Vector (null vector): A vector whose initial and terminal points are same, is called Zero vector. For exampleAA.Such

vector has zero magnitude and no direction, and denoted by0  . AA CA BC AB+ + = Or AB BC CA 0  = + + C B A

(iii) Like and Unlike Vectors: Two vectors are said to be (a) Like, when they have same direction.

(b) Unlike, when they are in opposite directions.

• and –a



are two unlike vectors as their directions are opposite, a



and 3a 

are like vectors.

(iv) Unit Vector: A unit vector is a vector whose magnitude is unity. We write, unit vector in the direction of a



as aˆ . Therefore .

(v) Parallel vectors: Two or more vectors are said to be parallel, if they have the same support or parallel support. Parallel vectors may have equal or unequal magnitudes and direction may be same or opposite. As shown in figure a b c O _A B C E D

(vi) Position Vector: If P is any point in the space then the vectorOP is called position vector of point P, where O is the origin of reference. Thus for any points A and B in the space,

| a | a aˆ   =

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AB=OB−OA

(vii) Co-initial vectors: Vectors with same initial point are called co-initial vectors.

As shown in figure OA,OB, OC and OD are Co-Initial vectors. O d _a c C B D _A b _. ADDITION OF TWO VECTORS:

Let OA a, AB b   = = and OB c  = _. Here c 

is sum (or resultant) of vectors a 

and b 

. It is to be noticed that

the initial point b



of coincides with the terminal point of a



and the line joining the initial point of a  to the terminal point of b  represents vector a b  

+ _{in magnitude and direction.}

PROPERTIES:

(i) a b b a+ = +

 

(So, Vector addition is commutative) (ii) a (b c) (a b) c       + + = + +

(So, Vector addition is associative) (iii) |a b| |a| |b|     + ≤ +

(equality holds when a  and b  are like vectors) (iv) |a b| ||a| |b||     − ≥ +

(equality holds when a 

and b 

are unlike vectors)

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(v) a 0 a 0 a      + = = + (vi) a ( a) 0 ( a) a      + − = = − + .

MULTIPLICATION OF VECTOR BY SCALARS : If a



is a vector and m is a scalar, then ma 

is a vector parallel toa  whose modulus is | m | times that ofa



. This multiplication is called Scalar Multiplication. If a

 and



b_{are vectors and m, n are scalars, then:} _{m (}a₎ = (  a_{)m = m}a _m(n)a_{= n(m)}a_{= (mn)}a (m + n)  a_{= m}a_{+ n}a _m(a₊b_{) = m}a_{+ m}b_. LINEAR COMBINATIONS:

Given a finite set of vectors    a b c, , ,...

then the vector

   

r xa= + yb + zc +...

is called a linear combination of    a b c, , ,...

for any x, y, z... .

We have the following results: (i) If

  a b,

are non-zero, non-collinear vectors then xa yb x a y b+ = '+ ' ⇒ =x x y y' ; = '

. (ii) Fundamental Theorem: 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 unique x, y∈R such that xa yb r    + = (iii) 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       + + = ' + ' + ' ⇒ = ' , = ' , = ' .. (iv) If    x x₁, ₂,...x_n

are n non zero vectors, & k

1, k2, ...kn are n scalars

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k x_{1 1} +k x_{2 2} +...k x_{n n} = ⇒0 k₁ =0,k₂ =0...k_n =0

then we say that vectors are Linearly Independent Vectors .

(v) If

  

x x₁, ₂,...x_n

are not Linearly Independent then they are said to be

Linearly Dependent vectors. i.e. if.

k x_{1 1} +k x_{2 2} +...k x_{n n} = ⇒0 k₁ =0,k₂ =0...k_n =0

& if there existsKr ≠0 at least one then

  

x x₁, ₂,...x_n

are said to be Linearly Dependent.

Example: Show that the vectors 5a 6b 7c, 7a 8b 9c

      + − + + & 3a 20b 5c    + + _are coplanar (Wherea, b, c   

are three non-coplanar vectors.). Solution: Let A 5a 6b 7c    + + = _,B=7a−8b+9c and C 3a 20b 5c    + + = _.

A

,

B

and Care coplanar. This indicates that xA+yB+zC=0must have a real solution for x, y, z other than (0, 0, 0).

Now, x(5a 6b 7c) y(7a 8b 9c) z(3a 20b 5c) 0           = + + + + − + + + (5x 7y 3z)a (6x 8y 20z)b (7x 9y 5z)c 0     = + − + + − + + + 5x + 7y + 3z = 0 6x – 8y + 20 z = 0 7x + 9y + 5z = 0 (Asa, b, c   

are non-coplanar vectors)

Now D = 0 5 9 7 20 8 6 3 7 5 = −

So the three linear simultaneous equation in x, y and z have a non-trivial solution.

Hence the given vectors are coplanar vectors.

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(a) The necessary and sufficient condition for three points with position vectorsa, b,c

  

to be col linear is that there exist three scalars x, y, z, not all zero such thatxa yb zc 0+ + =



 

Wherex y z 0+ + = .

(b) The necessary and sufficient condition for four points with position vectorsa, b,c    andd 

to be coplanar is that there exist scalars x, y, z, u, not all zero, such thatxa yb zc ud 0

     = + + + where x y z u 0+ + + = .

Example: Let 'O' be the point of intersection of diagonals of a

parallelogram ABCD. The points M, N, K &P are the mid points of OA, MB, NC and KD respectively. Show that N, O and P are collinear.

Solution: D _C B A O M N P K Now, 4 b 2 a 2 b 2 a N , 2 a M      + = + ≡ ≡ 8 a 3 b 2 2 a 4 b a K      − = − + ≡ 16 a 3 b 6 2 8 a 3 b 2 b P      − − = − + − ≡ 3 (2 ) 16 OP= − b a+ uuu  _

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( )

1 1 ( 2 ) 4 6 ONuuu= a+ b = − OPuuu .

Hence the points N, O& P are collinear.

SECTION FORMULA:

Let A, B & C be three non collinear points in space having the position vectors a,b   andr  . B A C O a r b ( x , y , z )1 1 1 ( x , y , z ) ( x , y , z )2 2 2 m n Let m n CB AC ₌ mAC=nCB

m

AC

=

n

CB

.

(As vectors are in same direction) Now, OA AC OC AC r a   − = ⇒ = + r CB b CB b r     − = ⇒ = +

Using (i), we get m n b n a m r + + =    .

