3 vectors - _solutions_fina

VECTORS IIT MATHS

23 I I T AKASH MULTIME

Daily Work Sheet-IProperties of Vectors

Solutions

01. Since 14,OP OQ OPQ

is

isoseceles.

Hence the internal bisector OM isperpendicular to PQ and M is the midpointof P and Q .

1 ˆˆ ˆ2 22

OM OP OQ i j k

02. 2 2 2 2 2 2 0a b c a b c ab bc ca

0ab bc ca any two of ,a b andc are zero

03. Let the origin of reference be the circumcentreof the triangle.

Let ,OA a ,OB b

OC c and OT t

Then a b c R

(circunradius)

Again 2OA OB OC OA OD

OA AH OH

Therefore, the P.V. of H is .a b c Since D

is the midpoint of HT, we have

2 2a b c t b c t a

2 2 2 2 .AT a AT a a R

But BC-2R sin A=R,

ATT = 2BC

04. We have DA a, AB b

and CB ka

Given, X and Y are the mid points of DB and ACthen

OB OD OA OCOX , OY2 2

OA OC OB OD DA BC a kaXY OY OX2 2 2

1 k aXY

2

1 kXY a 4

2

(given)

1 k 17 42

(taking +ve sign)

8 91 k k17 17

and 1 k a 4

2

25k 1 17 8 k17

05. 7i 4j 4k 2i j 2kc

9 3

c i 7j 2k9

(i)

2 22

c 1 49 4 x 5481 81

2 x 5454 981

Putting 9 in Eq. (i),

Then, c i 7j 2k

06. Since, a

and b

are equally inclined to c

, therefore

c

must be of the form

a bta b

Now, b a a b a ba b

a b a b a b a b

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VECTORS IIT MATHS


Also

b a a b a ba b

2 a b 2 a b 2 a b a b

Other two vectors cannot be written in the form

a bta b

07. Since, vectors x, x 1, x 2 , x 3, x 4, x 5

and x 6, x 7, x 8 are coplanar..

x x 1 x 2x 3 x 4 x 5 0x 6 x 7 x 8

Applying 2 2 1C C C and 3 3 1C C C

x 1 2x 3 1 2 0x 6 1 2

2 30 0 C C

x R 0 8. Let O be the origin

OA i j , OB i j

, OC pi qj rk

AB OB OA 2j

and AC OC OA p 1 i q 1 j rk

A, B, C are collinear

AC AB

p 1 i q 1 j rk 2j

On comparing p – 1 = 0, q – 1 = – 2l and r = 0 p = 1, r = 0 and q = 1 - 2l

For 1, q 22

and for 0 , q = 1

p = q = 1, r = 0And p = 1, q = 2, r = 0

09. Let a,b,c,p,q,r be the position vectors A,B,C,P,Q,R

Now p= OP OQOA OB

a q b

OROC

r c but a+b=c

0OA OB OCOP OQ OR

p q r

But P,Q,R are collinear

0O A O B O CO P O Q O R

or

OA OB OCOP OQ OR

10. Let OA a,OB b

and OC c

, then

AB OB OA b a

and

1 1 1OP a, OQ b, OR c3 2 3

and OS b a

P, Q, R and S are coplanar, then

PS PQ PR

(PS

can be written as a linear

combination of PQ

and PR )

OS OP OQ OP OR OP

OS OQ OR 1 OP

b c ab a 12 3 3

On comparing, 1

,3 2

3 1 2 1

11. The point which divides 5i and 5j in the ratio k : 1

is 5j k 5i .1

k 1

5i 5k jbk 1

Also b 37

21 25 25k 37k 1

25 1 k 37 k 1

2225 1 k 37 k 1

212k 74k 12 0 26k 37k 6 0

6k 1 k 6 0

k , 6 or 1k ,6

1k , 6 ,6

(b), (d) hold.12. sol:

2 3 4a b c a b c b c a c a b

3 , 4




VECTORS IIT MATHS


17 / 2 12

3, 3, 4 .

13. Sol:

ˆˆ ˆ 3a i xj k , ˆˆ ˆ4 4 2 2b i x j k ,

2b a , 2x , 2

3

14. Sol:

Let 0,0A 3,0B ,C h k

3 3 3,2 2

C

or 3 3 3,2 2

CB AB AC

= 3 3 3ˆ ˆ32 2

i i j

= 3 3 3ˆ2 2

i j

similan 3 3 3 ˆ2 2

CD i j

15. The given vectors are collinear if a = kb for some k,i.e.

xi – 2j + 5k = ki + kyj – zkk Þ (x – k)i + (-2 – ky)j + (5+ zk)k = 0

Hence x = k, y = -2/k and z = -5/k. Taking k = 1, 1/2, -1/2 and – 1, we get the values given by (a), (b), (c)

and (d), respectively.

16. Equation r(t) = R(u) we obtain

i j k1+ t - 2u + -10 + 2t - u + 1+ t - 2u = 0

Thus 1 + t – 2u = 0; - 10 + 2t – u = 0Solving, we obtain t = 7, u = 4, sor(7) = 8i + 8j + 9k = R(4)The two lines intersect at the tip of this vector.

17. 18. 19.1. The vector along the bisector of the givenangle is

b c 7i 4j 4k 2i j 2k i 7j 2kd981 9b c

(i)

54 3 6d 5 6 .81 9

15

Then, from Eq. (i), 5d i 7j 2k3

2. Let A be the initial point and the PV of B and C be

b

and c

respectively. Hence, let AB = c and AC = b.

