Comprehensive Vector Analysis Assignment with Step-by-Step Solutions

This document presents a detailed solution to a physics assignment focused on vector analysis. The assignment covers a range of problems, including finding vector components, performing vector addition and subtraction, calculating dot and cross products, and determining the angle between vectors. The solutions are presented in a step-by-step manner, clearly explaining each calculation and the underlying principles of vector algebra. The problems involve vectors given in both magnitude-angle and component forms, requiring conversions and manipulations to arrive at the final answers. The document provides comprehensive coverage of vector operations in two and three dimensions, making it a valuable resource for students studying physics and related fields. Desklib offers this and other resources to help students with their studies, including past papers and solved assignments.

Ques. No. 1
(a) 10 N @ 20°
(b) 25 m/s @ 98°
(c) 18 km@ 200°
Ques. No. 2
(a) 10N@20°
X component is 10 cos 20°

And Y component is 10sin 20°
cos 20°=0.9397
x=10 ×0.9397
x=9.397 N
sin 20° =0.342
y=10 × 0.342
y=3.42 N
(b) 25 m/s @ 98°
X component is 25cos 98°
And Y component is 25 sin 98°
cos 98°=−0.139
x=25 ×−0.139
x=−3.47 m/ s(−signindicate – x direction)
sin 98° =0.99
y=25× 0.99
y=24.75m/ s
(c) 18 km@ 200°
X component is 18 cos 200°
And Y component is 18 sin 200°
cos 200°=−0.9397
x=18 ×−0.9397
x=−16.91 km(−signindicate – x direction)
sin 200° =−0.342
y=18 ×−0.342
y=−6.156 km(−sign indicate – y direction)

Ques. No. 3
(a) ( 4 cm@ 40° ) +(3.5 cm @120° )
Resolving in x and y components
x1=4 cos 40°
cos 40°=0.766
x1=4 ×0.766
x1=3.064 cm
y1=4 sin 40°
sin 40°=0.642
y1=4 ×0.6427
y1=2.57 cm
x2=3.5 cos 120°
cos 120° =−0.5
x2=3.5 ×−0.5
x2=−1.5 cm(−signindic ate – x direction)
y2=3.5 sin 120°
sin 120° =0.866
y2=3.5 × 0.866
y2=3.03 cm
x=x1 +x2
y= y1 + y2
x1=3.064 cm , x2=−1.5 cm
y1=2.57 cm, y2=3.03 cm
x=3.064 cm+(−1.5)cm
x=(3.064−1.5)cm
x=1.564 cm
y=2.57 cm+3.03 cm
y=5.6 cm

R=1.564 cm ^x +5.6 cm ^y
(b) ( 5 cm @200° ) +(4.3 cm @140° )
x1=5 cos 200°
cos 200°=−0.9397
x1=5 ×−0.9397
x1=−4.698 cm(−sign indicate – x direction)
y1=5 sin 200°
sin 200° =−0.342
y1=5 ×−0.342
y1=−1.71 cm(−signindicate – y direction)
x2=4.3 cos 140°
cos 140° =−0.766
x2=4.3 ×−0.766
x2=−3.29 cm(−signindicate – x directio n)
y2=4.3 sin 140°
sin 140° =0.6427
y2=4.3× 0.6427
y2=2.76 cm
x=x1 +x2
y= y1 + y2
x1=−4.698 cm , x2=−3.29 cm
y1=−1.71 cm, y2=2.76 cm
x=−4.698 cm+(−3.29)cm
x=(−4.698−3.29)cm
x=−7.988 cm
y=−1.71cm+2.76 cm

