Publications Pvt. Ltd. Target Chapter 02: Scalars and Vectors

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γγg

1. F→

= 4 i + 3 j − 2 k ,

r→

= 1 i + 1 j + 0 k

→

τ = r F→ →

×

=

ˆ ˆ ˆi j k

1 1 0

4 3 2−

= [ i (−2) − j(−2) + k (3 − 4)]

= −2 i + 2 j − k

2. | a→

| = ( )22 21 2 2+ + − = 3 and

| b→

| = ( )22 22 1 1+ + − = 6

∴ | a→

| ≠ | b→

|

But two unequal vectors may have same

magnitude.

eg.: if P→

= i + j − k and Q→

= i − j + k , then

two vectors are unequal but | P→

| = | Q→

| 3. For the given two forces, magnitude of

resultant is maximum if 2 forces act along

same direction, i.e., max| R | | A B |→ → →

= + and

magnitude of resultant is minimum if 2 forces

act in opposite direction, i.e., min| R | | A B |→ → →

= −

For all other directions,

R = 2 2A B 2ABcos+ + θ where, θ is the

angle between A→

and B→

.

Therefore the magnitude of the resultant

between 3 N and 5 N will be between 8 N and

2 N.

4. A→

= 3 units due east.

∴ −4 A→

= −4 (3 units due east)

= −12 units due east = 12 units due west.

5. The displacement is along the Z direction, i.e.,

s→

= 10 k

Work done W = F s→ →

⋅

W = (−2 i + 15 j + 6 k )⋅10 k = 60 J 6.

R2 = P2 + Q2 + 2PQ cos θ

From this relation, it is clear that

R2 = P2 + Q2, when θ = 90°

R2 > P2 + Q2, when θ < 90°

R2 < P2 + Q2, when θ > 90°

7. 2 21 a b= + ⇒ a2 + b2 = 1 ….(i)

and ( ) ( )ˆ ˆ ˆ ˆai bj 2i j 0+ ⋅ + = ⇒ 2a + b = 0

⇒ b = −2a

Substituting for b in (i)

a2 + (−2a)2 = 1 ⇒ 1

a5

= and 2

b5

= −

8. ˆ ˆr i j→

= −

Torque at that point, ( ) ( )ˆ ˆ ˆr F i j 4F k→ → →

τ = × = − × −

ˆ ˆi k× = − j and ˆ ˆ ˆj k i× =

∴ →

τ = ( ) ( )ˆ ˆ ˆ ˆ4F i k 4F j k− × + ×

= ( ) ( )ˆ ˆ4F j 4F i− − +

= ˆ ˆ4Fi 4F j+

= ( )ˆ ˆ4F i j+

Scalars and Vectors02

R = P Q→ → →

+

D

A B

C

θ

Q→

P→

Publications Pvt. Ltd. Target Std. XI : Triumph Physics

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9. ( )

( ) ( )1/ 2 1/22 2 2 2

ˆ ˆ ˆ ˆi j k jcos

ˆ ˆ ˆ ˆi j k j

+ + ⋅θ =

+ + ×

= ( )1/2

ˆ ˆ ˆ ˆ ˆ ˆi j j j k j 0 1 0 1

3 31 1 1 1

⋅ + ⋅ + ⋅ + += =

+ + ×

( ˆ ˆ ˆ ˆi j k j 0⋅ = ⋅ = and ˆ ˆj j 1⋅ = ) 10. To find the net force, we vectorially add the

three vectors. The x-component is

Fnet x = −F1 − F2 sin 60° + F3 cos 30° = − 3 − 4 sin 60° + 10 cos 30°

= − 3 − 4 × 3

2+ 10 ×

3

2 = 2.196 N

and the y-component is

Fnet y = −F2 cos 60° + F3 sin 30° = − 4 cos 60° + 10 sin 30° = 3 N

The magnitude of net force is

Fnet = 2 2

net x net yF F+ = 2 2(2.196) 3+

= 3.72 N

The work done by the net force is

W = Fnet x ∆x = (3.72) (5) ≈ 18.6 J

11. A.B

| A B |

→ →

→ →

× =

ABcos

ABsin

θ

θ = cot θ

Given, A.B

| A B |

→ →

→ →

× =

1

3

∴ cot θ = 1

3

⇒ θ = cot−1 1

3

= 60° = c

3

π

12. The three vectors not lying in one plane

cannot form a triangle, hence their resultant cannot be zero. Also, their resultant will

neither be in the plane of P→

or Q→

nor in the

plane of R→

. Hence option (D) is correct.

13. Net force on the body F→

= 1 2F F→ →

+

= ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ5i j 2k 2i j 2k+ − + + −

= ˆ ˆ ˆ7 i 2 j 4k+ −

s→

= ( )ˆ ˆ ˆ ˆ ˆ ˆ6i 4 j 2k 2i 2 j 4k+ − − + −

= ˆ ˆ ˆ4i 2 j 2k+ +

W = F. s→ →

= ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ7i 2 j 4k 4i 2 j 2k+ − ⋅ + +

= 28 + 4 − 8 = 24 units.

14. Let, x yˆ ˆA iA jA

→

= + , x yˆ ˆB iB jB

→

= +

∴ A B→ →

+ = ( ) ( )x y x yˆ ˆ ˆ ˆiA jA iB jB+ + +

= ( ) ( )x x y yˆ ˆi A A j A B+ + +

Given Ax = 4 m, Ay = 6 m

Ax + Bx = 12 m , Ay + By = 10 m

∴ Bx = 12 m − Ax = 12 m − 4 m = 8 m

By = 10 m − Ay = 10 m − 6 m = 4 m. 15. The angle subtended is

sin θ = 2 2 2

3 3

616 3 4=

+ +

θ = sin−1 3

61

16. p→

= x yˆ ˆip jp+

= ˆ ˆi[3cos t] j [3sin t]+

According to Newton’s second law of motion,

F→

= d p

dt

→

∴ F→

= d ˆ ˆi3cos t j 3sin tdt +

= ˆ ˆi ( 3sin t) j(3cos t)− +

∴ | F |→

= 2 2( 3sin t) (3cos t)− + = 3

17. A→

⋅B→

= | A→

| | B→

| cos θ ⇒ cos θ = A.B

| A | . | B |

→ →

→ →

The component of A→

in the direction of B→

= A.B

|A| cos |A|

|A||B|

→ →→ →

→ →θ = ⋅

= 3 2 5

2 2

+= along B

→

18. The vector product of two non-zero vectors is

zero if they are in the same direction or in the

opposite direction. Hence vector B→

must be

parallel to vector A→

, i.e. along ± x-axis.

Publications Pvt. Ltd. Target Chapter 02: Scalars and Vectors

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19. Area of the triangle = 1

| A B |2

→ →

×

=

ˆ ˆ ˆi j k1

2 3 42

1 0 2

−

−

= 1 ˆ ˆ ˆi(6 0) j ( 4 4) k( 3)2 − − − − + +

= 1 ˆ ˆ ˆ6i 8j 3k2

+ +

= 1

36 64 92

+ +

= 1

1092× = 5.22 units

20. x y zˆ ˆ ˆA a i a j a k

→

= + +

Magnitude of vector 2 2 2

x y zA | A | a a a→ →

= = + +

where, ax, ay and az are the magnitudes of

projections of A→

along three coordinate axes

x, y and z respectively.

ˆ ˆ| j k |− = 2 21 ( 1) 2+ − =

∴ Component of vector A→

along the direction of

ˆ ˆ( j k)− = y za a

2

−