The internal bisector is ( b cr tc b

and the

equation of the line BC is r b c b

) and

hence the position vector of D is bb cc

b c

.

c c bbb ccBD bb c b c

c caBD BD c b

b c b c

Similarly, c caBE BE c b

c b c b

1 1 c b b c 2c 2 2 2 1 1BE BD ca ca ca a BC BC BE BD

3. AB b a

, BC c b

, CA a c

thebisectors of the angular B and C are

b a c br bc a

(i)

And b c c ar c

a b

(ii)

So that for the point P

b a c b b c c ab cc a a b




VECTORS IIT MATHS


c b

and 1 c a a

Then we get, ac

b c a

From Eq. (i),

b b c a ab aa cc cbac b a c ar b

b c a c b b c a

bb cc aa aa bb ccb c a a b c

20. 21. 22.

1. O'B O'C 2O'D 2 O' A AO OD

2O'A 2AO AO' AO' 2OD

2AO' 2AO AO'

AO' AO'

AO' O'B O'C 2AO

2.

NA NB NC NG3

NA NB NC 3NG

O'O O'O 1NO GO 3 O'O2 3 2

3. SA SB SC SG

3

SA SB SC 3SG

23. p AB AO OB a OB thus positionvector of B is p + a

q BC BO OC p a OC OC q p a .So position vector of E is

a q p a a p q 1 1 22 2OD OA AD OA BC a q Therefore position vector of mid point of CD is

/ a q a p q / a p q 1 2 1 2 2 2

24. Let the position vector of A, B, C, D, E, F be a, b, c, d, e, f. So

b c c a a bd ,e ,f

2 2 2

b c c a a bOD OE OF d e f a b c

2 2 2

AD BF CF d a e b f c d e f a b c O

OE OF DO e f d a OA

AD BE CF d a e b f c 2 1 2 13 3 3 3

25. Let O be the origin.

A a , 5B , C c are coplanar

0xa yb zc

yb zc xa

5y zc x ay z y z

LHS represent a point on BC

RHS represent a poin on OA

xOD ay z

x y zAD a

y z

1OD x ODAD x y z AD

26. ABCD is a tetrahedraon let , , ,a b c d be the

position vectors of , , ,A B C D since every system of 4 vectors is linearly dependens threexistsscalars , , ,x y z t such it at

0xa ya zc td

yb zc td x ay z t y z t

LHS is a point which lies in BCD planeRHS is a point which lies on OA It represents point P

x y z txaOP AP ay z t y z t

OP xx y z tAP

1OPAp




VECTORS IIT MATHS


27. Let C be the origin

CA a CB b25

CD b

Eq of AD os 25

r a b a

Eq of BE is 14

r b a b

Equations

we get 56,x 2

3

26

a bCH

Sinilarly 2

3a bCF

21

AFFB

2k

28. 3A 4B 5C

. .A B C B . .B C A C

. 0.C A B

2 2 2 2A B C A B C

5 2A B C

5k

29. ai j k x i b j k y i j ck

a x y 1 bx y

1 x cy

1 1 1 1 1 111 1 1 1 11 1

ya b c x y xx y

1

30. Let ,A a B b C c bethe vertices

4 3 25 5

c b a cOP OQ ,

3 7 310 4

b a b aOR OS

3 7 10 3 410 4

b a c b a cCR CS ,

3 410

a c bAP BQ CR

4 5 3 2 55 5

C b A a c bAP BQ ,

4 210 5

AP BQ CR

CS

,

Daily Work Sheet-II01. Sol:

.b aAMa

2

.b aAM

a

DM AM AD =

2

.a b a b

a

02. Sol: 2

0a b a

22 2 2 . . . 0a b c a b b c c a

22 2 3a b c

2 2 22 2 2 3a b b c c a a b c

2 2 2a b b c c a

= 22 22 . . . 9a b c a b b c c a

03. Sol: ˆ ˆ4 3b i j ˆ ˆ3 4c t i j

Let ˆ ˆr xi yj




VECTORS IIT MATHS


. 1r bb

2r cc

4 3 1x 3 4 1x y

2 11ˆ ˆ ˆ ˆ2 ,5 5

r i j r i j

04. Sol A vector coplaner with ,b c is

r b c

ˆˆ ˆ1 2 1 2r i j k

Projection of r on r aa

a

2 1 2 1 1 2 1 236

2 2 1 2 2

3 1 2

3 1 2 , 3 1 2 , 1 , 13

05. . 3OA OB Let , 0a b

2 . 3 cos 2 , 3 3a b a b ,

1cos2

, / 3a b

. 2 2 . 3 2CA CB a a b b a b

= 26 . 4a b a b =

6 342 4

= 0

090c

ortho centre = p v of c = 2 a b

06. a 1,3,sin2

makes an obtuse angle with thez-axis, therefore its z-component is negative.

sin2 0

1 sin2 0 (i)

But b.c 0

( orthogonal)2tan tan 6 0

tan 3 tan 2 0

tan 3, 2

Now, tan 3

22tan 6 3sin2 0

1 9 51 tan

Impossible sin2 0 Now, if tan 2

22tan 4 4sin2 0

1 4 51 tan

which is true.tan2 0

2 is in third quadrant also sin / 2 is

meaningful, if 0 sin / 2 1 , then

14n 1 tan 2

and 14n 2 tan 2

07.

Area of 12

ABC AB AC

Let DE n be the altitude.

Volume = 1 2 2.3 3

n area of ABC

2n

Let E divide

A 1,1,1 and 2,0,0M in the ratio :1




VECTORS IIT MATHS


: :1AE EM

ˆˆ ˆ ˆ21

i i j kOE

22 2 22

31

AE AD DE

2

23 12 2, 2 / 3

1

1,3,3 3, 1, 1OE or .