ORTHOGONAL SYSTEM OF UNIT VECTORS :

Let OX, OY and OZ be three mutually perpendicular straight lines.

Given any point P(x, y, z) in

space, we can construct the rectangular parallelepiped of which OP is

a diagonal and

OA = x, OB = y, OC = z.

Here A, B, C are (x, 0, 0), (0, y, 0) and (0, 0, z) respectively and L, M,

N are (0, y, z), (x, 0, z) and

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B _N X A M C Z O P ( x , y , z ) Y

Let iˆ,ˆ,j kˆdenote unit vectors along OX, OY and OZ respectively. We have r =OP=xiˆ+yjˆ+zkˆ  as OA=xiˆ, OB=yjˆandOC=zkˆ. AN OA ON = +

(

)

. So, , OP ON NP OP OA OB OC NP OC AN OB = + = + + = =

uuu uuu uuu

uuu uuu uuu uuu uuu uuu uuu uuu

kˆ n jˆ m iˆ z y x kˆ z jˆ y iˆ x | r | r r 2 2 2₊ ₊ = + + + + = = _    ⇒ r =riˆ+mrjˆ+nrkˆ . MULTIPLICATION OF VECTORS:

SCALAR PRODUCT OF TWO VECTORS (DOT PRODUCT): The scalar product, of two non-zero vectors a

 and b  is defined as θ cos | b || a |  .where θ is

angle between the two vectors, when drawn with same initial point. Note that 0≤θ≤π.

If at least one of a 

and b 

is a zero vector, thena.b   is defined as zero. PROPERTIES : (i) a.b b.a    

= _{(scalar product is commutative)}

(ii) 2 2 2 _a_._a _|_a_| _a a = =  = 

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(iii) (ma).b m(a.b) a(mb)       = = (where m is a scalar) (iv)     = θ − | b | . | a | b . a cos 1 _    (v) a.b=0⇔   Vectors a  and b 

are perpendicular to each other. [a 

,b 

are non-zero vectors].

(vi) .iˆjˆ= .jˆkˆ=kˆiˆ.=0 (vii) a.(b c) a.b a.c        + = + (viii) 2 2 2 2 _|_b_| _a _b | a | ) b a ).( b a (+ − =  −  =  −

(ix) Leta =a1iˆ+a2jˆ+a3kˆ, b=b1iˆ+b2jˆ+b3kˆ 



, Thena.b=(a1iˆ+a2jˆ+a3kˆ).(b1iˆ+b2jˆ+b3kˆ)

  . (x) Maximum value of  a_.b_{= |}a_{| |}b_|

(xi) Minimum value of



a_.b_{= – |}a_{| |}b_|

(xii) Any vector 

a_{can be written as,}a₌

   

a i i.  a j j.  a k k. 

e j e j d i

+ + .

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b . aˆ | b || a | b . a | b | cos OB ON _      = = θ = b O B θ _A a

Example: Prove that the angle in a semi-circle is a right angle.

Solution: Let O be the centre and AB the bounding diameter of the

semi-circle. Let P be any point on the

circumference. With O as origin. Let OA=a,OB=−a and OP=r.

P

B O A

As OA = OB = OP, each being equal to radius of the semi-circle. AP = r−aand BP=r−(−a)=r+a 2 2 _a r ) a r ).( a r ( BP . AP = − + = − = OP2 – OA2 = 0 AP and BP are perpendicular to each other, i.e.

0

90 APB=

∠ _.

VECTOR (CROSS) PRODUCT:

The vector product of two non-zero vectors a 

andb 

, whose modules are a and b respectively, is also a vector whose modulus is absinθ,

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where is θ(0≤θ≤π)the angle between vectorsa  andb  . The direction is that of a vectorn  perpendicular to both a  &b  , such that a, b,n    are in right-handed orientation. θ O a b a×b=|a ||b|sinθnˆ     . PROPERTIES: (i) a b (b a)     × − = × (ii) (ma) b m(a b) a (mb)       × = × = × (Where m is a scalar) (iii) a×b=0⇔    vectorsa  andb  are parallel. (provided a  andb  are non-zero vectors). (iv) iˆ jˆ jˆ jˆ kˆ kˆ 0  = × = × = ×

(v) iˆ×jˆ=kˆ=−(jˆ×iˆ),jˆ×kˆ=iˆ=−(kˆ×jˆ),kˆ×iˆ= jˆ=−(iˆ×kˆ) (Vi) a (b c) a b a c        × + × = + ×

(vii) Let a=a1iˆ+a2jˆ+a3kˆ  andb=b1iˆ+b2jˆ+b3kˆ  , then 1 2 3 3 2 1 b b b a a a kˆ jˆ iˆ b a×= = 2 3 3 2 3 1 1 3 1 2 2 1 ˆ_{i(a b} ₋_{a b ) j(a b}₊ˆ ₋_{a b ) k(a b}₊ˆ ₋_{a b )}

. (Viii) . |a ||b| | b a | sin _    × = θ

(ix) Area of triangle =

| b a | 2 1 sin ab 2 1 ap 2 1   × = θ =

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A B b P a O θ

(x) Area of parallelogram =ap absin |a b|   × = θ = O A b B C p θ a _. (xi)     a x b ≠ b x a (not commutative)

(xii) Unit vector perpendicular to the plane of

a x b a & b is a x b ±       .

(xiii) A vector of magnitude ‘r’ & perpendicular to the plane of

(

)

      a b is r a x b a x b & ± .

(xiv) Area of any quadrilateral whose diagonal vectors are

  d₁ & d₂ is given by 1 2 1 2   d x d .

(xv) 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 ₌ .

Example: If a, b, c be three vectors such that a + b + c = 0, prove that a

× b = b × c = c × a and deduce c C sin b B sin a A sin ₌ ₌ (Sine rule). Solution: Let

BC

,

CA

,

AB

represent the vectors a, b, c respectively.

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Then, we have A π- B B a c b C π- C π- A a + b + c = 0, ==> c = - (a + b) ==> b × c = b × (- a - b) = - b × a = a × b Similarly, c × a = a × b

Hence, bcsin(π−A)=casin(π−B)=absin(π−C) ==> b × c = c × a = a × b

==> bc sin A = ca sin B = ab sin C.

Hence, bcsin(π−A)=casin(π−B)=absin(π−C).

