# Vectors Work

August 10, 2017 | Author: Raghuram Seshabhattar | Category: Euclidean Vector, Mathematical Analysis, Mathematical Objects, Space, Geometry

Vectors +2...

#### Description

 p × [( x − q ) × p ] + q × [( x − r ) × q ] + r × [( x − p ) × r ] = 0 p = q = r = k r × [ x − p ) × r ] = 0    p × [( x − q ) × p ] + q [( x − r ) × q ] a = b = c = 1 66 (c × a ).(b × a ) (c × b).( a × b) l= 2 ( a × b) 2 (b × a )  a × b ) + β (b × c ) + γ ( c × a ) = 0

            2 c b a a × (b × 2c ) = b a , b , c β × γ ) = pα + qβ + rγ α γ 3

     β = 2i − J + k α O a × b 2 a × b 2 ( BD × AC ).(OD × OC ) DB OA OB OA           OA = a , OB = b , OC = 2a + 3b , OD = a − 2b p + q < 1 p and q

A(1, 0, 2)

B(–2, 1, 3)

TORS

D

C(2, -1, 1)

1 1 1 1   1 − 2 cos 1 AD = − i − J − k , AD =  , ,  3 3 3 3  1 + 2 cos θ 1 + 2 cos 2 1 + 2 cos       r = a + λb 1 2 cosθ −1  − , ,  1 + 2 cos θ 1 + 2 cos θ 1 + 2 cos θ   a a.a b.a a1 b a.b b.b = b1 c a.c b.c c1

a2 b2 c2

a 3 i a1 b3 × j a 2 c3 k a3

 b1 a b c − 11iˆ + 10 Jˆ + 2kˆ c c a cˆ = b2 = 0 a.a a.b a.c = 0 15 b3 b.a b.b b.c

 iˆ + cos( β − γ ) ˆj + akˆ iˆ + ˆj + cos(γ − β )kˆ iˆ + cos( β − α ) ˆj + cos(γ − α )kˆ 11 r .(3iˆ − ˆj + k            2 1 1   R n t n t r R t r R n r R t r R R b : c O, b , c  1 − r 2i − J − k r +    (t × n ) 2 2 r .(t × n )                        a , b and c kˆ kˆ kˆ a × b ) 2 c = (c × b ).( a × b )a + (a × c ).( a × b )b b a a , b , c a × (b × c ) a = b + c

      iˆ + 2 ˆj + kˆ r .a = 0 b r − c r a.b ≠ 0 a.a  b .a  c .a →

a x b)

  a.b  b .b  c .b

           a.c c × d ) + (b × c ).( a × d ) + (c × a ).(b × d ) = 0  b .c  c .c

u ⊥ v (u × v ).w ≤ 1 / 2 v w + w × u w u & v a a OE .CD = 0 (u1 × u 2 ).w ≤ 1 / 2

    w + w × u1 = u 2 ω u 2 u1 b a c c b a c c b a a × b + b × c = pa + qb + rc ˆ − 11i − 10 J + 2k c c= 15

   w + w × u1 = u 2 ω u 2 u1 b a c c b a c c b a a × b + b × c = pa + qb + rc ˆ − 11i − 10 J + 2k c= 15  cc  c '×a' a '×b' 1  a.b     b'×c' b= c= r= 2 a + kb + a × b  2  [ a ' b' c ' ] [ a' b' c' ] k +a  k  [ a ' b' c ' ] b × ( a × b) , ( a × b) 2

( a × b) × a b' = , ( a × b) 2

 a × b a × b c × a b × c AB PQ CD AB bˆ − (aˆ.b ) aˆ c' = (a × b) 2 [abc ] [abc] [abc ]

                →→  3 (aˆ × bˆ) aˆ b aˆ b & c (e .c )a = c c b a × (b × c ) + (a.b )b  (a − b ).c   r = b + tc a → b a b b+ 2 c − a c → →

a b = 4iˆ − ˆj + 3kˆ a = 3iˆ + 6 ˆj − 2kˆ

→ → → ˆ ˆ ˆ 1 1 1 + + = 1 a # b # c #1 i + j + ck 1− a 1− b 1− c

→ aiˆ + ˆj + kˆ, iˆ + bˆj + kˆ 5 iˆ + 2 ˆj + 2 kˆ 2iˆ + 2 ˆj + kˆ n→ + n→ + n→ + n→ = 0 → 4iˆ + 3 ˆj → c b b 1 2 3 4 3 3 3

( x − a) 2 ( y − a) 2 ( z − a) 2

( x − b) 2 ( y − b) 2 ( z − b) 2

→ → → → → → → → → → → → → → ( x − c) 2 → a, b, c 5 a + 3 b + 5 c 3 a + b + 2 c a − b − c a + 2 b + c ( y − c) 2 ( z − c) 2

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 coordinates [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 a (a) There are different ways of denoting a vector:  or or a are   a , b, c

different ways. We use for our convenience vectors, and a, b, c to denote their magnitude. (b) gives

A vector may be represented by a line segment OA and arrow direction of this vector. Length of the line segment gives the magnitude of the

vector. A

O

H e r e O is th e in itia l p o in t a n d A is th e te r m in a l p o in t o f O A

CLASSIFICATION OF VECTORS: (i)

etc. to denote

Equal vectors

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

(ii)

A

B

C

D

A B = C D A s w e ll a s d ire c tio n is s a m e

Zero Vector (null vector): A vector whose initial and terminal points are same, is called

Zero vector. For example

AA

.Such 

vector has zero magnitude and no direction, and denoted by AB + BC + CA = AA

Or

 AB + BC + CA = 0

B

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   a

(iv)

 a

 a as aˆ =  |a|

. Therefore .

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

O C E

(vi)

a

directions are opposite, and 3 are like vectors. Unit Vector: A unit vector is a vector whose magnitude is unity. We

write, unit vector in the direction of (v)

.

C

A

(iii)

0

A b c

B D

OP

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

AB = OB − OA

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

As shown in figure

and

OD

are Co-Initial vectors.

D

A d

C

a

O

c

B

b

.

  OA = a , AB = b

Here the

 c

and

 OB = c

.

is sum (or resultant) of vectors initial

a

to the terminal

point of

b

represents vector

and 

 b

. It is to be noticed that

b

point

coincides with the terminal point of  point of

 a

  a+b

 a

of

and the line joining the initial

in magnitude and direction.

PROPERTIES:   (i)

(ii) (iii) vectors) (iv) vectors)

  a +b = b+a

      a + ( b + c ) = (a + b ) + c     | a + b |≤| a | + | b |

    | a + b | ≥ || a | − | b ||

(So, Vector addition is commutative) (So, Vector addition is associative) (equality holds when

(equality holds when

 a

 a

and

and

 b

 b

are like

are unlike

     a +0 =a =0+a      a + ( −a ) = 0 = ( −a ) + a

(v) (vi)

.