08. DB DA AB

or DA DB AB

2 2 2DA DB AB 2DB.AB

(i)

In parallelogram, 2 2 2 22 a b c DB

2 2 2 2DB 2a 2b c

From Eq. (i),

2 2 2 2 2b 2a 2b c a 2AB.DB

2 2 23a b cAB.DB

2

09. Let A be the origin of reference and the positionvector of B, C, D be b, c, d, w.r.t A. So AB = b,

CD = d – c, AD = d, BC = c – b, AC = c and BD= d – b. The L.H.S. is equal to b. (d – c). The R. H. S. is

K d c b c d b 2 2 2 2

= K d.d c.c b.b c.b c.c d.d b.b d.b 2 2

K b. d c 2 . Hence K = 1/2.

10. Putting x = 1 in 2y x x 10 , we get y = 12

A is (1, 12)

dy dy2x 1dx dx

at A (1, 12) = 2(1) + 1 = 3

Tangent at A is y – 12 = 3 (x – 1)It meets the x-axis (where y = 0)

12 3 x 1 x 4 1 3

B is (-3, 0)

OA i 12j

, OB 3i

AB OB OA 4i 12j

OA.AB 4 144 148

11. 1p 3i 4j,q 5i,r p q 2i j4

and 1s p q i 2j2

Alternate (a)

2 2 2r ks 2i j ki 2kj 2 ki 1 2k j 2 k 1 2k 5k 5

and

2 2 2r ks 2i j ki 2kj 2 ki 1 2k j 2 k 1 2k 5 5k

Hence, r ks

= r ks

Alternate (b)

r.s 2i j . i 2j 2 2 0

r is perpendicular to s

Alternate (c)

2 2r s . r s r s 5 5 0

r s

is perpendicular to r s

.Alternate (d)

p 5, q 5, r 5

and s 5

p q r s

12. u v w 1

u v w a

3 7a.u a.v a.w a.a a.w 42 4

3a.w4

2 2u v a w

2 2 2 2u v 2u.v a w 2a.w

31 1 2u.v 4 12

3a.w

4

3u.v4

and 2 2v w a u

2 2 2 2v w 2v.w a u 2a.u

1 1 2v.w 4 1 3 3a.u

2

v.w 0

Also, 2 2w u a v

2 2 2 2w u 2w.u a v 2a.v




VECTORS IIT MATHS


71 1 2w.u 4 12

7a.v

4

7 12w.u 32 2

1w.u4

13. Let c = (c1, c2, c3). Then |c| = |a| = |b| =

2 2 21 2 32 = c + c + c . It is given that the angles

between the vectors are identical, andequal to f (say). Then

a.ba b

0 1 0 1cos22 2 ,

a.ca c

1 2c + c 122 2 and

b.cb c

2 3c + c 12 2 .

Hence c1 + c2 = 1 and c2 + c3 = 1. That is, c1 = 1 – c2and c3 = 1 – c2.

2 22 2 2 2 21 2 3 2 2 2 2 22=c +c +c = 1-c +c + 1-c =3c +2-4c

Therefore, c2 = 0 or c2 = 4/3. If c2 = 0, then c1 = 1 andc3 = 1, and if c2 = 4/3, then c1 = -1/3, c3 = -1/3. Hencethe coordinates of c are (1, 0, 1) or (-1/3, 4/3, -1/3).

14. Sol: Normal to the face OAB

1ˆˆ ˆ5 3m OA OB i j k

2m normal to the face

ˆˆ ˆ5 3ABC AB AC i j k

Angle below the faces OAB, ABC =

Angle between 11 2

19, cos35

m m =

1 12 619

Tan .

15. Equation of plane containing the line r a kb

and

perpendicular r.n q

b

and n are parallel

vector to n and b

is n x b

Hence, required plane is r a . n x b 0

16. Given, a 1

, c 1

and b 4

and a x b 2a x c

Since, angle between a

and c

is 1cos 1/ 4

1 1 1a.c a c 1.1.4 4 4

and

b 2c a

or b 2c a

Squaring 2 2 2 2b 4c a 4 c.a

216 4 2 12 0

4 3 0

4,3 17. 18. 19.

Equation of the face ABC is

r. b x c c x a a x b a b c and that ofedge OA is r = ta. If q denotes theangle between the face and the edge, then

b c c a a b .ab c c a a . a

x x x

sinx x x b =

a b cb c c a a . a x x x b

Now

6 6

0 0

2 0 0

0 0

1 cos60 cos60 1 12 12k cos60 1 cos60 k 12 1 12

12 12 1cos60 cos60 1

a.a a.b a.c / /

abc b.a b.b b.c / /c.a c.b c.c / /

/ / / / 6 6k 3 4 1 8 1 8 1 2 k

Also b c c a a b x x x is twice the area of thetriangle ABC which is equilateral with each side k so

that this is 23 k2

.

Hence 3

2

k / 2 2sinθ = =63 / 2 k .k

1cosθ =3

We have a b c k

a.b b.c c.a / 2 0 2k cos60 k 2 . (ABCis an equilateral triangle)

Equation of OAB and OBC are

r. a b , r. b c x 0 x 0 . The angle betweentwo planes is equal to the angle between the normals.If q is the angle then

a b b c.a b b c

x xcosx x =

a b . b cx x =a.b a.cb.b b.c

41 k4

a b a b b c 0 23x sin60 k x2

\ cosq = 1/2. Thus the angle between any two plane

faces is -1cos 1/ 2 .




