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ADDITION OF VECTORS OBJECTIVES 1.

2.

The point having position vectors

(b) Isosceles triangle

(c) Equilateral triangle

(d) Collinear

If

a = i + 2 j + 3 k, b = −i + 2 j + k

(c)

3i + 5 j + 4 k

(b)

3i + 5 j + 4 k

5.

then the unit vector along its resultant is

3i + 5 j + 4 k 50

(b) 3

(c)4

A unit vector a makes an angle (a)

i j k + + 2 2 2

(c)

−

(b)

i j k − + 2 2 2

π 4

AB + AC + AD + AE + AF = λ AD,

then

λ=

(d)6 with z-axis. If

a +i +j

is a unit vector, then a is equal to

i j k + − 2 2 2

(d) None of these

The perimeter of the triangle whose vertices have the position vectors

(a)

(2i + 5 j + 9 k ),

(b) 15 −

15 − 157

(d)

p = µ AD,

157

15 + 157

In a trapezium, the vector If

(i + j + k), (5 i + 3 j − 3 k)

is given by

15 + 157

(c)

7.

c = 3 i + j,

If ABCDEF is a regular hexagon and

and

6.

and

are the vertices of

(d) None of these

5 2

(a)2 4.

3i + 4 j + 2k, 4 i + 2 j + 3 k

(a) Right angled triangle

(a)

3.

2i + 3 j + 4 k,

BC = λ AD.

We will then find that

p = AC + BD

is collinear with

AD,

then

(a)

µ = λ +1

(b) λ = µ + 1

(c)

λ +µ =1

(d)

If OP = 8 and

OP

µ = 2+λ

makes angles

45 o

and

60 o

with OX-axis and OY-axis respectively, then

OP =

(a)

8 ( 2 i + j ± k)

(b)

4 ( 2 i + j ± k)

(c)

1 ( 2 i + j ± k) 4

(d)

1 ( 2 i + j ± k) 8

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www.sakshieducation.com 8.

The position vectors of two points A and B are

i + j−k

and

2i − j + k

respectively. Then

| AB | =

9.

(a) 2

(b) 3

(c) 4

(d) 5

The direction cosines of the resultant of the vectors

(i + j + k), (−i + j + k), (i − j + k)

and

(i + j − k),

are

10.

(a)

 1 1 1   , ,   3 6  2

(b) 

(c)

 1 1 1  − ,− ,−   6 6 6 

(d) 

 6  1  3

,

1

,

1   6 

,

1   3

6 ,

1 3

The position vectors of A and B are magnitude of

11.

 1

AB

2i − 9 j − 4 k

and

6i − 3 j + 8 k

respectively, then the

is

(a) 11

(b) 12

(c) 13

(d) 14

If the position vectors of A and B are

i + 3 j − 7k

and

5 i − 2 j + 4 k,

then the direction cosine of

AB

along y-axis is (a)

4

(b)

162

(c) – 5 12.

5

−

162

(d) 11

The position vectors of the points A, B, C are

(2i + j − k ), (3 i − 2 j + k )

and

(i + 4 j − 3 k )

respectively.

These points (a) Form an isosceles triangle (b) Form a right-angled triangle (c) Are collinear (d) Form a scalene triangle 13.

14.

3 OD + DA + DB + DC =

(a)

OA + OB − OC

(b) OA + OB − BD

(c)

OA + OB + OC

(d) None of these

The vectors

AB = 3 i + 4 k,

and

AC = 5 i − 2 j + 4 k

are the sides of a triangle ABC. The length of the

median through A is (a)

18

(b)

72

(c)

33

(d)

288

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www.sakshieducation.com 15.

The magnitudes of mutually perpendicular forces a, b and c are 2, 10 and 11 respectively. Then the magnitude of its resultant is

16.

(a) 12

(b) 15

(c) 9

(d) None

ABC is an isosceles triangle right angled at A. Forces of magnitude BC, CA

and

AB

(a)

21.