MULTIPLICATION OF VECTOR BY SCALARS : If

 a

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

 a

is a vector parallel to

 a

a

whose modulus is | m | times that of . This multiplication is called Scalar   a b a Multiplication. If and are vectors and m, n are scalars, then:  m( )    a a a a a = ( )m = m m(n) = n(m) = (mn)       b a a a a b a (m + n) =m +n m( + ) = m + m . LINEAR COMBINATIONS: Given

a

finite

set

    r = xa + yb + zc + ........

of

vectors

 a , b , c ,......

then

is called a linear combination of

the

vector

 a , b , c ,......

for

any x, y, z..... . We have the following results:  (i)

(ii)

If

 a,b

are

non-zero,

    xa + yb = x' a + y' b ⇒ x = x' ; y = y'

(iv)

vectors

then

.   a,b

Fundamental Theorem: Let be non-zero , non collinear vectors   a,b r . Then any vector coplanar with can be expressed uniquely as a  linear combination of 

(iii)

non-collinear

 a,b

  xa + yb = r that    a, b,c If are non-zero, non-coplanar vectors then:      xa + yb + zc = x' a + y' b + z' c ⇒ x = x' , y = y' , z = z' ..    x1 , x 2 ,...... x n

If

&

i.e. There exist some unique x, y R such

are n non zero vectors, & k , k , .....k are n scalars 1 2 n if the linear combination

   k1x1 + k 2 x 2 +........ k n x n = 0 ⇒ k1 = 0, k 2 = 0..... k n = 0

then

we

say

that

vectors are Linearly Independent Vectors .   (v)

x1 , x 2 ,...... x n

If Linearly

are not Linearly Independent then they are said to be Dependent vectors. i.e. if.

   k1x1 + k 2 x 2 +........ k n x n = 0 ⇒ k1 = 0, k 2 = 0..... k n = 0

at least one then

   x1 , x 2 ,...... x n

Example: Show that the vectors coplanar  (Where Solution: Let

  a , b, c

& if there exists

Kr ≠ 0

are said to be Linearly Dependent.

      5a + 6b + 7c, 7a − 8b + 9c

&

   3a + 20b + 5c

are

are three non-coplanar vectors.).

      A = 5a + 6b + 7 c B = 7a − 8b + 9c

,

and

   C = 3a + 20b + 5c

.

A, B

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

Now,

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

5x + 7y + 3z = 0 6x – 8y + 20 z = 0 7x + 9y + 5z = 0 (As 5

7

   a , b, c

are non-coplanar vectors)

3

6 − 8 20 = 0 7 9 5

Now D = So the three linear simultaneous equation in x, y and z have a nontrivial solution. Hence the given vectors are coplanar vectors. COLLINEARITY AND COPLANARITY OF POINTS:

(a)

The necessary and sufficient condition for three points with position

vectors

  a , b, c

to be col

not all zero such that Where (b)

is that there exist three scalars x, y, z,  linear   xa + yb + zc = 0

x+y+z=0

.

The necessary and sufficient condition for four points with position

vectors

  a , b, c

and

 d

to be coplanar is that there exist scalars x, y, z, u,

     xa + yb + zc + ud = 0

x+ y +z+u = 0

not all zero, such that where . 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 P K O

M

A

N

 a     +b a a + 2b M≡ , N≡ 2 = 2 2 4 Now,   a+b    −a 2 b − 3a K≡ 4 = 2 8   2b − 3a   −b+ − 6 b − 3a 8 P≡ = 2 16 uuu 3   OP = − (2b + a ) 16

B

uuu 1   1 uuu ON = (a + 2b ) = − OP 4 6

( )

. Hence the points N, O& P are collinear. SECTION FORMULA: Let A, B & C be three non collinear points in space having the   a, b r position vectors and . ( x 1, y 1, z 1)

A

n C

a

m

r O

Let

(x , y, z )

B ( x 2, y 2, z 2) b

AC n = CB m

mAC=nCB

m AC = n CB

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

Using (i), we get

.

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 (x, y, 0) respectively.

Y B

N P (x, y, z)

O

X

A

Let

ˆi , ˆj, kˆ

We have

M

C

Z

denote unit vectors along OX, OY and OZ respectively.  r = OP = xˆi + yˆj + zkˆ

as

OA = xˆi , OB = yˆj

and

OC = zkˆ

.

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

(

)

  r xˆi + yˆj + zkˆ r=  = = ˆi + mˆj + nkˆ  2 2 2 |r| x +y +z ⇒ r = rˆi + mrˆj + nrkˆ

.

MULTIPLICATION OF VECTORS: SCALAR PRODUCT OF TWO VECTORS (DOT PRODUCT):   a

b

The scalar product, of two non-zero vectors and is defined as   | a | | b | cos θ θ .where is angle between the two vectors, when drawn with same initial point. Note that If at least one of PROPERTIES :   (i) (ii)

0≤θ≤π

 a

 b

.

and is a zero vector, then

  a.b = b.a (scalar product is commutative) 2    2 2 a = a.a = | a | = a

 a.b

is defined as zero.

(iii) (iv)

     (ma ).b = m(a.b) = a (mb)   a.b  −1 θ = cos      | a |.| b |    a.b = 0 ⇔ a

(v) Vectors non-zero vectors]. (vi) (vii) (viii) (ix)

(xi)

and

 b

  a b

are perpendicular to each other. [ ,

ˆi.ˆj = ˆj.kˆ = kˆ.ˆi = 0      a.( b + c) = a.b + a.c         (a + b).(a − b) = | a |2 − | b |2 = a 2 − b 2

Let

  a = a1ˆi + a 2 ˆj + a 3 kˆ, b = b1ˆi + b 2 ˆj + b 3 kˆ

Then (x)

(where m is a scalar)

,

 a.b = (a 1ˆi + a 2 ˆj + a 3 kˆ ).(b1ˆi + b 2 ˆj + b 3 kˆ )

.

Maximum value of

    b a b a . =| | | |

Minimum value of

    b a b a . =–| | | |

 a

(xii) Any vector can be written as,

 a

=

ej ej d i

   a. i i + a. j j + a. k k

.

Algebraic projection of a vector along some other vector:

are

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

O

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 semicircle. 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 − a

and

BP = r − (−a ) = r + a

AP.BP = (r − a ).(r + a ) = r 2 − a 2

2 2 = OP – OA = 0

AP and BP are perpendicular to each other, i.e. VECTOR (CROSS) PRODUCT: The vector product of two non-zero vectors

 a

and

 b

∠APB = 900

.