VECTORS IIT MATHS


20. 21. 22.

a.u b.u c.u

a b c u v w a.v b.v c.v

a.w b.w c.w

2a.a b.a c.a

a b c a b c a b c a.b b.b c.b

a.c b.c c.c

Now, 22a.a a a 16

1a.b b.a a b cos / 3 4.4. 82

1a.c c.a a c cos / 3 4.4. 82

22b.b b b 16

1b.c c.b b c cos / 3 4.4. 82

2 2c.c c 4 16

From Eq. (i),

3 316 8 8 2 1 1

a b c 8 16 8 8 1 2 1 8 .4 64 x 328 8 16 1 1 2

1. Volume of paralellopiped = (base area) (height)

132 2 2x x4x4x sin h 8 3xh2 3

h 4 2 / 3

2. Volume of the tetrahedron = 1 a b c6

=

1 x32 26 =

16 23

3. Volume of the triangular prism = 1 a b c2

=

1 x 32 2 16 22

23. If a b 1 then a b . a b 1So

a b a.b a.b / 2 2 2 1 1 2

/ / cos 1 2 2 3

If a b a b then a.b / 0 2

If a b 2 then cosq < 0 which is true if

/ 2

If a b 2 then cosq > 0 which is true if

/ 0 224. A plane containing r = a + tb passes through a and

vector r – a and b are collinear. Hence the equationof plane passing through c and containing line r =a + tb has r – a, c – a and b as coplanar vector asthus the required equation is r = a + t (c – a) + pb.

Again for the plane containing r = a + tb andperpendicular to r . c = a, the vectors r – a, b and care coplanar so we have r = a + tb + pc.

For (iii) the vectors r – a, b – a, c – a being coplanar,the equation of the plane is

For (iv) again r – a, b – a and c are coplanar, so therequired equation is

r a b a c a b c t p 1 t t p .

25. 221 .a b a b a b

21 . 2 .a b a b

2 2 25 2 .a b a a b

2 2 2 21 a b a b

4

26. 1 2 3. 0 0a b x x x

1 2 3, ,x x x No. of ordered Triples

0,0,0 12,1,1 32,0,2 61,0,1 8

2, 1, 1 33,1,2 6

25 5

27.2 2 2 2

d a c b c a d b

.k b a d c

2k 28. P Area of quadrilateral OABC

1 62

OB AC a b

q Area of parallelogra with ,a b as adja




VECTORS IIT MATHS


cent sides a b

6P q 29.

30. ˆ ˆˆ ˆ ˆ ˆ î j j k i j k so that unit vector

perpendicular to the plane of ˆ î j and ˆj k is

1 ˆˆ ˆ .3

i j k

Similarly, the other two unit vectors are

1 ˆˆ ˆ3

i j k and 1 ˆˆ ˆ3

i j k

The requiered volume

1 1 11 41 1 1

3 3 3 31 1 1

Daily Work Sheet-III01. Sol: 1 2 3c a b a b

Taking dot product with a

.a c

Taking dot prouce with and addtion

1 2 3.a b a b c

.

02. Sol: 12

a b c b

1. .2

a c b a b c b

since b

and c are not collinear 1.2

a c and

. 0a b

0, 60a c

, 0, 90a b

03. Sol: Let 1ˆˆ ˆ ˆm i i j k

Let 2ˆ ˆˆ ˆ ˆ ˆ ˆm i j j k i j k

Let 1 2 3ˆˆ â a i a j a k is r to 1 2,m m

1 2a m m

Let be the angle below a and ˆˆ ˆ2 2i j k

Then ˆˆ ˆ ˆ ˆ. 2 2

cos ˆˆ ˆ ˆ ˆ2 2

i j i j k

i j i j k

= 12

045 or 0135 .

04. Sol:

2. .a a a b ab a aa b a b

2a a a a b a a a b

2 . .a a b a a a b

= 2 2a a b

= 4a b

05. Sol:

sin 12 11 1 1

pp

2 2 2 sin 0p p descriminant , 0

sin ,12

21 0p 1p

06. Sol:

. 0d a b c 0a b c x y z

0x y z

3 3 3 3x y z xyz

and




VECTORS IIT MATHS


2 2 2 2102

x y z xy yz zx x y z

0xy yz zx .

07. Sol:

2 3 41 2 51 2

1 , 9 / 2

08. Sol:

. . 1b c

p a aa b c

,

. . 1c aq b bb c a

,

. . 1a bc ca b c

09. Sol: .a b c a c b b c a

Given vector is r to both b c and a

10. 2a b b a a b b a

2 2a b a b b a a b b a a b

b a

2 . 1a b b a b a

either 0b a or 1 . 1a b

either b a or . 0a b

either or / 2

11. Sol: 3 3 33 0

a b cb c a abc a b cc a b

0a b c

. . . . 0v v v v

12. Sol: . cosu v

w w u v

. 0 .w u v u . cosu w

. . .w v w u v v v

. 1w v

. 1v w

22

. . .

. . .. . .

u v v v u wu v w v u v v v w

w u w v w w

=

1 cos coscos 1 1cos 1 1

2 21 cos = 2sin

= 1 sin2

13. 1i

j

pp

1k

j

pp

= 0,2, 2i k

j j

p pp p

14. Sol: a b

a a a b

3 .a a a b

a b

sinb b ,

2 2 2sin3

, 2 2sin3

2 1cos3

3 cosa a a b




VECTORS IIT MATHS


3 a a b , 13

a b a

15. 1 2V V

a b c a b c

a c b a b c a c b b c a

a b c b c a

either c and a are collinear or b

is

perpendicular to both a and c a c

16. Given, 1 1 1 ˆˆ ˆ6 3 3

i j k a b c d

a b d c a b c d

a b d c

[ ,a b and c are coplanar]

.a b d a b d

a b d

cos

, ||d a d b d a b

= 0sin 30 .1. 1 0 orab

1 11.1. .1 12 2

From

(i), 1 2 2 ˆˆ ˆ3 3 3

c i j k

ˆˆ ˆ2 23

i j k

17. 18. 19.