(d)30 is

is collinear with a, then ( λ being some non-zero scalar)

(b)

λb

(c) λc

AC + AF + AB

(b)

(c)

AC + AB − AF

(d) None of these

a = 2i + 5 j

and

b = 2i − j,

i−j

AC + AF − AB

then the unit vector along

(b) i + j

2

In the triangle ABC,

(d)0

AE =

(a)

If

a + 2b

is equal to

λa

(c)

2 (i + j)

AB = a , AC = c , BC = b

(a)

a +b +c =0

(b) a + b − c = 0

(c)

a −b+c =0

(d)

a+b

will be

(d) i + j 2

, then

−a + b + c = 0

If the position vectors of the point A, B, C be i, j, k respectively and P be a point such that AB = CP ,

22.

b + 3c

In a regular hexagon ABCDEF,

(a) 20.

2

If a, b and c be three non-zero vectors, no two of which are collinear. If the vector

a + 2b + 6 c

19.

(c) 11 + 2

(b) 5

collinear with c and

18.

and 6 act along

respectively. The magnitude of their resultant force is

(a) 4 17.

2 2, 5

then the position vector of P is

(a)

−i + j + k

(b)

(c)

i + j−k

(d) None of these

If in the given figure A

P

−i − j + k

OA = a , OB = b

and

AP : PB = m : n,

then

OP =

B

O

(a)

m a + nb m +n

(b)

na + m b m +n

(c) m a − n b

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(d) m a − n b m −n

www.sakshieducation.com 23.

24.

If

a = 3 i − 2 j + k, b = 2 i − 4 j − 3 k

(a)

3i − 4 j

(b)

3i + 4 j

(c)

4i − 4 j

(d)

4i + 4 j

If

A, B, C

∆ABC ,

(a) 0 (c)

c = −i + 2 j + 2 k ,

a +b +c 3

then

GA + GB + GC

(b)

A+B+C

(d)

a +b −c 3

a +b+c

is

is

If a and b are the position vectors of A and B respectively, then the position vector of a point C on AB produced such that

26.

then

are the vertices of a triangle whose position vectors are a, b, c and G is the

centroid of the

25.

and

(a)

3a − b

(b)

3b − a

(c)

3a − 2b

(d)

3 b − 2a

AC = 3 AB

is

If the position vectors of the points A, B, C be

i + j, i − j

and

a i +bj+ck

respectively, then the

points A, B, C are collinear if (a)

a =b =c =1

(b) a = 1, b and

27.

28.

29.

(c)

a=b =c=0

(d)

c = 0, a = 1

c

are arbitrary scalars

and b is arbitrary scalars

In a triangle ABC, if

2 AC = 3 CB,

(a)

5 OC

(b)

(c)

OC

(d) None of these

If

ABCDEF

then

2OA + 3 OB

equals

− OC

is regular hexagon, then

(a) 0

(b)

2 AB

(c) 3 AB

(d)

4 AB

AD + EB + FC =

If O be the circumcentre and O' be the orthocentre of the triangle ABC, then O' A + O' B + O' C =

(a) 30.

(b)

OO'

(c) 2 OO'

2 O' O

(d)0

If ABCD is a parallelogram and the position vectors of A, B, C are 7 i + 7 j + 7 k,

(a)

i + 3 j + 5 k, i + j + k

then the position vector of D will be

7i + 5 j + 3k

(b) 7 i + 9 j + 11 k

(c) 9 i + 11 j + 13 k

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(d)

8i + 8 j + 8k

and

www.sakshieducation.com 31.

32.

33.

34.

If

AO + OB = BO + OC ,

(a) Equilateral triangle

(b) Right angled triangle

(c) Isosceles triangle

(d) Line

If

D, E, F

(a)

DC

are respectively the mid points of (b)

(a)

2 GG ' 3

(b) GG'

(c)

2 GG'

(d)

and

BC

∆ABC

in

, then

BE + AF =

(d) 3 BF

(c) 2 BF

2

A ' B' C '

respectively, then

AA '+ BB ' + CC ' =

3 GG'

If D, E, F be the middle points of the sides BC, CA and AB of the triangle ABC, then is

(a) A zero vector

(b) A unit vector

(c) 0

(d) None of these

A and B are two points. The position vector of A is the ratio 1 : 2. If

36.

1 BF 2

AB, AC

If G and G' be the centroids of the triangles ABC and

AD + BE + CF

35.

then A, B, C form

a −b

6 b − 2a .