, whose modules

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

ab sin θ

,

where is

θ(0 ≤ θ ≤ π)

 n

 b

 a

the angle between vectors and . The direction is 

that of a vector perpendicular to both right-handed orientation.

 a

 b

& , such that

  a , b, n

are in

b θ

O

a

    a × b = | a | | b | sin θ nˆ

.

PROPERTIES:   (i) (ii) (iii)

  a × b = −( b × a )       ( ma ) × b = m(a × b) = a × (mb) (Where m is a scalar)      a×b = 0 ⇔ b a

vectors

and

are parallel.

(provided

 a

and

 b

are non-zero

vectors). (iv) (v) (Vi)

 ˆi × ˆj = ˆj × ˆj = kˆ × kˆ = 0 ˆi × ˆj = kˆ = −(ˆj × ˆi ), ˆj × kˆ = ˆi = −(kˆ × ˆj), kˆ × ˆi = ˆj = −(ˆi × kˆ )        a × (b + c) = a × b + a × c   a = a 1ˆi + a 2 ˆj + a 3 kˆ b = b1ˆi + b 2 ˆj + b 3 kˆ

(vii) Let

and

ˆi ˆj   a × b = a1 a 2 b1 b 2

(Viii) . (ix)

  |a×b| sin θ =   | a || b |

Area of triangle =

, then

kˆ a3 b3

=

ˆi(a b − a b ) + ˆj(a b − a b ) + k(a ˆ b −a b ) 2 3 3 2 3 1 1 3 1 2 2 1

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

.

B b O

(x)

P

θ

A

a

Area of parallelogram =

C

B

b

    axb≠bxa

p

θ

O

(xi)

  ap = ab sin θ =| a × b |

A

a

.

(not commutative)

    axb a & b is ±   axb

(xii) Unit vector perpendicular to the plane of . (xiii) A vector of magnitude ‘r’ & perpendicular to the plane of   r axb   a & b is ±   axb

(

)

. (xiv) Area of any quadrilateral whose diagonal vectors are

  d1 & d 2

1   d1 x d 2 2

by . (xv) LaGrange’s

Identity:

     2 2   a .a a & b ;(a x b) 2 = a b − (a . b) 2 =   a .b

For

 a .b  b.b

any

two

is given

vectors

.

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

BC, CA , AB

deduce

sin A sin B sin C = = a b c

(Sine rule).

represent the vectors a, b, c respectively.

Then, we have A

π- A

b

c

π- C

B C

a

π- B

a + b + c = 0, ==> c = - (a + b) ==> b × c = b × (- a - b) =-b×a=a×b Similarly, c×a=a×b Hence,

bc sin(π − A ) = ca sin( π − B) = ab sin(π − C)

==> b × c = c × a = a × b

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

bc sin(π − A ) = ca sin( π − B) = ab sin(π − C)

.

SCALAR TRIPLE PRODUCT (BOX PRODUCT): The scalar triple product of three vectors

   (a × b).c

the between Let

=

  | a | | b | | c | sin θ cos φ

   axb & c

  a, b

θ

 c

& is defined as  

where is the angle between angle 

. It is also denoted as

  [a b c]

.

   a = a1iˆ + a2 ˆj + a3 kˆ, b = b1iˆ + b2 ˆj + b3kˆ, c = c1iˆ + c2 ˆj + c3kˆ

Then

ˆi ˆj   a × b = a1 a 2 b1 b 2

kˆ a a 3 = ˆi 2 b2 b3

a 3 ˆ a1 a 3 ˆ a1 a 2 −j +k b3 b1 b 3 b1 b 2

.

a&b

φ

& is

a    (a × b).c = c1 2 b2

a1 a 2 a3 a1 a 3 a1 a 2 − c2 + c3 = b1 b 2 b3 b1 b 3 b1 b 2 c1 c 2

a3 b3 c3

.

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

Note that

, hence in scalar triple product dot and cross are     (a × b).c [a b c ] interchangeable. Therefore we indicate by . PROPERTIES:  (i)

  | (a × b).c |

represents the volume of the parallelepiped, whose adjacent

sides are represented by vectors Therefore three vectors a1 a 2 b1 b 2 c1 c 2

a3 b3 = 0 c3

  a , b, c

,

(ii)

Volume of the tetrahedron =

(iii)

      [a + b c d ] = [a c d ] + [ b c d ]  [a a b ] = 0

  a , b, c

in magnitude and direction.  [a b c ] are coplanar if = 0. i.e.

1   | [( a b c] | 6

.

(IV) . (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 .( cx b) i. e. [ a b c ] = − [ a c b ]    b a c

(vii) If = a i+a j+a k; 1 2 3

= b i+b j+b k & = c i+c j+c k then 1 2 3 1 2 3

a1 a 2 a 3  [a b c] = b1 b 2 b 3 c1 c2 c3

In general, i

        a = a 1 l + a 2 m + a 3 n b = b1 l + b 2 m + b 3 n

;

a1  a b c = b1

a2

a3

b2

b3

c1

c2

c3

[ ]

.

[ l m n]

&

    c = c1 l + c2 m + c3 n

then

; where 

   a,b,c

   , m & n

  ⇔ [a b c] = 0

are non-coplanar vectors .

(viii) If are coplanar . (ix) Scalar product of three vectors, two of which are equal or parallel is 0    [a b c] = 0 i.e. .    a,b,c Note :  If are non-coplanar then   [a b c] > 0 &  [a b c] < 0

for left handed system

(x) (xi) (xii)

for

handed

system

[ a + b

]

[ a b c]

.

[i j k] = 1.   [ K a b c ] = K[ a b c ]

.

     [(a + b) c d ] = [ a c d ] + [ b c d ]

(xiv) Remember that: .

[ a − b

.

]

     b− c c−a

=0

     b+ c c+a

&

VECTOR TRIPLE PRODUCT: If

right

   a,b,c

The vector triple product of three vectors at least    a,b,c

is

a

zero

vector

 b

 c

and are

=2

is defined as

collinear

   a × ( b × c)

vectors

. one  a

or is

perpendicular  b

    a × (b × c) = 0

 c

and , only then non-zero vector

to

   a × ( b × c)

. In all other cases in the

both will be a  a

plane of non-collinear vectors and perpendicular to the vector .      a × (b × c) = λb + µc λ µ Thus we can take , for some scalars & . Since     a ⊥ a × ( b × c) ,         a.(a × (b × c)) = 0 ⇒ λ(a.b) + µ(a.c) = 0 ⇒ λ(a.c)α, µ = −(a.b)α

α

.

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

 a

Example: For any vector , prove that Solution:

for same scalar

ˆi ×(a × ˆi) + ˆj×(a × ˆj) + kˆ × (a×k) ˆ = 2a

.