Since by definition of reciprocal systems

b.a ' c.a ' 0 , a' is perpendicular to both b and

c, and hence parallel to b x c. so a ' t b c x .

Also a.a ' 1, so a.a ' ta. b c 1 x

t

a. b c a b c

1 1x

Thus b ca'a b c

x

. Similarly c ab'a b c

x

and

a bc'a b c

x

a b c b c a c a ba a' b b' c c'

abc

x x x x x x

x x x

a b c a.c b a.b c,b c a b.a c b.c a x x x x

and c a b b.c a a.c b x x

a b c b c a c a b x x x x x x 0

So a a' b b' c c' x x x 0

Next a' b' c ' a ' b ' .c ' x

= b c c a . a b

a b c 3

1 x x x x

= e.a c e.c a . a b

a b c 3

1 x

e b c x

= b c .a c . a b

a b c 3

1 x x

= b c .a . c. a b

a b c 3

1 x x

= a. b c . a b .c

a b c 3

1 x x

= a b c1

We can also show that

b' c ' c ' a ' a ' b 'a ,b ,ca ' b ' c ' a ' b ' c ' a ' b ' c '

x x x

20. 21. 22.

1. O'B O'C 2O'D 2 O' A AO OD

2O'A 2AO AO' AO' 2OD

2AO' 2AO AO'

AO' AO'




VECTORS IIT MATHS


AO' O'B O'C 2AO

2.

NA NB NC NG3

NA NB NC 3NG

O'O O'O 1NO GO 3 O'O2 3 2

3. SA SB SC SG

3

SA SB SC 3SG

23. (A)

a b . b c x c a a b, b c, c a 2 a b c

A – S

(B) a b, b c, c a 0

B – Q

(C) i x a x i j x a x j k x a x k 2a

C – P(D)

a b c . a x b x a c a b c . a x c b x a b x c

a x a 0

a.b x c b.a x c c.b x a a.b x c a.b x c b.c x a

a b c

D – R (a) holds

24. (A) a x a ' b x b ' c x c '

= a x b x c b x c x a c x a x b

0a b c

A – S

(B) a' b' c ' a b c 1

B – R(C) a.a ' b.b' c.c ' 1 1 1 3

C – P(D)

i j k1 1 2

3 2i k1 2 23b x c3a ' 3 2i k2 3 1 4 1a b c1 1 21 2 2

D = Q (d) holds

25. .c c a c . . .c c c a c c

. 1c a , -----1 and `

. . .a c b a b c a b b

24 2 1Sin b C

Since ,b c are non collinear

. . 4 2 sina c a b -----22 1 .a b -------3

from 1,2,321 1 4 2 sin

21 1 sin 0

1,sin 1

26. Eq of plane OQR is 2 2 0x y z r

distance from 3, 2, 1 is equal to 3

27. Let OABCDEF be a parallelpiped with

concurrent edges 1, ,a b c V abc Let

, ,OABC OCDE OEFA be three faceswhose diagonals are

, ,OB OD OF

,OD b c OF a b volume of parallelopiped

2V a b b c c a

2 12V V

28. . 0r a b c

0b c a m c ab n abc

0a b c l m n ,




VECTORS IIT MATHS


l m n o

29. u v 1x sec u v2 2u v

Similarly for vectors y

and z

As

2

x x y y x z z x x x y z

22 2 21 sec sec sec u v v w w u

64 2 2 2

Now,

u v v w w u u v w v w w u

w.v w.w u

x x y y x z z x x

22 2 24 sec sec sec u v w

64 2 2 2

As

u v v w w u 2 u v w

22 2 21 sec sec sec u v w

16 2 2 2

30. x y z a

. .a x y z a a 3.4

a z

1. . . .2

x x y z x a x y x z

3. . . .4

y x y z y a x y y z

1. . . .4

z x y z z a z x y z

. 0y z , 1.4

x z , 3.4

x y

x y z c

. .x z y y z x c

14

y c .

Regular Class WorkExercise-I

01. c b d and p AC BD AD

Hence p c d b d (using

c b d ) or 1

02. Let ,OA a OB b and ,OC c

then

AB b a and

1 1 1, , .3 2 3

OP a OQ b OR c

Since , ,P Q R and S are coplanar, then

PS PQ PR

( PS

can be written as a

linear combination of PQ

and PR )

OQ OP OR OP

i.e., 3 2 3aOS OP b c

13 2 3aOS b c

.......................(i)

Given

OS AB b a

..............................(ii)

From (i) and (ii), 10,

3

and

2

2 1 3 1




VECTORS IIT MATHS


03.

Point P lies on 2 23 3.x y Now from thediagram, according to given conditions,AP AB

or 2 23 0 4x y or

223 4x y

Solving (i) and (ii) we get 0x and 1y

Hence point P has position vector ˆ.j04.

From the diagram, it is obvious that locus iscone concentric with the positive x-axishaving vertex at the origin and the slant heightequal to the magnitude of the vector.

05. The position vector of any point at t is

2 2 ˆˆ ˆ2 4 5 2 6r t i t j t k

ˆˆ ˆ2 4 4 6dr t i j t kdt

22

ˆˆ ˆ4 4 2tt

dr d ri j k anddt dt

16 16 4 4 Hence, the required unit tangent vector at

1 ˆˆ ˆ2 2 2 .3

t is i j k

06. --------07.