A point P divides the line AB in

is the position vector of P, then the position vector of B is given by

(a)

7 a − 15 b

(b) 7 a + 15 b

(c)

15 a − 7 b

(d) 15 a + 7 b

The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12N. The magnitude of the two forces are

37.

(a) 13, 5

(b) 12, 6

(c) 14, 4

(d) 11, 7

If three points A, B, C are collinear, whose position vectors are 11 i + 3 j + 7 k

39.

(b) 2 : 3

The vectors

3i + j−5k

and

(c)2 : 1

a i + b j − 15 k are

(a)

a = 3, b = 1

(b) a = 9, b = 1

(c)

a = 3, b = 3

(d) a = 9, b = 3

(d)1 : 1

collinear, if

If a, b, c are three non-coplanar vectors such that a +b +c +d

(a) 0

and

respectively, then the ratio in which B divides AC is

(a) 1 : 2 38.

i − 2 j − 8 k, 5 i − 2 k

a + b +c =αd

is equal to (b) α a

(c) β b

(d) (α + β ) c

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and

b + c + d = β a,

then

www.sakshieducation.com 40.

41.

42.

43.

If

(x , y, z ) ≠ (0, 0, 0)

and

(i + j + 3 k) x + (3 i − 3 j + k) y +(−4 i + 5 j) z = λ (x i + y j + zk),

(a) – 2, 0

(b) 0, – 2

(c) – 1, 0

(d) 0, – 1

The vectors

i + 2 j + 3 k, λi + 4 j + 7 k, −3 i − 2 j − 5 k

(a) 3

(b) 4

(c) 5

(d) 6

The points with position vectors (a) – 8

(b) 4

(c) 8

(d) 12

are collinear, if λ equals

10 i + 3 j, 12 i − 5 j

and

If three points A, B and C have position vectors if they are collinear, then

a i + 11 j

(b) (– 2, 3)

(c) (2, 3)

(d) (– 2, – 3)

are collinear, if

(1, x , 3), (3, 4 , 7 )

(x , y) =

(a) (2, – 3)

then the value of λ will be

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and

(y, − 2, − 5)

a=

respectively and

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ADDITION OF VECTORS HINTS AND SOLUTIONS 1.

(c)

AB = i + j − 2k, BC = i − 2 j + k , CA = 2i − j − k

Clearly |

AB | =| BC | =| CA | = 6

ˆ = 3i + 5 j + 4 k . R = 3i + 5 j + 4 k ⇒ R 5 2

2.

(c)

3.

(b) By triangle law,

AB = AD − BD, AC = AD − CD

E

D

F

C

A

Therefore, = Hence 4.

(c) Let

B

AB + AC + AD + AE + AF 3 AD + ( AE − BD) + ( AF − CD) = 3 AD

λ=3,

a = li + m j + n k,

where π

a makes an angle ∴n =

1 2

, l2 + m 2 =

∴ a = li +m j+

AE = BD, AF = CD] .

[Since

with

4

….. (i)

k 2

k 2

Its magnitude is 1, hence From (i) and (ii),

5.

(a)

a =−

z − axis.

1 2

a + i + j = (l + 1)i + (m + 1)j +

Hence

l 2 + m 2 + n 2 = 1.

2lm =

i j k − + 2 2 2

(l + 1)2 + (m + 1)2 =

1 2

.....(ii)

1 1 ⇒l=m =− 2 2

.

a = 4 i + 2 j − 4 k ⇒ | a | = 16 + 16 + 4 = 6 b = −3 i + 2 j + 12 k ⇒ | b | = 144 + 4 + 9 = 157 c = −i − 4 j − 8 k ⇒ | c | = 64 + 16 + 1 = 9

Hence perimeter is

15 + 157 .

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www.sakshieducation.com 6.

(a) We have,

p = AC + BD = AC + BC + CD = AC + λ AD + CD

= λ AD + ( AC + CD) = λ AD + AD = (λ + 1)AD.

Therefore

p = µ AD ⇒ µ = λ + 1.

7.

(b) Here is the only vector 4(

8.

(b)

9.

(d) Resultant vector

whose length is 8.

AB = i − 2 j + 2 k ⇒| AB | = 3 . = 2i + 2 j + 2 k.

Direction cosines are 10. (d)

2 i + j ± k) ,

 1 1 1     3 , 3 , 3 .  