   [ˆi × (a × ˆi )] + [ˆj × (a × ˆj)] + [kˆ × (a × kˆ )]

= =

     [( ˆi.ˆi )a − (ˆi.a )ˆi ][( ˆi.ˆj)a − (ˆj.a )ˆj] + [( kˆ.kˆ )a − ( kˆ.aˆ )kˆ ]      a − (ˆi.a )ˆi + a − (ˆj.a )ˆj + a − (kˆ.aˆ )kˆ [

=

ˆi.ˆi = ˆjˆj = kˆkˆ = 1]

   3a − [( ˆi.a )ˆi + (ˆj.a )ˆj + (kˆ.aˆ ) kˆ ]

Let

 a = a 1ˆi + a 2 ˆj + a 3 kˆ

. Then

ˆi a = ˆi (a ˆi + a ˆj + a kˆ ) = a ˆi 2 + a (ˆi.ˆj) + a (ˆi.kˆ ) = a (1) + a (0) = a 1 2 3 2 2 3 1 2 1

Similarly,

ˆj.aˆ = a , kˆ.a = a 2 3

    3a − (a 1ˆi + a 2 ˆj + a 3 kˆ ) = 3a − a = 2a

.

L.H.S. = R.H.S. RECIPROCAL SYSTEM OF VECTORS:    a ′, b′

Let and

 a, b

 c

& be three non-coplanar vectors. Then the system of vectors  c′

which satisfies

    a.a ′ = b.b′ = c.c′ = 1

called

      a.b′ = a.c′ = b.a ′ = b.c′ = c.a ′ = c.b′ = 0

and the 

, is

  a , b, c reciprocal system to the vectors  .        b× c c×a  a×b a ′ =    b′ =    c ′ =    [a b c ] [a b c ] [a b c ]

,

,

.

PROPERTIES: (I) (II) is

a.b′ = a.c′ = b.a ′ = b.c′ = c.a ′ = c.b′ = 0

The scalar triple product [a b c] of three non-coplanar vectors a, b, c the reciprocal of the scalar triple product formed from reciprocal system.

Example: Solve the vector equation: 

     r × b = a × b , r .c

b

is not perpendicular Solution: We are given;  

to .

  r ×b = a×b    (r − a) × b = 0     ( r − a) b Hence and are parallel    r −a = t b   r .c

And we know 

Or 

. . . (i)

= 0,

Taking dot product of (i) by    r .c − a.c

=

 c

 t (b.c)

  a.c t (b.c ) – =   a.c      b.c 

t=–

 r

from (i) and (ii) solution of is  r

=

 a

  a.c       b.c  b

= 0 provided that

.

we get

. . . (ii)

 c

KEY CONCEPTS · If

are any `n' vectors and

are scalars, then

known as linear combination (LC) of the vectors · If

are any `n' vectors and

is

.

are scalars. IF

and

atleast one of is not equal to zero, then the vectors be linearly dependent(L-D)

are said to

· 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

· If the vectors

are said to be coplanar.

are coplanar then

are also coplanar.

STATEMENTS AND REASONING EXAMPLES:   1.

STATEMENT -1: If

x&v

are unit vectors inclined at an angle  and  

is a unit vector bisecting the angle between them, then

 x

 x+v x= α 2 cos 2

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

=

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 &

uuu x + v AD = 2

uuu 1 α | AD | = 2 cos 2 2 2

, So

= cos

α 2

 unit vector along AD i.e. x is given by 2.

=

STATEMENT -1: The points with position vectors    4a − 7b + 7c

      a − 2b + 3c, − 2a + 3b − c

,

are collinear.

because STATEMENT-2: The position vectors are linearly dependent vectors.

  a − 2b + 3c,

-2

      a + 3b − c, 4a − 7b + 7c

(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 (

         a − 2b + 3c) + λ 2 (−2a + 3b − c) + λ 3 (4a − 7b + 7c) = 0

  a, b & c

equating coefficients of

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

then the angle between

are three unit vectors such that

  a &b

   1 a × (b × c) = b 2

is /2

because STATEMENT-2: If

   1 a × (b × c) = b, 2

then

 a .b

= 0.

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

whereas

   a × (b × c)

comparing

 a.b

=

=0

1 b 2

So

 a

is perpendicular to

 b

‘a’ is correct. 4.

STATEMENT -1: In ABC, because STATEMENT-2: If

uuu uuu uuu AB + BC + CA

uuu  uuu  OA = a, OB = b

=0

uuu   AB = a + b

then

(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. (C) In ABC

uuu uuu uuu uuu AB + BC = AC = − CA

uuu uuu uuu OA + AB = OB

uuu AB

+

uuu BC

-

uuu CA

=

u O

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

5.

STATEMENT -1: p = 9/2 and q = 2.

    a = 3i + pj + 3k

because STATEMENT-2:

If

and

    a = a1 i + a 2 j + a 3 k

    b = 2i + 3 j + qk

and

are parallel vectors it

    b = b1 i + b 2 j + b3k

are parallel

a1 a 2 a 3 = = b1 b2 b3

(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. (A) 3 p 3 = = 2 3 q

from

      a = a1 i + a 2 j + a 3k, b + b1 i + b 2 j + b 3k

    a = 3i + pj + 3k

 6.

and

 b

=

   2i + 3 j + qk

are parallel 

a1 a 2 a 3 = = b1 b 2 b 3

2

3 p 3 = = 2 3 q

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

(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. (A) 7.

STATEMENT -1:    r = a + αb

and

because

The shortest distance between the skew lines 

   r = c + βd

is

  | [a − c bd] |  | b×d |

STATEMENT-2: Two lines are skew lines if three axist no plane passing through than (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. (B) A – Defn B – Defn 8.

STATEMENT -1: IF because STATEMENT-2:

 a.b

 a.b = 0

 a

= 0  either

 b

 a =0

or

 b

 a

= 0 or 

 b

=0

(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. (D)

9.

STATEMENT -1:

    A× B = B× A

because     A × B =| A || B |

STATEMENT-2: fingers curls from A to B

sin

 n

, when  is angle, when your

(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. (D) 10.

STATEMENT -1: If the vectors coplanar, then ||2 is equal to 16. because STATEMENT-2: The vectors

  a,b

2iˆ − ˆj + kˆ

and

 c

,

ˆi + 2ˆj − 3kˆ

and

3iˆ − λˆj + 5kˆ

   a, (b × c

are coplanar iff

are

)=0

(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. 2 −1 1 1 2 −3 3 λ 5

= 0   = -4

 ||2 = 16 Ans. (A) 11.