2 2 3, 2 3 5, 3

: 3, 1, 2Solving .

08.

As Q tends to B, 'P' becomes midpoint ofAC.

AP PC and 0BQ

0 1 0 1.APPC QC

09. ˆˆ ˆ ˆa i j j k

ˆ ˆˆ ˆ ˆ, 2i k i j k

10. ˆ ˆˆ ˆ ˆ4 3 ; 14 2 5OA i k OB i j k

ˆ ˆˆ ˆ ˆ4 3 14 2 5ˆˆ ;5 15

i k i j ka b

ˆ ˆˆ ˆ ˆ12 9 14 2 515

r i k i j k

ˆˆ ˆ2 2 415

r i j k

2 ˆˆ ˆ 215

r i j k

11. Let A (1, 1, -2), B(0, 5, -5), d.c.s of AB :

1 4 3, ,26 26 26

and

265

r

Req. points =

1 1 1, ,

1 26 4 26 3 261 . ,1 . , 2 .5 5 526 26 26

x lr y mr z nr

12. If ˆ ˆ. . .x y a x a y This equality must

hold for any arbitrary a

13.2

1a b c

22 2 2a b c a b

1 2 3cos 2 cos 2 cos 1b c c a

1 2 3cos cos cos 1

One of 1 2 3, & should be obtuse anlge.




VECTORS IIT MATHS


14. The angle between a and b

is obtuse.Therefore, 0a b

214 8 0x x x 7 2 1 0x x

0 1/ 2x ----------- (i)

The angle between b

and the z axis is

acute and less than / 6 . Therefore,

cos / 6b kb k

/ 6 cos cos/ 6

2

3253

xx

2 24 3 159x x

2 159x

159 or < 159x x ------------------(ii) Clearly, (i) and (ii) cannot hold together.

15. ˆˆ cos ,a b ˆ ˆ cos ,b c ˆ ˆ cosc a

Now ˆˆ ˆ 1a b c

2ˆˆ ˆ 1a b c

22 2ˆˆ ˆa b c ˆ ˆˆ ˆ ˆ ˆ2 1a b b c c a

1 1 1 2 cos cos cos 1

cos cos cos 1

16. ˆ ˆ ˆ ˆ cos6

i j c i j c

3 3ˆ ˆ2 2

i j c

17. 2 3a b a b

2 24 9 6 13a b a b

18.2

163a b c

22 29a b c

1 2 32cos 6cos 6cos 16,

3 / 6,2 / 3

1 2 32cos 6cos 5 6cos

1 2 maxcos 3cos 4

19. Let P.V. of P, A, B and C be , ,p a b and c

respectively, and O ( 0

) be the circumcentre

of equilateral triangle .ABC Then

13

p b a c

Now 2 2 2 2 2 .PA a p a p p a

Similarly, 2 2 2 2 .PB b p p b

and 2 2 2 2 .PC c p p c

22 26. 2 . 23lPA p a b c l

as / 3 0a b c

20. c xa xb y a b

. . cosa c b c (as ,a b

and c are unitvectors)

2. cosa c x a x

Also, 2

. cosb c x b

cosc a b y a b

Now, 2 1 cosc a b y a b

cos a b y a b

2 21 cos a b a b y a b a b

2 22 22 21 cos a b y a b

2 21 2cos y 2

2 1cos2 2

y

2 10 cos2

1 1cos2 2

3,4 4




VECTORS IIT MATHS


21. Let r be the new position. Then

ˆ ˆr k i j

. Also 1 1.2 2

r k

Also, 2 2 2 1 12 1 22 2

ˆ1 ˆ ˆ

2 2kr i j

22. Sol:

Q P S R R Q S P P R S Q

= Q S Q R P S P R R S R P Q S

Q P P S P Q R S R Q

2 Q S S R R Q

= 4 Area of QRS

23. Sol: 2 3 0a b c ,

3 0a b c b 3a b b c

2 0a c b c 3c a b c

6a c b c c a b c

similary 2 a b 3 c a

24. Taking cross product with a on both the sides of a xr = b, we get

ax axr axb a.r a a.a r axb a axb a r 2

a a x br

a

2

Clearly this r satisfy a x r = b and r . a = a

24. a b c 1

Given thatAngle between a

and c

= angle between b

and c

a.c b.ca.c cos b.ca c b c

Since, a.b 0

c a b a x b

a.c a.a a.b a. a x b .1 0 0

i.e. cos

Similarly, b.c cos

cos Again,

22 21 c.c 2 a x b

22 22 2 2 22 a b a.b 2

2 2 21 2 1 2cos cos2

22 2 1 1 cos2

2 2

25. x y a y a x

(i)

x x y b

(ii)

x.a 1

(iii)By (i) and (ii),

x x a x b x x a x x x b x x a b a x x xa a x b

2

2

a axba.a x 1.a a xb a x 1.a a xb x

a

2

2 2

a 1 a a x ba a x by a x a

a a

26. Sol: r a b v a b

2.r b b

, 2.a r b b a

,

2c pr b a

-------------- (i)

22 .1 . . 0 . 0 bcbc p rb bc p bbp

2bcr a

p p

27. Sol: sinv a b v a b

.u a a b b

cosu a b




VECTORS IIT MATHS


22 cosu a b

=

2 2 21 cos 2cos sin

u v

. . . .u b a b a b b b

. 0u b

28. Sol: Let 3p a b

.q b a b a . .a a b a b a

a b a

p q are r to each other

3 a bBCTan ACA a b a

03 60TanA A

030B

29. Sol:

090C , AB P

Let CBA , 090CAB

. . .AB AC BC BA CA CB

=

0cos 90 cos 0p AC BC p

= sin cosAC p BC p

= 2 2

AC BC

= 2 2AB p

30. Sol: 2

1 1a b a b

2-2 2 . 1a a b

11 cos 22

2 1sin4

31. a a a a b

=

a a a b a a a b

= 4a a b

= 4 a a b

= 4 a b a a a b

= ˆ4 4 4 4 16 48b b b b

32. 4 5 9 0a b c Vectors ,a b

and c arecoplanar.

b a and c a are collinear

0b c c a

.