AB = (6 − 2)i + (−3 + 9)j + (8 + 4 )k = 4 i + 6 j + 12 k | AB | = 16 + 36 + 144 = 14 .

11. (b)

AB = 4 i − 5 j + 11 k

Direction cosine along

y − axis

=

−5 16 + 25 + 121

=

−5

.

162

12. (c) AB = i − 3 j + 2k BC = −2i + 6 j − 4k CA = i − 3 j − 2k | AB | = 1 + 9 + 4 = 14 | BC | =

4 + 36 + 16 =

56 = 2 14

| CA | = 1 + 9 + 4 = 14 | AB | + | AC | =| BC |

Hence A, B, C are collinear. 13. (c)

3 OD + DA + DB + DC = OD + DA + OD + DB + OD + DC = OA + OB + OC .

14. (c) P.V. of

AD =

B

(3 + 5 )i + (0 − 2) j + (4 + 4 )k = 4i − j + 4k 2 A

D

C

| AD | = 16 + 16 + 1 = 33

.

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www.sakshieducation.com 15. (b)

R = 4 + 100 + 121 = 15 . C

2 2

5

A

16. (b)

B

6

R cos θ = 6 cos 0 ° + 2 2 cos(180 o − B) + 5 cos 270 o

ABC is a right angled isosceles triangle i.e.,

∠B = ∠C = 45 °

∴ R 2 = 61 + 8 (1) − 24 2 ⋅

1

− 20 2 ⋅

2

1

= 25

2

∴R =5 . 17. (d) Let So,

a + 2b = xc

then

1 y=− . 2

and

a

c

and

a + 2 b + 6 c = (1 + 2 y)a

are non-zero and non-collinear, we have

in either case, we have a + 2 b + 6 c

=0

x +6 =0

and

1 + 2y = 0

.

AE = AC + CD + DE E

D

F

C

A

= AC + AF − AB

{∵ CD = AF and DE = − AB }. B

,

19. (d) a + b = 4 i + 4 j, therefore unit vector 20. (b)

a + 2 b + 6 c = ( x + 6)c

(x + 6)c = (1 + 2 y)a

Since

18. (b)

b + 3c = y a ,

and

4 (i + j) 32

=

i+j

.

2

AB + BC + CA = 0 ⇒ a + b − c = 0.

21. (a) Let the position vector of P is

x i + y j + z k,

then

By comparing the coefficients of

i, j

and

Hence required position vector is

−i + j + k .

AB = CP ⇒ j − i = x i + y j + (z − 1)k

k , we

get

x = −1,

22. (b) Concept

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y = 1 and z – 1 = 0 ⇒ z = 1

i.e.,

x = −6

and

www.sakshieducation.com 23. (c)

a + b + c = (3 + 2 − 1)i + (−2 − 4 + 2)j + (1 − 3 + 2)k = 4 i − 4 j .

24. (a) Position vectors of vertices A, B and C of the triangle ABC = a, b and c. We know that position vector of centroid of the triangle (G) = a + b + c . 3

Therefore, GA + GB + GC =0 25. (d) Since given that

AC = 3 AB.

it means that point

C

divides

AB

externally. Thus

AC : BC = 3 : 2

A

a

b

O

Hence 26. (d) Here

B

3 .b − 2 .a = 3 b − 2a . 3−2

OC =

AB = −2j, BC = (a − 1)i + (b + 1)j + ck

The points are collinear, then

AB = k (BC )

−2 j = k {(a − 1)i + (b + 1) j + ck}

On comparing, Hence 27. (a)

k (a − 1) = 0 , k (b + 1) = −2, kc = 0 .

c = 0, a = 1

and b is arbitrary scalar.

2OA + 3 OB = 2(OC + CA ) + 3 (OC + CB ) = 5 OC + 2CA + 3 CB = 5 OC

,

{∵ 2 CA = −3 CB } .

28. (d) A regular hexagon ABCDEF. E

D

F

C B

A

We know from the hexagon that EB = 2 FA ,

Thus

and

FC

is parallel to

AD

or

AB

is parallel to FC = 2 AB

BC

AD = 2 BC ; EB

or

.