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

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

5 2

,−

4 5 2

,

1 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.lx + my + nz = P be the equation of a plane in the normal form.

 D.N. of the plane 3x – 4y + 5z = 10 3

be < 3, -4, 5 >  D.C <

5 2

,

−4 1 , 5 2 2

>

Ans. (A). 12.

STATEMENT -1: The value of expression

ˆi(ˆj × k) ˆ + ˆj(kˆ × ˆi) + k(i ˆ ˆ × ˆj) = 3

because STATEMENT-2:

ˆi(ˆj × k) ˆ = [i.j.k] ˆ ˆ ˆ =1

(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. ˆi(ˆj × k) ˆ + ˆj(kˆ × ˆi) + k(i ˆ ˆ × ˆj)

=

13.

ˆi.iˆ + ˆj.jˆ + k.k ˆˆ

= 1 + 1 + 1 = 3.

Ans. (A)

STATEMENT -1: A relation between the vectors 

   a×b r =  a.a

because STATEMENT-2:

  r, a and b

is

   r ×a = b

 r.a = 0

(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.

Since Q

We have

=( 

   b= r×a      a × b = a × (r × a)

     a.a) r − (a.r)a    a×b r =  a.a

=

   (a .a) r Q a. r

=0

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 . 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. 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. 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. 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 If = 0 And the vectors X = (x2, x,1), Y = (y2,y,1), Z = (z2,z,1) are non coplanar, then the vectors (a2,a,1), (b2,b,1), (c2,c,1) are co-planar.

3.

4.

5.

6.

7.

8.

9.

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

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 10.

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

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 . 23.

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. 24.

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. 25.

Reciprocal System

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

a = , b = , c = 27.

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

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

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

a=. 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 30.

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

along . Ans: 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 . 32.

______ (1) Find p, q, r interms of . 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. 1.

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. 4.

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 . 6.

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.

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. 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. (i)

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/.

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.

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 then is A. 7

B. 42

C.

is  to

and

D. None

PRACTICE QUESTIONS Angle between Vectors 11. If

, then

(A) (B) (C) Projection along a Vector 12. Let of

and

and

whose projection on

(A) (B) Dot Product: 13. Let then

(D)

and

(C)

be three vectors, A vector in the plane is of magnitude

is

(D)

be two non-collinear unit vectors. If

and

is

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

(D)

14. If and c are unit vectors, then (A) 4 (B) 9 (C) 8 (D) 6 Parallel Vectors:

does not exceed

,

15. If

and

(A) (B) Parallel Vectors:

. If

(C)

, then d will be

(D)

16. The point of intersection of is (A) (B) (C) Section Formula:

and

and

, where

and

(D) None of these

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 then may belong to

are inclined at an angle

(A) (B) (C) Modulus of Vector: 19. Let value

and

. If

,

(D) None of these

be the vectors such that

, if

and

then the

of is [IIT - 95] (A) 47 (B) (C) 0 (D) 25 Vector Triple product: 20. Let

and

. If

is a vector such that

angle between and is , then (A) 2/3 (B) 3/2 (C) 2 (D) 3 Vector Product of four vectors:

[IIT - 99]

and the

21. Let the vectors

and

be such that

planes determined by the pairs of vectors

. Let and

and

be

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 then

(B) If

,

(p) 1

is

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

(D) If

are three non zero vectors such that

(r)

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 (A)

units is (B) (C)

(D) LEVEL-II MODEL QUESTIONS

1. Let to , then (A)

and a unit vector

(B)

(C)

2. If the vectors triangle ABC, then (A)

(D)

form the sides BC, CA and AB respectively of a

(B)

(C) 3. If

(D) and

are two unit vectors such that

to each other then the angle between (A)

be coplanar. If perpendicular

(B)

(C)

and

and is

(D)

4. Let there be two points A and B on the curve satisfying (A)

and

(B)

are perpendicular

in the plane OXY

then the length of the vector

(C)

is

(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 (A)

and (B)

(C)

and (D)

7. Given that (A)

and

(B)

8. The vector to (A)

(C)

(A)

and

and

is perpendicular

is

(D) None of these is twice its Y-component. If the magnitude of the vector

and it makes an angle of (B)

is :

(D) None of these

is perpendicular to

9. X-component of is

then the angle between

(C)

. the angle between (B)

, then the angle between a and b is

(C)

with z-axis then the vector is : (D) None of these

10. If (A)

then the angle between (B)

(C)

and

is :

(D)

11. If the non zero vectors

are perpendicular to each other then the

solution of the equation,

is

(A)

(B)

(C) 12.

(D) None of these and

are mutually perpendicular unit vectors .

satisfying (A)

and (B)

13. If (A)

then

(C)

is equal to

(D)

then the value of (B)

(C)

is a vector

is

(D)

14*. If in a right angle triangle ABC, the hypotenuse then (A)

is equal to (B)

(C)

15. If

(D) 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 (A) (C)

to both a and b. If (B) (D) None of these

then

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 and (C)

(q)

is

and

for some non

(r)

zero vector , then the value of is (D) The volume of parallelopiped whose sides are given , and 23. Column - I Column -II

is

(t) 1

(A) For given vectors and

the projection of the vector

the vector (B) If

and

(s) 0

(p) on

is are unit vectors satisfying

(q)

and

is equal to

(C) If , then k is (D) The value of c for which the vector LEVEL -III MODEL QUESTIONS

(r) (s)

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)

3. The vector (A)

(B)

(C)

(D) None of these

, is given by (D) None of these

II A new operation * is defined between two non antiparallel vectors as

, where

is the angle between

4. The condition for which (B)

(C)

and

are perpendicular is

(A) 5.

is

(A)

(B) not defined (C) 0 (D) None of these

(D) None of these

6. For (A) (B) (C)

and .

= 0 is a necessary condition is a necessary condition is a sufficient condition, where

and

(D) None of these 7. Let

are non zero mutually perpendicular unit vectors. There is a

vector coplanar with (A) 8.

(B)

(C)

and

is

are unit vectors along the coordinate axes and

(B)

(C)

then q is

(D)

9. If the vectors only one real x then (A)

then minimum value of

(D)

makes angle q with (A)

and

and

are coplanar vectors for

is

(B) 2 (C) 4 (D)

10. If are unit vectors such that is equal to (A) 11. Let

(B)

(C)

then

(D) None of these

be three non-coplanar vectors and

the relations

are vectors defined by

, 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 equals

if

(C)

= is a unit vector such that

(A)

(B)

14. If

are three non coplanar vectors, then

, then

(D) equals

(A) 0 (B) 15. If

(C)

(D)

be three mutually perpendicular vectors of the same magnitude. If

a vector satisfies the equation given by (A)

(B)

16. If

,then is

(C)

(D)

are unit coplanar vectors, then the scalar triple

product

=

(A) 0 (B) 1 (C)

(D)

17. Let and . Then depends on [IIT2002] (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 (A)

and (B) 3 (C)

20. If (A)

, (B)

becomes minimum is [IIT-2004] (D) and

, then

is

(C) (D)

21. The unit vector which is orthogonal to the vector coplanar with the vectors (A) 22. If

(B)

(C)

and

and is

is [IIT-2004]

(D)

are three non zero, non coplanar vectors and [IIT-2005]

then the triplet of pairwise orthogonal vectors is (A)

(B)

(C)

(D)

23.