SCALAR TRIPLE PRODUCT ( BOX PRODUCT) :

The scalar triple product of three vectorsa, b   &c  is defined as c ). b a (×  =|a ||b ||c|    sin cosθ φ

whereθ is the angle between

  a & b &φis the angle between    a x b & c . It is also denoted as[ ]    a b c . Let 1 2 3 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ _, ˆ ˆ _, ˆ ˆ a a i a j a k b b i b j b k c c i c j c k= + + = + + = + + Then 2 1 2 1 3 1 3 1 3 2 3 2 3 2 1 3 2 1 b b a a kˆ b b a a jˆ b b a a iˆ b b b a a a kˆ jˆ iˆ b a×= = − +  .

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3 2 1 3 2 1 3 2 1 2 1 2 1 3 3 1 3 1 2 3 2 3 2 1 c c c b b b a a a b b a a c b b a a c b b a a c c ). b a (× = − + = . Therefore(a b).c (b c).a (c a).b (b a).c (c b).a (a c).b                   × − = × − = × − = × = × = × . Note that(a b).c (b c).a a.(b c)          × = × = ×

, hence in scalar triple product dot

and cross are

interchangeable. Therefore we indicate(a b).c    × by[a b c]    . PROPERTIES: (i) |(a b).c|    ×

represents the volume of the parallelepiped, whose adjacent sides are represented by vectorsa,b,c

  

in magnitude and direction. Therefore three vectorsa,b,c

   are coplanar if[a b c]    = 0. i.e. 0 c c c b b b a a a 3 2 1 3 2 1 3 2 1 = ,

(ii) Volume of the tetrahedron =

| ] c b a [( | 6 1  . (iii) [a bc d] [a c d] [bcd]           + = + (IV) [a a b]=0    .

(v) 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] (vi)             a. (b x c)= −a c x b i e.( ) . . [a b c] = −[a c b] (vii) If  a = a 1i+a2j+a3k;  b = b 1i+b2j+b3k &  c = c 1i+c2j+c3k then

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[ a b c] a a a b b b c c c = 1 2 3 1 2 3 1 2 3 . In general, i     a =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 . (viii) If    a b c, , are coplanar⇔[ ]=    a b c 0 .

(ix) Scalar product of three vectors, two of which are equal or parallel is 0 i.e.[ ]   a b c = 0 . Note : If    a b c, ,

are non-coplanar then for right handed system [a b c  ]> 0

&

for left handed system[ ]    a b c <0 . (x) [i j k] = 1. (xi) [K a b c] K a b c[ ]       = . (xii) [( ) ] [ ] [ ]           a + b c d = a c d + b c d .

(xiv) Remember that:

[

]

      a−b b c− c a− = 0 &

[

]

      a+b b c+ c a+ = 2

[ ]

   a b c .

VECTOR TRIPLE PRODUCT:

The vector triple product of three vectors

   a b c, , is defined asa (b c)    × × . If at least one    a b c, , is a zero vector b  andc 

are collinear vectors ora 

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perpendicular to both b  andc  , only thena (b c) 0     = × ×

. In all other casesa (b c)    × × will be a

non-zero vector in the

plane of non-collinear vectors and perpendicular to the vectora 

. Thus we can takea (b c) b c

     µ + λ = × ×

, for some scalarsλ&µ. Since ) c b ( a a     × × ⊥ , 0 ) c . a ( ) b . a ( 0 )) c b ( a .( a × × = ⇒λ  +µ  =  α − = µ α λ ⇒ (a.c) , (a.b)

for same scalar

α_. a (b c) (a.c)b (a.b)c          − = × ×

Example: For any vectora



, prove thatˆi ×(a × i) + j×(a × j) + k ×(a×k) = 2aˆ ˆ ˆ ˆ ˆ

   

. Solution: [iˆ×(a×iˆ)]+[jˆ×(a×jˆ)]+[kˆ×(a×kˆ)]

 



= [(iîˆ.)a−( .iâ)iˆ][( .iˆjˆ)a−( .jâ)jˆ]+[(kˆ.kˆ)a−(kˆ.aˆ)kˆ]

 



= a−( .iâ)iˆ+a−( .jâ)jˆ+a−(kˆ.aˆ)kˆ      [ iîˆ.= jˆjˆ=kˆkˆ=1] = 3a−[( .iâ)iˆ+( .jâ)jˆ+(kˆ.aˆ)kˆ]

   Leta =a1iˆ+a2jˆ+a3kˆ  . Then 1 2 1 3 2 2 2 3 2 1iˆ a jˆ a kˆ) a iˆ a (ˆ.ijˆ) a (ˆ.ikˆ) a (1) a (0) a a ( iˆ a iˆ= + + = + + = + = Similarly, .jaˆ a2, kˆ.a a3 ˆ ₌ ₌ a 2 a a 3 ) kˆ a jˆ a iˆ a ( a 3− ₁ + ₂ + ₃ = −=  . L.H.S. = R.H.S.

RECIPROCAL SYSTEM OF VECTORS: Let a, b

 

&c 

be three non-coplanar vectors. Then the system of vectors b , a′ ′  and c′ 

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which satisfiesa.a′=b.b′=c.c′=1       anda.b′=a.c′=b.a′=b.c′=c.a′=c.b′=0             , is called the

reciprocal system to the vectorsa,b,c    . ] c b a [ c b a __    × = ′ , [abc] a c b __    _× = ′ , [abc] b a c __    × = ′ . PROPERTIES: (I) a.b′=a.c′=b.a′=b.c′=c.a′=c.b′=0

(II) The scalar triple product [a b c] of three non-coplanar vectors a, b, c

is the

reciprocal of the scalar triple product formed from reciprocal system.

Example: Solve the vector equation: r b a b r c× = × , .

      = 0 provided thatc  is not perpendicular tob  . Solution: We are given;

b a b r     × = ×  (r−a)×b=0    Hence(r a)   − andb  are parallel  r a tb    = − . . . (i) And we know r.c   = 0,

 Taking dot product of (i) byc  we get c . a c . r   − ₌t(b.c) –a.c   =t(b.c)   Or t = –       c . b c . a     . . . (ii)  from (i) and (ii) solution of r

 is r  =a  –       c . b c . a     b .

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KEY CONCEPTS

· If are any `n' vectors and are scalars, then is known as linear combination (LC) of the vectors .

· If are any `n' vectors and are scalars. IF and atleast one of is not equal to zero, then the vectors are said to be linearly dependent(L-D)

· If are any `n' vectors and are scalars. If and if all are zero then the vectors are said to be linearly independent (L I).