VECTORS IIT MATHS


33. a b a d

a b a c d

a b a d c c d a

a d a b c

34. a b c

a a b a c

2a b a a b a c

2a a cb

a

a b

35. The given equation reduces to2 2 2 1 0a b c x a b c x

0D

36.

b c a d b a c d b d c a

c a b d c b a d c d a b

c b a d a c b d a d b c

0b c a d c a b d a b c d

37. ˆ ˆˆ ˆ ˆ ˆ5 2 2 6 2

a b c a c b a b c

i j k i j k

ˆ ˆˆ ˆ ˆ ˆ1 1 1 1 4 2i j k i j k

1 1, 4 and 1 3 2

23

38.

, 0, 0a b c b d c d

a d b d c d c d

d a d d c d

d d c c d d

2d c

2

d a dc

d

39.

a b r a a b c r a b r c

a b c r a b r c

Similarly,

b c r c b c a r b c r a

and,

c a r b c a b r c a r b

a b r c b c r a c a r b

3 abc r bcr a car b abr c

3 2a b c r a b c r a b c r

40. Given that ,a b and c are non-copalnar..

0a b c

Again 0a b c a c

0a c b a b c a c

0a c

a and c are perpendicular..

a b c a c b a b c




VECTORS IIT MATHS


0a b c c

41. Consider a tetrahedron with verties O(0, 0, 0),A(a, 0, 0), B(0, b, 0), and C(0, 0, c).

Its volume 16

V a b c

Now centroids of the faces OAB, OAC, OBCand ABC are

1 2 3/ 3, /3,0 , / 3,0, / 3 , 0, / 3, / 3G a b G a c G b c

and 4 / 3, / 3, / 3 ,G a b c respectively..

4 1 4 42 3/ 3, / 3, / 3.G G c G G b G G a

Volume of tetrahedron by centroids

1 16 3 3 3 27

a b cV V

27K

42. .a b b c b c c a c a a b

a b c b a b c c a b c a

3 4a b c b c a a b c

43. 1

2

3

V la mb nc

V na lb mc

V ma nb lc

when ,a b and c are

non-coplanar. Therefore,

1 2 3

11 0

1

m na b cV V V n m

m n

But

10 1 0.

1

m na b c n m

m n

Therefore,

1 m n 2 2 21 1 0m m n n

0l m n

Obviously, 2 0lx mx n is satisfied by

1x due to (i).

3 3 3 3l m n lmn

2 2 2 0,l m n l m n lm mn ln

which is true.

44. It is given that ,b and

are coplanar

verctors. Therefore, 0

0a b cb c ac a b

3 3 33 0abc a b c

3 3 3 3 0a b c abc

2 2 2 0a b c a b c ab bc ca

0a b c 2 2 2 0a b c ab bc ca

. . . 0v v v

v is perpendicular to ,

and .

Exercise-II01. 02. 03.

13

BL b

13

AL a b

Let AP AL

and P divides DB in the ratio




VECTORS IIT MATHS


:1

Then 3AP a b

Also 1AP a b

Form (i) and (ii), 13

a b ba

and 13

34

Hence, P divides AL in the ratio 3 : 1 and Pdivides DB in the ratio 3 : 1.

Similarly Q divides DB in the ratio 1 : 3.

Thus 14

DQ DB and 14

PB DB

12

PQ DB i.e. : 1: 2PQ DB .

04. 05. 06.1. The line of intersection of the plane

r. 3i j k 1 and r. i 4j 2k 2

is

common to both the planes. Therefore, it is perpendicular to normals to the two planes i.e.

1n 3i j k and

2n i 4j 2k . Hence, it is parallel to the vector

1 2n x n 3i j k x i 4j 2k 2i 7j 13k

Equation of the plane passing througha i 2j k

and normal to the vector

1 2n n x n

2i 7j 13k is r.a .n 0 r.n a.n 0

r. 2i 7j 13k 0 or r. 2i 7j 13k 1

2. Since, required plane contains the line

r 2i t j k and is perpendicular to the plane

r. i k 3 , then its passes through the point

a 2i and is parallel to the vectors b j k

and

c i k

Hence, it is perpendicular to the vector

ˆ ˆˆ ˆ

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ0

n b c j k i k

j i j k k i i j k

Therefore, the equation of the required plane is

r a .n 0 r.n a.n r. i j k 2

3. The equation of the plane is

r. 2i j k 6

PM is parallel to normal to plane (i), then equationof PM is

P ’

P

r 2i j k 6

M

r 3i 5j 7k t 2i j k 3 2t i 5 t j 7 t k

(ii) Solving Eqs. (i) and (ii), we get

3 2t i 5 t j 7 t k . 2i j k 6

2 3 2t 5 t 7 t 6

6t 18 6 t 2

Then from Eq. (ii), r i 3j 5k

Let the image (or reflection) is a

, then

a 3i 5j 7k i 3j 5k2

a 3i 5j 7k 2i 6j 10k

a 5i j 3k

07. 08. 09.