. A vector coplanar to

along

of magnitude

and

has a projection

, then the vector is [IIT-2006]

(A) (B) (C) (D) None of these 24. The edges of a parallelepiped are of unit length and are parallel to noncoplanar unit vectors such that parallelepiped is [IIT-2008] A)

. Then, the volume of the

(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 given by of

and

(A)

. Where P is farthest from origin O, let M be the length be the unit vector along and

. Then [IIT-2008]

(B)

(C) 26. Let

. (where O is the origin ) is

(D) 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

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

29. If

(B) (D) None of these are non coplanar unit vectors such that

angle between (A)

(B)

and

(C)

then the

is :

(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) 34. Let

and

(D) are non-zero vectors such that

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

,

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 . Q. a  b = c, b  c = a, then

Q. P, Q, R be 3 mutually  vector of same maq. If a vector x satisfies the eqn. + .then x = _____. Soln: p × [ x × p − q × p] + q × [ x × q − r × q ] + r × [ x × r − p × r ] = 0

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 – [( 3x – x = x=

pˆ .x ) pˆ + (qˆ.q )qˆ + ( rˆ.x )r ] = − x

p+q +r

(p + q + r) 2

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. 1.

2.

If

a = pi + sin θJ + k b= c

2i + PJ + k =i+J+k

a,b ,c

If + 1) /2. 3.

Let

are coplanar find all possible values of P& Q, P – 1, Q = (2n

a, b and d

be non coplanar vectors equally inclined to one

another at an angle . If

a ×b + b ×c

. (taking dot product with

=

a,b ,c,b × c

pa + qb + rc

).

. Find p, q, r in terms of

Ans: (p, q, r) =

 1 − 2 cos θ 1  , ,  1 + 2 cosθ 1 + 2 cos θ 1 + 2 cos θ

 −1 2 cosθ −1  , ,  1 + 2 cos θ 1 + 2 cos θ 1 + 2 cos θ

  

or

  

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. 1.

r

Find in terms of one parameter

3.

2

r=

r .n1 = 1 & r .n2 = 1

.[

2

n 2 − n1 .n2 n − n1 n 2 + 1 2 n1 (n1 × n2 ) (n1 × n 2 ) 2 n2

+t

n1 × n2

. r × a = b and r × c = d

Find the condition for the equations to be consistent assuming the condition for consistency to be satisfied, solve the eqns. 4.

b .d + a .d = 0 r =

[

1 ( d .b .a + d .a b + b 2 c ) [a b c ]

5. Prove that the projection of the line r=

q − a .n   b .n    . n + t b − .n  a +   n2 n 2    

if [

ab c

]  0.

r = a + bt

normal

on the plane

r = a + b + pn

r.n

= q is

solving with the

plane. 6. In a tetra hedran OABC, OA  BC, show that OB 2 + CA2 = OC2 + 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 , A=

a ×b + a a2

then

B=

1.

b × a + a (a 2 − 1) a2

Show that the eqn. of a line passing through the pt. with p. v.

equally in a lined to the vectors

a,b ,c

d

and

whose moduli are a, b, c is

 a (b × c ) + b ( c × a ) c ( a × b )  r = d +α  abc   2.

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

D1 respectively. If

AB1 = λ1 AB

,

then prove that

1 1 1 = + λ3 λ1 λ 2

. 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 n 2. 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)

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). 2.

3.

Let

xˆ , yˆ

and be unit vectors such that

xˆ + yˆ + zˆ = a , xˆ × ( yˆ × zˆ ) = b .( xˆ × zˆ ) × zˆ = c

terms of

a,b ,c

a .xˆ =

&

3  7 , a. yˆ = 2 4

,

a =2

, find

xˆ , yˆ , zˆ

in

.

1 xˆ = (3a + 4b + 8c ) 3 yˆ = −4c zˆ =

1.

4 (c − b ) 3

cos α

sin α

cos α

sin α

0

cos β cos γ

sin β 0 × cos β sin γ α − 1 cos γ

sin β sin γ

0 =0 1

0

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

2i+7J-5k intersects vectors

1.f: R  R, f(x) =

e x − e−x 2

A&B

B = −3i + 2 j + 4k

. Find p. .v of pts. of intersection.

1-1, on to, inverse graph.

2. Solve the eqn: f(x) = x3 – [x] = 5

Vector

,

Angle bisector AB 1.

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

= 6,

AC = 8

and A = 90; AM and BN are bisectors of the angles A and B. Find the cosine of the angle between the vectors A (0) 6

 N b 8

 a

M   b −c 10

B

C

  a b AM = λ  +  6 8    3b − 8a   BN = µ   30   AM =

a2 b2 + = 2λ 36 64

BN = µ

4 5 5

, consider

1

Ans: C 6J A

10 M (…..) N 8i (…..)

B

AM .BN

AM

and

BN

.

1.

QUESTION BANK level-i The vector is 1) Null vector 2) unit vector 3) parallel to 4) a vector parallel to

2. If equal to

, 1)

then

is

2)

3) 4) 3. If A=(2,7,-4) and AB=(-2,-5,3)then B= 1.(0,1,2) 2.(0,-1,2) 3.(0,2,-1,) 4.(0,2,1) 4. If and , then is true for 1) a = -1 2) a = 1 3) all real values of `a' 4) for no real values of `a' 5. Let A=2i+4k, B= , D =2i+k.Then the value of CD in terms of AB is 1. AB 2. AB 3. AB 4. AB 6. If the position vectors of the points A,B,C,D are (0,2,1),(3,1,1),(-5,3,2),(2,4,1) respectively and if PA+PB+PC+PD = 0 then the position vector of Pis 1. 7.

2. and

3.