AD + EB + FC = 2 BC + 2 FA + 2 AB

= 2(FA + AB + BC ) = 2(FC) = 2(2 AB ) = 4 AB

. A

29. (b)

O ′A = O ′O + OA O ′B = O ′O + OB

O′

O ′C = O′O + OC

O B

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C

is parallel to

FA

or

www.sakshieducation.com ⇒ O ′A + O ′B + O ′C = 3 O ′O + OA + OB + OC

Since

OA + OB + OC = OO ′ = −O ′O

∴ O ′A + O ′B + O ′C = 2 O′O

.

30. (b) Let position vector of D is

x i + y j + z k,

then

AB = DC ⇒ −2 j − 4 k = (7 − x )i + (7 − y)j + (7 − z )k

⇒ x = 7, y = 9, z = 11 .

Hence position vector of 31. (c)

AB = BC

32. (a)

BE + AF = OE − OB + OF − OA

will be 7 i + 9 j + 11 k .

D

(As given). Hence it is an isosceles triangle.

C

E

F

A

=

B

D

OA + OC OB + OC − OB + − OA 2 2

= OC −

OA + OB = OC − OD = DC . 2

33. (d) GA + GB + GC = 0 and

G ′A ′ + G ′B ′ + G ′C ′ = 0

⇒ (GA − G ′A ′) + (GB − G ′B ′) + (GC − G ′C ′) = 0 ⇒ (GA + G ′G − G ′A ′) + (GB + G ′G − G ′B ′) + (GC + G ′G − G ′C ′) = 3 G ′G

⇒ (GA − GA′) + (GB − GB′) + (GC − GC′) = 3 G′G ⇒ A ′A + B ′B + C ′C = 3 G ′G ⇒ A A ′ + B B ′ + C C ′ = 3 GG ′ .

AD = OD − OA =

34. (a)

b +c b + c − 2a −a = 2 2

,

(Where Similarly,

BE = OE − OB =

O

is the origin for reference)

c +a c + a − 2b −b = 2 2

and

CF =

a + b − 2c 2

35. (a) Standard problem. 36. (a)

P + Q = 18 , R = 12, θ = 90 o ,

(say)

tan θ = tan 90 o = ∞

⇒ P + Q cos α = 0 , ∴ cos α =

−P Q

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.

www.sakshieducation.com Also,

(12) 2 = P 2 + Q 2 + 2 PQ cos α

Or

144 = P 2 + Q 2 + (2 P )(− P)

⇒ 144 = Q 2 − P 2 = (Q + P)(Q − P)

Or

144 = 18 (Q − P)

Q−P =8

or

After solving Q = 13, P = 5. 37. (b) Let the B divide AC in ratio 5i − 2k =

then

λ(11 i + 3 j + 7 k ) + i − 2 j − 8 k λ +1

⇒ 3λ − 2 = 0 ⇒ λ =

38. (d)

λ :1 ,

2 3

i.e., ratio = 2 : 3.

3 1 −5 = = ⇒ a = 9, b = 3 . a b − 15

39. (a) We have

a +b +c =α d

∴ a + b + c + d = (α + 1)d

and

and

b +c +d = β a

a + b + c + d = (β + 1) a .

⇒ (α + 1) d = (β + 1) a

If

α ≠ − 1,

then

(α + 1) d = (β + 1) a ⇒ d =

β +1 a α +1

 β +1  ⇒ a + b + c = α d ⇒ a + b + c = α a α +1 

α (β + 1)   ⇒ 1 − a + b + c = 0 α +1   ⇒ a , b, c

are coplanar which is contradiction to the given condition,

a + b + c + d = 0.

40. (d) From given equation (1 − λ )x + 3 y − 4 z = 0 x − (λ + 3)y + 5 z = 0

3 x + y − λz = 0

⇒

(1 − λ ) 3 −4 1 − (λ + 3) 5 = 0 ⇒ λ = 0, − 1 . 3 1 −λ

1

41. (a)

2 3 λ 4 7 = 0 ⇒ λ = 3. −3 −2 −5

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∴ α = −1

and so

www.sakshieducation.com 42. (c) If given points be Also,

A, B, C

then

AB = k (BC )

or

2 i − 8 j = k [(a − 12 )i + 16 j] ⇒ k =

2 = k (a − 12 ) ⇒ a = 8 .

43. (a) If A, B, C are collinear. Then

AB = λ BC

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−1 2