· Two collinear vectors are always linearly dependent. · Three coplanar vectors are always linearly dependent. · If are any two vectors then

· If are any two vectors then

· If are any two vectors then is a vector perpendicular to both and

· If are any three vectors then scalar tripple product between the vectors is defined as and denoted by

· ·

· In general ·

· If =0 then the vectors are said to be coplanar.

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STATEMENTS AND REASONING EXAMPLES:

1. STATEMENT -1: If x & v  

are unit vectors inclined at an angle and  x is a unit vector bisecting the angle between them, then

x v x 2cos 2 + = _α    Because

STATEMENT-2: If ABC is an isosceles triangle with AB = AC = 1,_ then vector representing bisector of angle A is given by AD

uuu = AB AC 2 +  

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False (D) Statement -1 is False, statement-2 is True.

1. Option (a) is correct

In an isosceles triangle ABC is which AB = AC, the median and bisector from A must be same line  statement 2 is true.

Now x v AD 2 + =   uuu & 2 1 | AD | 2cos 2 2 α = uuu , So | AD | uuu = cos 2 α

 unit vector along AD i.e. x is given by

AD x | AD | = uuu  uuu =

2. STATEMENT -1: The points with position vectors a 2b 3c, 2a 3b c− + − + −

      , 4a 7b 7c− +  are collinear. because

STATEMENT-2: The position vectors a 2b 3c,− +

 

- 2a 3b c,+ − 

 

4a 7b 7c− + 

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(C) Statement-1 is True, Statement-2 is False (D) Statement -1 is False, statement-2 is True. 1 ( 2 3 a 2b 3c)− + + λ − +( 2a 3b c)− + λ (4a 7b 7c) 0− + =       equating coefficients of a, b & c  

both sides we will get values of _1, 2 & 3 such that 1 + 2 + 3 = 0. Which is the condition for linearly dependent vectors & all for collinearity of the points. ‘a’ is correct.

3. STATEMENT -1: If a, b, c 

 

are three unit vectors such that

1 a (b c) b

2

× × = 

 

then the angle between a & b   is /2 _ because STATEMENT-2: If 1 a (b c) b, 2 × × =    then a .b   = 0.

a (b c) b (a.b)c× ×  −   whereas a (b c)× ×    = 1 b 2  comparing a.b   = 0

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So a  is perpendicular to b  ‘a’ is correct. 4. STATEMENT -1: In ABC,  AB BC CA+ + uuu uuu uuu

= 0 because STATEMENT-2: If OA a, OB b= = uuu _ uuu  then AB a b= + uuu _ 

(C) Statement-1 is True, Statement-2 is False (D) Statement -1 is False, statement-2 is True. (C)

In ABC _ AB BC AC+ = = −CA uuu uuu uuu uuu

 AB uuu + BC uuu - CA uuu = O u

OA AB OBuuu uuu uuu+ =

is triangle law of addition

Hence statement-1 is true statement-2 is false.

5. STATEMENT -1: a 3i pj 3k= + +

  



and b 2i 3 j qk= + +

   

are parallel vectors it p = 9/2 and q = 2. because STATEMENT-2: If a a i a j a k= 1 + 2 + 3     and b b i b j b k= 1 + 2 + 3     are parallel 3 1 2 1 2 3 a a a b =b =b

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(D) Statement -1 is False, statement-2 is True. (A) 3 p 3 2 3 q= = from a a i a j a k, b b i b j b= 1 + 2 + 3 + 1 + 2 + 3k         are parallel  3 1 2 1 2 3 a a a b =b = b a 3i pj 3k= + +   and b  = 2i 3 j qk+ +    2  3 p 3 2= =3 q

6. STATEMENT -1: The direction ratios of line joining origin and point (x, y, z) must be x, y, z

because

STATEMENT-2: If P is a point (x,y, z) in space and OP = r then directions cosines of OP are

x y z , , r r r

(C) Statement-1 is True, Statement-2 is False (D) Statement -1 is False, statement-2 is True. (A)

7. STATEMENT -1: The shortest distance between the skew lines r a= + αb   and r c= + βd    is | [a c bd] | | b d | − ×     because

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STATEMENT-2: Two lines are skew lines if three axist no plane passing through than

(C) Statement-1 is True, Statement-2 is False (D) Statement -1 is False, statement-2 is True. (B) A – Defn B – Defn 8. STATEMENT -1: IF a.b 0=    a   b  because STATEMENT-2: a.b   = 0  either a 0=  or b  = 0 or a   b  = 0

(D) 9. STATEMENT -1: A B B A× = ×     because STATEMENT-2: A B | A || B |× =     sin  n 

, when is angle, when your fingers curls from A to B

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(D)

10. STATEMENT -1: If the vectors 2i j kˆ ˆ ˆ− + , ˆi 2j 3k+ −ˆ ˆ and 3iˆ− λ +ˆj 5kˆ are coplanar, then | | 2_{is equal to 16.}

because

STATEMENT-2: The vectors a , b  

and c 

are coplanar iff a,(b c× 

 

) = 0 (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

2 1 1 1 2 3 3 5 − − λ = 0  = -4 _  | |_ 2 _{= 16} Ans. (A)

11. STATEMENT -1: A line L is perpendicular to the plane 3x – 4y + 5z = 10

because

STATEMENT-2: Direct on co-sines of L be <

3 4 1

, ,

5 2 −5 2 2 .

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D.N. of the plane 3x – 4y + 5z = 10  be < 3, -4, 5 >  D.C < 3 5 2 , 4 1 , 5 2 2 − > Ans. (A).

12. STATEMENT -1: The value of expression ˆ ˆ ˆi( j k) j(k i) k(i j) 3× +ˆ ˆ ˆ× +ˆ ˆ ˆ× = because

STATEMENT-2: ˆ ˆ ˆi( j k) [i.j.k] 1× = ˆ ˆ ˆ =

ˆ ˆ ˆ_{i( j k) j(k i) k(i j)}_{× +}ˆ ˆ ˆ_{× +}ˆ ˆ ˆ_×

= ˆ ˆ ˆ ˆ ˆ ˆi.i j.j k.k+ + = 1 + 1 + 1 = 3. Ans. (A)

13. STATEMENT -1: A relation between the vectors r,a and b    is r a b× =     a b r a.a × =      because STATEMENT-2: r.a 0=  

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Since b r a= ×    Q _{We have}a b a (r a)× = × ×      = (a.a) r (a.r)a−       = (a .a) r    Q a. r   = 0  a b r a.a × =      Ans. (A) EXCERSISE 1. Show that the vectors

I – j – 2k, 2i + 3J + k and 7i + 3J – 4k are coplanar.