y=5.6 cm
R=−7.988 cm ^x +1.05 cm ^y
(c) ( 3.5 cm ^x + 4 cm ^y ) −(−5.5 cm ^x+ 3 cm ^y )
x=x1 +x2
y= y1 + y2
x1=3.5 cm , x2=−5.5 cm
y1=4 cm, y2=3 cm
x=3.5 cm−(−5.5 cm)
x=(3.5+5.5)cm
x=9 cm
y=4 cm−3 cm
y=1 cm
R=9 cm ^x +1 cm ^y
(d) ( 5 cm ^x +2 cm ^y ) +(−2 cm ^x +3 cm ^y)
x=x1 +x2
y= y1 + y2
x1=5 cm , x2=−2 cm
y1=2 cm, y2=3 cm
x=5 cm+(−2 cm)
x=(3.5−2)cm
x=1.5 cm
y=2cm+3 c m
y=5 cm
R=1.5 cm ^x+ 5 cm ^y
Ques. No. 4
(a) ( 4 cm@ 40° )∧ ( 3.5 cm @120° )
R= √ a2+ b2 +2 abcosθ
θ=120−40
θ=80°

R= √ 42 +(3.5)2+2 × 4 ×3.5 × cos 80°
R= √33.11
R=5.754 c m
(b) ( 5 cm @200° ) ∧(4.3 cm@ 140° )
R= √a2+ b2 +2 abcosθ
θ=200−140
θ=60°
R= √ 52+(4.3)2 +2 ×5 × 4.3× cos 60°
R= √64.99
R=8.061 cm
(c) ( 3.5 cm ^x + 4 cm ^y ) . (−5.5 cm ^x +3 cm ^y )
x1=3.5 cm , x2=−5.5 cm
y1=4 cm, y2=3 cm
a= √ x1
2 + y1
2
a= √3.52+ 42
a= √28.25
a=5.315 cm
b= √ x2
2 + y2
2
b= √ −5.52 +32
b= √39.25
b=6.265 c m
R= √a2+ b2 +2 abcosθ
cosθ= a .b
|a||b|
cosθ= x1 x2+ y1 y2
|a||b|

cosθ= ( 3.5 ) ( −5.5 ) + ( 4 ) (3)
|5.315||6.265|
cosθ= −7.25
33.298
θ=cos−1 (−0.2177)
θ=102.6°
R= √5.3152 +6.2652 +2 ×5.315 ×6.265 × cos 102.6°
R= √ 52.97
R=7.278 cm
(d) ( 5 cm ^x +2 cm ^y ) . (−2 cm ^x +3 cm ^y )
x1=5 cm , x2=−2 cm
y1=2 cm, y2=3 cm
a= √ x1
2 + y1
2
a= √ 52 +(−2)2
a= √29
a=5.385 cm
b= √ x2
2 + y2
2
b= √ 22+ 32
b= √ 13
b=3.6 c m
R= √a2+ b2 +2 abcosθ
cosθ= a .b
|a||b|
cosθ= x1 x2+ y1 y2
|a||b|
cosθ= ( 5 ) (−2 ) + ( 2 ) (3)
|5.385||3.6|
cosθ= −4
19.386
θ=cos−1 (−0.206)
θ=101.9°

R= √ 5.3852 +3.62+2 ×5.385 ×3.6 × cos 101.9°
R= √33.96
R=5.827 cm
Ques. No. 5
(a) ( 4 cm@ 40° ) . ( 3.5 cm @120° )
a . b=|a||b|cosθ
θ=120−40
θ=80°
a . b=|4 ||3.5|cos 80°
a . b=2.43 cm
(b) ( 5 cm @200° ) .(4.3 cm@ 140° )
a . b=|a||b|cosθ
θ=200−140
θ=60°
a . b=|5||4.3|cos 60°
a . b=10.75 cm
(c) ( 3.5 cm ^x +4 cm ^y ) . (−5.5 cm ^x +3 cm ^y )
a . b=x1 x2 + y1 y2
x1=3.5 cm , x2=−5.5 cm
y1=4 cm, y2=3 cm
a . b=3.5 × (−5.5 ) +4 ×3
a . b=−19.25+12
a . b=−7.25 cm
(d) ( 5 cm ^x +2 cm ^y ) .(−2cm ^x+3 cm ^y )
a . b=x1 x2 + y1 y2 x1=5 cm , x2=−2 cm
y1=2 cm, y2=3 cm
a . b=5 × (−2 ) +2 ×3