4.

are two vectors

where are two scalars. Then the lengths of the vector are equal for 1) all values of 2) only finite number of values of 3) infinite number of values of 4) No value of 8. If a is a vector so that its resultant with 3i+4j-2k is i, then a = 1. 2i+4j+2k 2. -2i-4j+2k 3. -2i-4j-2k4. -2i+4j-2k 9. Let A,B,C be the vertices of a triangle.If AB=6i-7j+k, AC=-i-6j+7kthen B C= 1. -7i+j+6k 2. 7i-j+6k 3. 7i-j-6k 4. -7i-j-6k 10. If AB = 3i-2j+k, BC = i+2j-k, CD= 2i+j+3k OA= i+j+k,then the position vector of D is 1. 7i-2j-3k 2. 7i-2j+3k 3. 7i+2j-3k 4. 7i+2j+4k 11. Let A,B,C be the vertices of the triangle ABC and let then 1. 0 2.ai 3. 3a 4. (i+j+k) 12. If the position vectors of the vertices A,B,C are i+2j+3k, 3i-4j+5k and -2i+3j+7k then the vector determined by the side BC is 1. 5i+7j+2k 2. -5i+7j+2k 3. 5i-7j+2k4. 5i+7j-2k 13. If the position vectors of P and Q arei+3j-7k and 4i+5j+2k then PQ= 1. 3i+2j+9k 2. 3i-2j+9k 3. -3i2j+9k 4.i+j+k 14. If a= 2i+4j-5k, b=i+j+k, c=j+2k, then the unit vector parallel to a+b+c is

1. 2. 3. 4 . 15. If r1 = (3,-2,1) r2 = (2,-4,-3) ,r3 = (-1,1,2) then the modulus of the vector 2r1 3r2 - 4r3 is 1. 12 2. 13 3. 14 4. 15 16. The unit vector parallel to i-3j-5k is 1.

(i-3j-5k) 2. (i-3j-5k)

3. (i-3j-5k) 4. (i-3j-5k) 17. The unit vector parallel to the resultant vector 2i+3j-k and 4i-3j+2k is

1. 6i+k 2. 3.

4. 2i+k

18. The positon vectors of the points A,B,C are i+2j-k , i+j+k , 2i+3j+2krespectively.If A is chosen as the origin then the position vectors of B and C are 1. i+2k, i+j+3k 2. j+2k, i+j+3k 3. -j+2k, i-j+3k 4. -j+2k, i+j+3k 19. If AB = 2i-3j+k ,CB = i+j+k,CD = 4i-7j then AD= 1. 5i+11j+k 2. 5i-11j 3. 5i+11j 4. -5i+11j 20. If A,B,C,D are the vertices of a quadrilateral taken in order thanAB+BC+CD+DA= 1. 4BD 2. 3BD 3. 0 4. 2BD 21. ABCDE is a pentagon.If forcesAB,AE,BC,DC,ED and AC act at a point then their resultant is 1. 3AC 2. AC 3. 2AC 4. 4AC 22. Let ABCD be a trapezium and let P,Q be the midpoints of the nonparallel sides AD,BC respectively.Then PQ = 1.AB+DC 2. (AB+BC) 3. (AB+DC) 4. (AB+BC) 23. Let A=(12,3),B=(3,-1,5).C=(4,0,-3).Then the triangle ABC is right angled at 1.

2.

3. 4. 24. The perimeter of the triangle whose vertices are the points 2i-j+k, i-3j5k, 3i-4j-4k is 1.

2.

3. 4. 25. Let A=2i+3j+4k,AB=5i +7j+6k. Then B= 1. -(7i+10j+10k) 2. (7i-10j+10k) 3. (7i+10j-10k) 4. (7i+10j+10k) 26. If AO+OB=BO+OC then 1. A is the midpoint of BC 2. B is the midpoint of CA 3. C is the midpoint of AB 4. C divides AB in the ratio 1:2

27. The resultant of two concurrent forces nOP and mOQ is (m+n)OR.Then R divides PQ in the ratio 1. m : n 2. n : m 3. 1 : n 4. m : 1 28. Let A= 2i+7j, B= i+2j+4k, c = The ratio in which C divides AB internally is 1 1 : 4 2. 2 : 3 3. 3 : 2 4. 5 : 1 29. Let A =-2i+3j+5k, B =7i-k, If C divides AB in the ratio 1:2 then the position vector of C is 1. 3i+2j+k 2. 2i+3j+k 3. 3i+j+2k 4. i+2j+3k 30. If a+2b ,2a+b be the position vectors of the points A and B, then the position vector of the point C which divides AB internally in the ratio 2:1 is 1. 2. 3. 4. 31. 2i-j+k,i-3j+4 k be the position vectors of two points then the position vector which divides theabove two points in the ratio 2:3 is 1. 2. 3. 4. 32. The point C =(12/5, -1/5, 4/5) divides the line segment AB in the ratio 3:2 If B=(2,-1,2) then A= 1. (3,1,1) 2. (3,1,-1) 3. (3,-1,-1) 4. (-3,1,-1) 33. If b=3i-4j and a are collinear and if amakes an obtuse angle with i and =10 then a = 1. 6i+8j 2. -6i+8j 3. -6i-8j 4. 6i-8j 34. Let a = (1,1,-1), b =(5,-3,-3), c =(3,-1,2) .The vectors which are collinear with c and whose lengths are equal to that of a+b are 1. (6,-2,4),(-6,2,-4) 2. (6,2,4),(-6,2,-4) 3. (-6,-2,-4),(-6,2,-4) 4. (6,-2,4),(-6,2,4) 35 Let A and B be points with position vectors If the point `C' on OA is such that then AD is 1.

2.

3.

4.

with respect to origin O.

is parllel to

and

36. P,Q,R,S have position vectors .Then 1. PQ and RS bisect each other 2. PQ and PR bisect each other 3. PQ and RS trisect each other 4. QSand PR trisect each other

respectively such that

37. The position vectors of the points A,B,C are respectively divides then

in the ratio 3:4 and Q divides

If P

in the ratio 2:1 both externally

= 1.

2.

3.

4.

38. If are the position vectors of A and B then one of the following points lie on 1.

2.

3. 4. RATIO 39. Let A =(-3,4,-8) ,B=(5,-6,4). Then the ratio in which the XOY plane divides the line segment AB is 1. 2 : 1 2. 1 : 2 3. 3 : 5 4. 2 : 3 40. Let A =(-3,4,-8) ,B=(5,-6,4). Then the coordinates of the point in which the XY- plane or XOY plane divides the line segment AB is 1.(7,-8,0)

2.

3. 4. 41. Let A =i+2j+3k , B= 4i+2j, C=2i+2j+2k. Then the ratio in which C divides AB is 1. 3 : 4 2. 1 : 3 3. 1 : 2 4. 1 : 1 42. If 3a+4b-7c =0 then the ratio in which C(c) divides the join of A(a) and B(b) is 1. 1 : 2 2. 2 : 3 3. 3 : 2 4. 4 : 3

43. If a,b,c,d are the position vectors of the points A ,B,C,D respectively such that 3a+5b-3c-5d =0 then AB intersectsCD in the ratio 1. 2 : 3 2. 3 : 2 3. 3 : 5 4. 5 : 3 DIRECTION COSINES & RATIOS 44. A straight line is inclined to the axes of Y and Z at angles respectively. The inclination of the line with the Xaxis is 1. 2. 3. 4. 45. The direction cosines of i+2j+2k are 1. (1,2,2) 2.