2. Show that the points P(), Q (), R (), S () are coplanar given that are non – coplanar .

3. The vertices of triangle are A (2,3,0) B (-3,2,1), C (4,-1,0). Find the area of the triangle ABC and unit vector normal to the plane of this triangle.

4. Let OACB be a //gm with O at the origin and OC a diagonal. Let D be the mid point of OA using vector methods, Prove that BD and CO intercept in the same ratio. Determine this ratio.

5. D, E divide side BC and CA of a triangle ABC in the ratio 2 : 3

respectively. Find the position vector of the point of interception of AD and BE and the ratio in which this point divides AD and BE. 10/9, 15/4.

6. If a, b and c be three nonzero vectors, number of two of which are collinear. If the vector a + 2b is collinear with c, and b + 3c is collinear with a then find the value of a + 2b + 6c = 0

7. If = 0

And the vectors X = (x2, x,1), Y = (y2,y,1), Z = (z2,z,1) are non co-planar, then the vectors (a2,a,1), (b2,b,1), (c2,c,1) are co-planar. 8. Let = and i be two vectors to each other in the xy plane. Find all 

vectors in the same plane having projections 1 and 2 along and respectively.

9. A line makes angle , , and with the diagonals of a cube,     prove that cos2 + cos2 + cos2 + cos2 = 4/3._ _ _ _

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10.Let Ar, (r = 1,2,3,4) be the areas of the faces of a tetrahedron. Let nr, be the out ward drawn normals to the respective faces with

magnitudes equal to corresponding areas. Prove that

11.Find a vector of magnitude 5 units, coplanar with vectors 3i – j – k and I + j – 2k and to the vector ._

12.If A  (1,1,1) and C  (0,1,-1) are given vectors, then find a vector B satisfying A x B C and A.B = 3. ( )

13. If the vectors, and are coplanar and (), then prove that .

14. Two sides of triangle are formed by the vectors , . Find the angles. 15. Let a, b, c be three vectors such that a + b + c = 0, |a| = 3, |b| = 5, |c| = 7. Find the angle between and .

16. |a| = 3, and |b| = 4. Find the value of for which the vectors a + b   and a - b are to each other. 

17. The area of a //gm whose diagonals are given by a = 3i + j – 2k and b = i – 3j + 4k.

18. If the unit vectors and are inclined at an angle 2 and |a – b| is less  than 1. then if 0   , find the range of ._{ } _

19. The volume of the tetrahedron whose vertices are the points with position vectors I 6j + 10k, I – 3j + 7k, 5i – j + k and 7i – 4j + 7k is 11 _ _ _ cubic then find .

20. Determine the value of c, so that for all real x the vector cxi – 6j + 3k and xi + 2j + 2cxk make an acute angle with each other.

21. 1. Show that the distance of a point A() to the line is 

22. 2. Find the scalars , iff = (4 - 2 - sin )+ ( 2 – 1) and     (where are non collinear.

23. 3. Let be a unit vector and be a non zero vector not parallel to . Find the angles of the triangle, two sides of which are represented by the vectors and .

24. 4. The p. v. of two pts A and C are 9i – j + 7k and 7i – 2j + 7k respectively. The pt. of intersection of vectors = 4i – j - 3k and = 2i – j + 2k is P. If vector is to and CD and _ PQ = 15 units. Find P. v. of Q.

25. 1. Show that the segments joining vertices to the centriod of opposite faces of a tetrahedrown are concurrent. Hence find the position vector of the point of concurrence.

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26. Let a, b, c are set of non – coplanar vectors, the set of vectors a, b, c reciprocal to it is given by

a = , b = , c =

27. Find the system reciprocal to a, b, a b. a =

28. Find the set of vectors reciprocal to – i + J + k, i – J + k and i + J – k.

29. If a, b, c are non-coplanar vectors, and a, b and c is the reciprocal system show that

a = .

30. Show that if k is a non-zero scalar and a and b are two vectors the soln. of the equation kr + r  a = b is

31. The vector – i + J + k bisects the angle between the vectors and 3i + 4j. Determine the unit vector

along . Ans:

32. Let a, b, c are unit vectors equally inclined at an angle to  each other and for p, q, r are some scalars s.t .

______ (1)

Find p, q, r interms of .

1. Let and be given non – zero and non collinear vectors and be a vector such that  = - . Express in terms of &.

2. If and be two given non – collinear unit vectors and be a vector such that . Prove that .

3. In the triangle ABC, a point P is taken on the side AB such that AP: BP = 1:2 and a point Q is taken on the BC such that CQ: BQ = 2:1. If ‘R’ be the point of inter section of lines AQ and CP, using vector method find the area of ABC, if it is known that area of the ABC is one unit._ _ 49/28 sq. unit.

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4. In ABC, D is the mid point of side AB and E is the centriod of   CDA. If where 0 is the circumcentre of ABC, using vectors prove that AB _ = AC.

5. A non zero vector is  to the line of inter section of the plane

determined by the vectors i, i + J and the plane determined by the vectors i, – J + k. Find the angle between and i-2J+ 2k. Q = cos-1 1/3.

6. P, Q are the mid points of the non – parallel sides BC and AD of a trapezium ABCD. Show that APD = CQP._ _

7. ABC and A1B1C1 are two coplanar triangles such that the perpendiculars from A, B, C to the sides B1C1, C1A1, A1B1 of the triangle A1B1C1 to the sides BC, CA, AB of the triangle ABC are also concurrent.

8. If two pairs of opposite edges of a tetrahedron are at right angles, then show that the third pair is also at right angle. Further show that for such a tetrahedron, the sum of the squares of each pair of opposite edges is the same.

9. If the from two vertices B & C to the opposite faces of a  tetrahedron ABCD intersect, then BC in to AD._

10. Let be unit vectors. If is a vector such that = . Then prove that and that the equality holds iff .

Q. 1. If the four points a, b, c, d are coplanar then show that [a b c] = [b c d] + [c a d] + [a b d].

Q.2. If a, b, c, d are any four vectors, then prove that (. (. Q.3. Show that, [a  b b  c c  a] = [a b c]2 =

Q. 4. If , find the vector which satisfies the eqns () = 0 and .

Q. 5. Find a vector of magnitude 5 units, coplanar with vectors 3i – j – k and i + J –2 k and to the vector _ 2.

Q.6. The value of [a-b b-c c – a] = 0.