a . b=−10+6
a . b=−4 cm
Ques. No. 6
(a) ( 4 cm@ 40° ) × ( 3.5 cm@ 120° )
x1=4 cos 40°
cos 40°=0.766
x1=4 ×0.766
x1=3.064 cm
y1=4 sin 40°
sin 40°=0.642
y1=4 ×0.6427
y1=2.57 cm
x2=3.5 cos 120°
cos 120° =−0.5
x2=3.5 ×−0.5
x2=−1.5 cm(−signindicate – x direction)
y2=3.5 sin 120°
sin 120° =0.866
y2=3.5 × 0.866
y2=3.03 cm
x1=3.064 cm , x2=−1.5 cm , x3=0 cm
y1=2.57 cm, y2=3.03 cm , y3=0 cm
a × b=
| ^i ^j ^k
x1 x2 x3
y1 y2 y3
|
a × b=
| ^i ^j ^k
3.064 −1.5 0
2.57 3.03 0|

a × b= ^i ( ( −1.5 ) ×0−3.03 ×0 ) − ^j ( 3.064 × 0−2.57 × 0 ) + ^k (3.064 ×3.03−(−1.5)× 2.57)
a × b=13.139 cm ^k
(b) ( 5 cm @200° ) ×(4.3 cm@140° )
x1=5 cos 200°
cos 200°=−0.9397
x1=5 ×−0.9397
x1=−4.698 cm(−sign indicate – x direction)
y1=5 sin 200°
sin 200° =−0.342
y1=5 ×−0.342
y1=−1.71 cm(−sign indicate – y direction)
x2=4.3 cos 140°
cos 140° =−0.766
x2=4.3 ×−0.766
x2=−3.29 cm(−signindicate – x direction)
y2=4.3 sin 140°
sin 140° =0.6427
y2=4.3× 0.6427
y2=2.764 cm
x=x1 +x2
y= y1 + y2
x1=−4.698 cm , x2=−3.29 cm, x3=0 cm
y1=−1.71 cm, y2=2.76 cm, y3=0 cm
a × b=
| ^i ^j ^k
x1 x2 x3
y1 y2 y3
|
a × b=
| ^i ^j ^k
−4.698 −3.29 0
−1.71 2.764 0
|

a × b= ^i ( (−3.29 ) ×0−2.764 × 0 )− ^j (−4.698 ×0−(−1.71)×0 ) + ^k (−4.698× 2.764−(−3.29)×(−1.71))
a × b=−18.6 cm ^k
(c) ( 3.5 cm ^x +4 cm ^y ) × (−5.5 cm ^x +3 cm ^z )
x1=3.5 cm , x2=−5.5 c m,x3=0 cm
y1=4 cm, y2=0 cm, y3=3 cm
a × b=
| ^i ^j ^k
x1 x2 x3
y1 y2 y3
|
a × b=
| ^i ^j ^k
3.5 −5.5 0
4 0 3
|
a × b= ^i ( ( −5.5 ) ×3−0 ×0 )− ^j ( 3.5 ×3−4 × 0 ) + ^k (3.5× 0−4 ×(−5.5))
a × b=−16.5 cm ^i−10.5 cm ^j+22 cm ^k
(d) ( 5 cm ^x +2 cm ^z ) ×(−2 cm ^x +3 cm ^y )
x1=5 cm , x2=0 cm, x3=−2 cm
y1=2 cm, y2=3 cm, y3=0 cm
a × b=
| ^i ^j ^k
x1 x2 x3
y1 y2 y3
|
a × b=
| ^i ^j ^k
5 0 −2
2 3 0 |
a × b= ^i ( 0 ×0−(−2) ×3 ) − ^j ( 5 ×0−(−2)×2 ) + ^k (5 ×3−0 ×2)
a × b=6 cm ^i−4 cm ^j+15 cm ^k