Given that 2x y z

and they are

inclined at an angle of 600 with each other.0. . . 2. 2 cos 60 1x y y z z x

. .x y z a x z y x y z a y z a

Similarly, y z x b z x b

(ii)

y a z

, x z b

(from (i) and (ii))

(iii) Now, x y c

z b z a c

z a b z b a c




VECTORS IIT MATHS


z a b c b a

(iv)

a b z a b a b c a b b a

2 2

. .a b z a b z a b a b c a b ab b a

(v)

Now, (i) 2 2

2 2 2 2a y z

Similary, (ii) 2

2b

Also (i) and (ii) 2

2a b y x a b

(vi) Also

. . . . 1 1 0a b z y x z y x x z

And 2

. . . 1a b y x z z x

Thus from (v), we have

2 2z a b c b a b a or

1/ 2z a b c b a

1/ 2y a z a b c b a

and

1/ 2x z b a b c a b

Exercise-III01.

,AB a BC b

AC AB BC a a

2AD BC b

(because AD is parallel to BC and twice its length).

2CD AD AC b a b

b a

FA CD a b

DE AB a

EF BC b

2AE AD DE b a

2CE CD DE b a a b a

02. (a) Taking cross product with c on both sides ofgiven equation, we have

r a c d cx x x c c ox

r.c a a.c r d c xTaking cross product with a, we get

a d ca.c a r a d c r a

a.c

x xx x x x

Again taking cross product with a, we have

a d c a d ca r a a a.a r a.r a a

a.c a.c

x x x xx x x x

a d c a d ca.rr a a a aa.a a.c a.a a.c a.a

x x x x

x x

where a.ra.a

(b) Suppose that r is a solution. Expressing r as alinear combination of non-coplanar vectors a, b, a xb, we have r a b a x b x y z Substituting inthe given equation, we get

a b a b b a a b a b x y z x y x z x x

a b a b b a a.a b a.b a b x y z x y x z 0

a.b a a.a b a x b λx-z y+z 1 yz- y 0

a.b , a.a , x - z 0 y + z 1 0 z - y 0(a, b, a x b are non-coplanar)

a.b

a aa

2 22 2 22

1x , y , z

Thus a.br a b a b

a 22

1 x

(c) The given equations can be written

r c x b 0r c and b are collinear so that

r c b r c b Where a is a scalar. Putting this valuer . a 0 , we get

c.ac . a b . aa.b

0

Thus c.ar c ba.b




VECTORS IIT MATHS


(d) As in part (ii), we can write

r a b a b x y z x

Substituting this value in r b ax , we obtain

a b o a b b a x x y z x x

a b a.b b b.b a a x x z

a b a.b b z b a a a 2x x z

, b a.b 2x 0 1 z 0 0

We can verify r satisfy the given equation for anyscalar y. Thus

r a b bb

21 x y We can verify r satisfy

the given equation for any scalar y.03. For (a) any vector coplanar with a and b is

a b i j kThis will be perpendicular to a if

. . 1 1 1 0 3Thus required vector is of the form

i j k i j k 4 2 2 2 2 one such vectoris in (c).

(b) If i j k x y z is the required vector in (ii) then2x – y + z = 0, x + y = 0

Which gives x = 0, y = zThus one such vector is in (d)(c) A vector in (c) is b c i j k x 2 3

A vector in (d) is a c i k x04. (A)

2 2a b a x b a x b . a x b a x b a x b 1

A – p, A – r, A – s.(C) a.a ' 1

C – p, C – r, C – s

(D) a.b' 0

D – q

05. (A) 11

xr.c x a b c

4

1x 4r.c

Similarly 2 3x 4r.a, x 4r.b

1 2 3x x x 4 r.a r.b r.c 4r. a b c

A – s

(B) 2a x a x b a.b a a.a b a b b

a x a x a x a x b a x a x b b b

B – r

(C) Since A.B 0

A B;A.C 0

A C

A B x C

A k B x C k B C sin6

1 = k(1) (1) (1/2) k = 2 C – q

(D) b ca x b x c2

1 1a.c b a.b c b c2 2

1a.c2

and 1a.b2

Since 1 3a.b cos

42

3a b cos cos4

31.1.cos cos4

34

D – p

Exercise-IV01. We have adjacent sides of triangle

3, 4.a b

The length of the diagonal is 5.a b

Since it satisfies the Pythagoras theorem,

a b .

Hence the parallelogram is a rectangle.

02.3 3 92 2 2

p pq

and 2q

Thus, both statement are true and statement2 is correct explanation for statement 1.

03. a b a b are the diagonals of a paral

lelogram whose sides are a and b

.

a b a b

diagonals of the parallelogram have thesame length. The parallelogram is a rectangle.




VECTORS IIT MATHS


a b

04. Any point on the line

ˆ ˆˆ ˆ ˆ ˆ2 3 2 3 6r i j k t i j k is

ˆˆ ˆ1 2 2 3 3 6r t i t j t k

Substituting in the equation of plane

ˆˆ ˆ. 5r i j k 1t Required distance is distance below the

points ˆˆ ˆ2 3i j k and

ˆˆ ˆ3 5 3i j k (i.e) 7

05. x y b

x a x b

x a b

a x a a b

2

a a bx

a

1

06. Let 1 1,P x y 2 2,Q X Y be two points on22xy

'OP i projection of OP on x-axis

1 1x 1 2y

'OQ i projection of OQ on x-axis

2 2x 2 16y

2OP i j 2 16OQ i j

4 3 42

OQ OP i j

4 52

OQ OP




3 vectors - _solutions_fina

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