Q. 7. If d = (a  b) + (b  c) + v(c  a) and [a b c] = 1/8. Then find the _ _ + + v in terms of a, b, c ,d.

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Q. If , then show that = abc sin ( + ). ( is angle between a and c, is_{  } _ angle between a + b.)

Q. If are co – planar vectors and is not parallel to . Then prove that (. Q. Using vector method, find the ratio in which the bisector of an angle in a triangle divides the opposite side.

Q. If the vectors i + cos( - ) j + cos( - d) where , , are different_{ } _ _{  } angles. If these vectors are coplanar. P. T. a is ind of , , .  

Cos( - )i + J + cos( - ) _{ } _{ } cos( - )i + cos( - )j + a_{ } _{ }

Q. If are unit vectors satisfying a + b + c = 0, then find a.b + b.c + c.a (a + b + c)2 = 0.

Ans: –3/2

Q. Three non zero vectors p, q, r are pair wise non collinear. Further the vector p + q is collinear with r and q + r is collinear with p, find the sum of the vectors p, q and r.

Q. In a ABC, are the position vectors of A, B, C.

(i) Prove that the internal bisector of angle A bisects the side BC internally in the ratio .

(ii) Find the p. v. of the point in which a from B meets the internal bisector of angle A.

Q. Solve the simultaneous vector eqns for . ( - ). = 0, (-). = 0, . = 1 &.n = 0 (All are non zero) consider - = + + ()  

Ans: =

Q. The vector i + 2J + 2k turns through a right angle and passing through the +ve x – axis on the way. Find the vector in its new position.

Ans: 2 .

Q. From the A(1, 2, 0), is drawn to the plane ) = 2, meeting it at the point P. Find the co – ordinates of point P and the distance AP. P  (14/11, 21/11, 1/11), AP = 1/.

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Q. cos( - ) _{  cos( - ),}_{ } where , _{ } and _ are different angles. If these vectors are coplanar show that is independent of _,

and .

 

Q. The vector – i + j + k bisects the angle between the vectors and 3i + 4j. Determine the unit vector

along .

Q. If the vectors a, b, c are coplanar, show that Soln:

Q. Show that the segments joining vertices to the centriod of opposite faces of a tetrahedron are concurrent. Hence find the position vector of the point of concumence.

Eqn. of a line: Ans: and

1. The vertices of a ABC are A(1, 0, 2), B(–2, 1, 3) and C(2, -1, 1). Find  the eqn. of the line BC, the foot of the from A to BC and the length of _ the .

2. If are two unit vectors inclined at an angle to each other. , if lies.  _ _ _ (0, 2 )._

Ans: (2 /3, 4 /3).  3. The vectors

(al + al)i + (am + am)J + (an+ an)k (bl + bl)i + (bm + bm)J + (cn + cn)k.

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form an equilateral triangle. Product of two determinants are coplanar

are collinear are mutually ._

4. If the length of is three times of the length of and is to _ then is

A. 7 B. 42 C. D. None

PRACTICE QUESTIONS Angle between Vectors

11. If , then

(A) (B) (C) (D) Projection along a Vector

12. Let and be three vectors, A vector in the plane of and whose projection on is of magnitude is

(A) (B) (C) (D)

Dot Product:

13. Let and be two non-collinear unit vectors. If and , then is

(A) (B) (C) (D) Modulus of Vectors:

14. If and c are unit vectors, then does not exceed (A) 4 (B) 9 (C) 8 (D) 6

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15. If and . If and , then d will be (A) (B) (C) (D)

Parallel Vectors:

16. The point of intersection of and , where and is

(A) (B) (C) (D) None of these Section Formula:

17. Statement -1 : If I is the incentre of then

Statement -2 : The position vector of centroid of is

(A) STATEMENT1 is True, STATEMENT2 is True; STATEMENT2 is a correct explanation for STATEMENT1

(B)STATEMENT1 is True, STATEMENT2 is True; STATEMENT2 is NOT a correct explanation for STATEMENT1

(C)STATEMENT1 is True, STATEMENT2 is False (D)STATEMENT1 is False, STATEMENT2 is True Angle between Vectors:

18. Unit vectors and are inclined at an angle and . If , then may belong to

(A) (B) (C) (D) None of these Modulus of Vector:

19. Let be the vectors such that , if and then the value

of is [IIT - 95] (A) 47 (B) (C) 0 (D) 25 Vector Triple product:

20. Let and . If is a vector such that and the angle between and is , then [IIT - 99]

(A) 2/3 (B) 3/2 (C) 2 (D) 3

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21. Let the vectors and be such that . Let and be planes determined by the pairs of vectors and respectively, then the angle between and is [IIT - 99]

(A) 0 (B) (C) (D) Basic concepts:

22. Matrix match type Column - I Column - II

(A) Let and , (p) 1

then is

(B) If are three vectors of equal magnitude (q) and the angle between each pair of vectors in

such that then is

(C) If , and then is (r) (D) If are three non zero vectors such that

then the value of (s) 0

may be` (t) 8

Vector equation:

PASSAGE: Consider a line and a plane

23. The angle between the line and the plane is (A) (B) (C) (D)

24. The position vector of the point of intersection of the line and the plane is

(A) (B) (C) (D)

25. The position vector of the point on the given line whose distance from the plane is units is

(A) (B) (C) (D)

LEVEL-II

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1. Let and a unit vector be coplanar. If perpendicular to , then

(A) (B) (C) (D)

2. If the vectors form the sides BC, CA and AB respectively of a triangle ABC, then

(A) (B)

(C) (D)

3. If and are two unit vectors such that and are perpendicular to each other then the angle between and is

(A) (B) (C) (D)

4. Let there be two points A and B on the curve in the plane OXY satisfying and then the length of the vector is (A) (B) (C) (D)

5. If and D are four points in space satisfying then the value of k is

(A) 2 (B) 1/3 (C) 1/2 (D) 1

6. If and and , then the angle between a and b is (A) (B) (C) (D)

7. Given that and then the angle between is : (A) (B) (C) (D) None of these

8. The vector is perpendicular to and is perpendicular to . the angle between and is

(A) (B) (C) (D) None of these

9. X-component of is twice its Y-component. If the magnitude of the vector is and it makes an angle of with z-axis then the vector is :

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10. If then the angle between and is : (A) (B) (C) (D)

11. If the non zero vectors are perpendicular to each other then the solution of the equation, is

(A) (B)

(C) (D) None of these

12. and are mutually perpendicular unit vectors . is a vector satisfying and then is equal to

(A) (B) (C) (D)