3.

46. A line makes an angle .Then 1. 1

2.2

4. with the coordinate axes

= 3. 3/2

4. 1/3

47. A line makes an angle with the X,Y,Z axes .Then = 1. 1 2. 2 3. 3/2 4. 4 48. The direction cosines of two lines are l1, m1, n1 ; l2, m2, n2 .Then the value of = 1.1 2. 0 3. 4 4. 2 49. If e = li+mj +nk is unit vector ,the maximum value of lm+mn+nl is 1.

2. 0

3. 1

4.

50. If l.m.n are the d.c's of a vector if ,then the maximum value of lmn is 1. ¼ 2. 3/8 3. 1/2 4. 3/16 51. If OA =3i+j-k,|AB| =

and AB has the direction ratios 1,-1,2 then OB =

1. 2. 3. 4. CENTROID 52. Let G and G | be the centroids of the triangles ABC and A| B| C | respectively . Then AA|+BB|+CC| = 1. 2GG| 2.3G|G 3. 3GG| 4. 3/2 GG| 53. Let A= 2i+4j-k ; B = 4i+5j+k .If the centroid G of the triangle ABC is 3i+5j-k then the position vector of C is 1. 3i-6j+3k 2. 3i-6j-3k 3. 3i-6j+2k 4.3i+6j-3k

54. If the position vectors of A,B,C are 3i+j-k,i-2j, 2i-j+3k then the position vector of the centroid of the triangle ABC is 1.2i-2/3j+2/3k 2.2i+2/3j+2/3k 3.3i-2j+2k 4. 2i-2/3j-2/3 55. If G is the centroid of the triangle ABC then GA+GB+GC = 1. AB 2.BC 3.4GA 4.0 56. If and be the vertices of a triangle whose circumcentre is the origin then orthocentre is given by 1)

2)

3)

4)

57. Let A = -i+6j+6k , B = -4i+9j+6k, .If G is the centroid of the triangle ABC is 1. A right angled triangle 2. A right angled isosceles triangle 3. An isosceles triangle 4. An equilateral triangle 58. The triangle ABC is defined by the vertices A = (0,7,10) B = (-1,6,6) and C = (-4,9,6) let D be the foot of the attitude from B to the side AC . then BD is 1) 2) 3) 4) 59. Let A(a), B(b),C(c) be the vertices of the triangle ABC and let D,E,F be the mid points of the sides BC,CA,AB respectively.If P divides the median AD in the ratio 2:1 then the position vector of P is 1.0 2.a+b+c 3. 4. 60. The midpoints D,E,F of the sides BC,CA,AB of vectors 1.

, 2.

Then the centroid of 3.

have position

is the point :

4.

61. If G is the centroid of triangle ABC, then = 1.1/4 2.1/3 3.2/3 4.4/9 TRIANGLE 62. If D is the mid point of the side BC of triangle ABC, then AB+AC=

1. AD 2.1/2AD 3.2AD 4.4AD 63. Let ABC be a triangle and AD,BE,CF be its medians then AD+BE+CF= 1.4AB 2.3BC 3.4CA 4.0 64. Let ABC be a triangle and let D,E,F be the midpoints of the sides BC,CA,AB respectively. Then

=

1. 2. 3. 4. 65. Let ABC be a triangle and let D,E be the midpoints of the sides AB,AC respectively, then BE+DC= 1. BC 2. 3. 4. 66. Let ABC be a triangle and let S be its circumcentre and O be its orthocentre. The SA +SB+SC= 1. 4SO 2. 3SO 3. 2SO 4.SO 67. Let ABC be a triangle and let S be its circumcentre and O be its orthocentre. The OA +OB+OC= 1. O 2. 2SO 3. 2OS 4. 3OS | 68. If O is the circumcentre and O is the orthocentre of a triangle ABC and if AP is the circumdiameter then AO|+O|B+O|C= 1. OA 2.O'A 3.AP 4.AO 69. If AB = -3i+4k and BC = the length of the median AM is

are the sides of the triangle ABC, then

1. 2. 3. 4. 70. If D,E,F the midpoints to the sides of a triangle ABC then area of 1. 71. Let

2.

3.

4.

be the distinct real numbers. The points with position vectors are

1.collinear 2.form an isosceles triangle 3.right angled triangle 4.equilatetral triangle 72. If the vectors

=

form a triangle then = 1.6 2.-6 3.12 4.-1 73. The in-centre of the triangle formed by the points i+j+k , 4i+j+k , 4i+5j+k is 1. 2.i+2j+3k 3.3i+2j+k 4.i+j+k 74. The position vectors of A,B,C are vector of the circumcentre of the triangle ABC is 1.

2.

3.

4.

Then the position

COLLINEAR 75. If and are the vectors whose directions are neither parallel non coincident, then 1)

0,

76. If

and

,

and

0 2]

implies -1,

-1 3]

1,

-1 4]

1,

1

are two non-collinear vectors, then the points are collinear if

1) 2) 3) 4) 77. Let a, b be two non-collinear vectors .If A = (x+4y)a +(2x+y+1)b, B=(y-2x+2)a + (2x-3y-1)b and 3A=2B ,then (x,y)= 1.(1,2) 2.(1,-2) 3.(2,-1) 4.(-2,-1) 78. Let a, b be two non-collinear vectors .If c = (x-2) a+b , d=(2x+1) ab are collinear then x= 1. 1 2. 1/2 3. 1/3 4. 1/4 79. If i+pj+k , 2i+3j+qk are like parallel vectors then (p,q)= 1. 2.(2,2) 3. 4. 80. If the points A(a),B(b),C(c) satify the relation 3a-8b+5c=0 then the points are 1.vertices of an equilateral triangle 2.collinear

3.vertices of a right angled triangle 4.vertices of a isosceles triangle 81. If the vectors 2i+aj+k , -5i+3j+bkare collinear , then (a,b)= 1. (6/5,5/2) 2.(-6/5,-5/2) 3.(-12,-25) 4.(12,25) 82. If the vectors 2i-3j+6k , b are collinear

=14 , then b=

1. 2. 3. 4. 83 If 4i+5j-k , 8i+aj-2k are parallel vectors then a = 1.10 2.5 3.-10 4.-5 84. If a, a are collinear and are in the same direction then is 1.=0 2.>0 3.