13. If then the value of is (A) (B) (C) (D)

14*. If in a right angle triangle ABC, the hypotenuse

then is equal to

(A) (B) (C) (D)

15. If and are linearly dependent vectors and then

(A) (B) (C) (D)

PRACTICE QUESTIONS

16. For three vectors which of the following expressions is not equal to any of the remaining three

(A) (B) (C) (D)

17*. Which of the following expressions are meaningful ? (A) (B) (C) (D)

18. Let the unit vector a and b be perpendicular and the unit vector c be inclined at an angle to both a and b. If then

(A) (B)

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19. The position vectors of the vertices of a quadrilateral ABCD are a,b,c and d respectively. Area of the quadrilateral formed by joining the middle points of its sides is

(A) (B)

(C) (D)

20. The volume of the tetrahedron, whose vertices are given by the vectors and with reference to the fourth vertex as origin, is (A) cubic unit (B) cubic unit (C) cubic unit (D) None of these

21. The co-ordinates of the foot of perpendicular drawn from point P (1 , 0 , 3) to the join of points and is

(A) (B) (C) (D) 22. Column -I Column - II

(A) The value of a for which the vectors (p) 4 and

are coplanar is

(B) The area of a parallelogram having sides (q)

and is

(C) and for some non (r) zero vector , then the value of is

(D) The volume of parallelopiped whose sides are (s) 0 given

, and is (t) 1 23. Column - I Column -II

(A) For given vectors (p)

and the projection of the vector on the vector is

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and is equal to

(C) If , (r)

then k is

(D) The value of c for which the vector (s) LEVEL -III

MODEL QUESTIONS

I. In a rhombus OABC, vector are respectively the position vectors of vertices A, B ,C with reference to O as origin . A point E is taken on the side BC which divides it in the ratio of 2:1 . Also, the line segment AE intersects the line bisecting the angle O intenally in point P. If CP, when extended meets AB in point F, then

1. The position vector of point P, is

(A) (B) (C) (D) None of these 2. The position vector of point F , is

(A) (B) (C) (D) None of these 3. The vector , is given by

II A new operation * is defined between two non antiparallel vectors and as , where is the angle between and .

4. The condition for which and are perpendicular is (A) (B) (C) (D) None of these 5. is

(A) (B) not defined (C) 0 (D) None of these 6. For

(A) = 0 is a necessary condition (B) is a necessary condition

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(D) None of these

7. Let are non zero mutually perpendicular unit vectors. There is a vector coplanar with and and then minimum value of is (A) (B) (C) (D)

8. are unit vectors along the coordinate axes and makes angle q with then q is

(A) (B) (C) (D)

9. If the vectors and are coplanar vectors for only one real x then is

(A) (B) 2 (C) 4 (D)

10. If are unit vectors such that then is equal to

11. Let be three non-coplanar vectors and are vectors defined by the relations , then the value of the expression

is equal to (A) 0 (B) 1 (C) 2 (D) 3

12. Let a,b,c be distinct nonnegative numbers. If vectors and lie in a plane then c is [IIT-88]

(A) the AM of a and b (B) the GM of a and b (C) the HM of a and b (D) equal to zero

13. Let if = is a unit vector such that , then equals

(A) (B) (C) (D)

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(A) 0 (B) (C) (D)

15. If be three mutually perpendicular vectors of the same magnitude. If a vector satisfies the equation ,then is given by

(A) (B) (C) (D)

16. If are unit coplanar vectors, then the scalar triple

product =

(A) 0 (B) 1 (C) (D)

17. Let and . Then depends on

[IIT-2002]

(A) only x (B) only y (C) Neither x Nor y (D) both x and y PRACTICE QUESTIONS

18. Let and . If is a unit vector, then the maximum value of the scalar triple product is [IIT-2003]

(A) (B) (C) (D)

19. The value of `a' so that the volume of parallelopiped formed by and becomes minimum is [IIT-2004]

(A) (B) 3 (C) (D)

20. If , and , then is (A) (B) (C) (D)

21. The unit vector which is orthogonal to the vector and is coplanar with the vectors and is [IIT-2004]

(A) (B) (C) (D)

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then the triplet of pairwise orthogonal vectors is (A) (B) (C) (D)

23. . A vector coplanar to and has a projection along of magnitude , then the vector is [IIT-2006]

24. The edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors such that . Then, the volume of the parallelepiped is [IIT-2008]

A) (B) (C) (D)

25. Let two non-coplanar unit vector and form an acute angle. A point P moves so that at any time t the position vector . (where O is the origin ) is given by . Where P is farthest from origin O, let M be the length of and be the unit vector along . Then [IIT-2008]

(A) and (B)

(C) (D)

26. Let and . If u is a unit vector, then for the maximum value of the scalar triple product

(A) (B)

(C) (D)

27. For unit vectors b and c and any non zero vector a , the value of ` is

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28. Three non-coplanar vectors a, b and c are drawn from a common initial point. The angle between the plane passing through the terminal points of these vectors and the vectors is

(A) (B) (B) (D) None of these

29. If are non coplanar unit vectors such that then the angle between and is :

(A) (B) (C) (D)

30. The number of vectors of unit length perpendicular to vectors and is

(A) 1 (B) 2 (C) 3 (D) 4

31. If is given that and , and If , then

(A) (B) (C) (D)

32. If are non null vectors such that then

(A) is equal to 1 (B) cannot be evaluated (C) is equal to zero (D)None of these

33. If are unit coplanar vectors, then scalar triple product is equal to

(A) 0 (B) 1 (C) (D)

34. Let and are non-zero vectors such that , then is equal to

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Q. = i + 2J + 3k

= 3i + 2J + k and  (

then p, q, r are Ans: 0, 10, –3.

Q. If are unit vectors such that then the angles which makes with and are /3 & /2.

 

Q. The vectors a, b, c are of same length taken pair wise, they form equal angles. If a = i+J b = J +k then C equals.

(1, 0,1) (1, 2, 3) (–1, 1, 2) (–1/3, 4/3, –1/3) Q. Let a = 2l – J + k, b = l + 2J – k, c = i+ j- 2k. A vector in the plane of b and c whose projection on a is of

magnitude is

A. 2i+ 3J – 3k 2i + 3J + 3k -2i+J+5k 2i+ J + 5k Q. If a = (0, 1, -1), c =(1, 1, 1), then a vector be satisfying a  b + c = 0 and a – b = 3 is Ans: (1, 1, -2)

Q. If ( and at least one of the numbers , , is non zero. Then the     vectors a, b, c are coplanar.

Q. The scalars l and m are such that la + mb = c where a & b are monocollinear vectors then m = .

Q. Let a, b, c be three vectors having magnitude 1, 1 &2, if a  (ac) + b = 0 then the angle between a & c is /6.

Q. The points A(–1, 3, 0) B(2, 2, 1) C(1, 1 ,3) determine a plane. The distance from the plane to the pt. D(5, 7, 8) is .

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Q. P, Q, R be 3 mutually vector of same maq. If a vector x satisfies the _ eqn. + .then x = _____. Soln: 0 ] [ ] [ ] [ × − × + × × − × + × × − × = × x p q p q x q r q r x r p r p k2_{[x−(pˆ.x)pˆ)−(q−(pˆ.q)pˆ)]+[(x−(qˆ.x)qˆ)−(r −(qˆ.r)qˆ)]}(0)=0 we know (a.i)i +(a− j)j+(a−k)k =a parallelly – [(pˆ.x)pˆ+(qˆ.q)qˆ+(rˆ.x)r]=−x 3x – x = p+q+r x = 2 ) (p+q+r

1. Through the middle point M of the side AD of a parallelogram ABCD, the st. line BM is drawn cutting AC at R and CD produced at Q; prove that QR = 2 RB.

2. If a = pi+sinθJ +k

b = 2i + PJ + k c = i + J + k

If a ,,b care coplanar find all possible values of P& Q, P – 1, Q = (2n + 1) /2.

3. Let a,b and d be non coplanar vectors equally inclined to one another at an angle . If _ a×b +b×c= pa+qb +rc. Find p, q, r in terms of

. (taking dot product with

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Ans: (p, q, r) =     + + − + θ θ θ θ 1 2cos 1 , cos 2 1 cos 2 , cos 2 1 1 or     + − + + − θ θ θ θ 1 2cos 1 , cos 2 1 cos 2 , cos 2 1 1

1. If one diagonal of a quadrilateral bisects the other then if also bisects the quadrilateral.

2. P, Q are the mid points of the non-parallel sides BC and AD of a trapezium ABCD. Show that APD = CQB._ _

3. Find r in terms of one parameter r.n1 =1& r.n2 =1. [

2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 2 ) ( ) ( . n n _n _n n n n n n n n n r × − + × − = + t n1×n2.

4. Find the condition for the equations r×a =b and r×c =d to be consistent assuming the condition for consistency to be satisfied, solve the eqn_s. [ ) . . . ( ] [ 1 0 . . _d _b _a _d _a_b _b2_c c b a r d a d b + = = + + if [abc]  0.

5. Prove that the projection of the line r =a+bton the plane r.n= q is

= r     ₋ +       ₊ − _n n n b b t n n n a q a ₂ . . .₂ .

normal r =a+b + pn solving with the plane.

6. In a tetra hedran OABC, OA BC, show that OB 2_{+ CA}2_{= OC}2₊ AB2_.

7. Two medians of a are equal, show that the triangle is isoscelious.

1. If A+B =a,A−a =1, A×B =b,then 2 a a b a A = × +

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2 2 ₁₎ ( a a a a b B = × + −

1. Show that the eqn. of a line passing through the pt. with p. v. d and equally in a lined to the vectors a ,,b cwhose moduli are a, b, c is

      × + × × + = abc b a c a c b c b a d r α ( ) ( ) ( ) .

2. A st. line ‘L’ cuts the lines AB, AC, AD of a gm ABCD at pts. B1, C1, D1 respectively. If

AB AB1 =λ₁

, AD1 =λ2AD,AC1 =λ3ACthen prove that 2 1 3 1 1 1 λ λ λ = + .

3. Consider a ABC, having AD as its median through the vertex A. let  P be any pt. on its this median using vector methods prove that for every position of P on AD area of APB is equal to of APC. 

4. ABCD is a gm and M be any pt. on the side AB. Using vector methods prove that for every position of M, AC & DM divide each other in the same ratio. Also locate the pt. ‘M’ such that AC and DM trisect each other.

5. If A1, A2,……..,An are n pts. on a circle one unit radius, prove that the sum of the squares of their mutual distances is not greater than n2_.

6. In a triangle ABC, D divides AC in the ratio 2:1. BD is produced to F such that DF = 2 BD. Prove that AF is  to BC & equal to 2BC & CF is  to median and is twice the median.

1. Consider the vectors

i + cos( - )J + cos( - )k, cos( - )i + J + cos( - )k and cos( -_{ } _{ } _{ } _{ } _ )i + cos( - )J + ak where . & are different angles. If these

     

vectors are coplanar show that a is ind. of , , ._{   (a = 1)}

2. In a PQR, S and T are pts. on QR and PR respectively such that  QS = 3SR and PT = 4 TR. Let M be the point of intersection of PS and QT. Determinethe ratio QM: MT (15:4).

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3. Let x ˆˆ,yand zˆbe unit vectors such that c z z x b z y x a z y xˆ+ ˆ+ ˆ= ,ˆ×(ˆ×ˆ)= .(ˆ×ˆ)×ˆ= & 4 7 ˆ . , 2 3 ˆ .x= ay = a  , a =2, find xˆ,yˆ,zˆ in terms of a ,,b c. ) 8 4 3 ( 3 1 ˆ a b c x= + + c yˆ =−4 ) ( 3 4 ˆ c b z= − 0 1 sin cos 0 sin cos 0 sin cos 1 sin cos 0 sin cos 0 sin cos = × − γ γ β β α α α γ γ β β α α

1. The p. v. of pts. P & Q are 5i+7j-2k and –3i+3j+ 6k respectively. The vector A =3i-j+k passes through the pt. P and the vector B =−3i+2j+4k, 2i+7J-5k intersects vectors A &B. Find p. .v of pts. of intersection.

1.f: R  R, f(x) = 2

x

x _e

e ₋ −

1-1, on to, inverse graph. 2. Solve the eqn_{: f(x) = x}3_{– [x] = 5}

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Angle bisector

1. In a triangle ABC the lengths of the to sides are known; AB

= 6, 8

= AC

and A = 90; AM and BN are bisectors of the angles A and B. Find the cosine of the angle between the vectors AM and BN .

A (0) B C 8 6 10 a N b  c b− M     + = 8 6 b a AM   λ     ₋ = 30 8 3b a BN   µ = AM  λ 2 64 36 2 2 = +b a 5 5 4 µ = BN , consider AM.BN Ans: - 10 1 A B C 6 J M (…..) 8i (…..) N

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