Two vectors have magnitudes 3 unit and 4 unit respectively . What should be the angle between them if the magnitude of the resultant is
(a) 1 unit
(b) 5 unit and
(c) 7 unit
Concept of Physics - 1 , HC VERMA , Chapter - "Physics and Mathematics"
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SOLUTION::
Magnitude of vector a , |a| = 3 m
Magnitude of vector b , |b| = 4 m
(a) If resultant , R = 1 unit then,
R = √(a²+b²+a.b.cos Ф)
1 = √(3² + 4² + cos Ф)
Ф = 180°
(b) If resultant , R = 5 unit then,
R = √(a²+b²+a.b.cos Ф)
5 = √(3² + 4² + cos Ф)
Ф = 90°
(c) If resultant , R = 7 unit then,
R = √(a²+b²+a.b.cos Ф)
7 = √(3² + 4² + cos Ф)
Ф = 0°
Hope it helps!
SOLUTION::
Magnitude of vector a , |a| = 3 m
Magnitude of vector b , |b| = 4 m
(a) If resultant , R = 1 unit then,
R = √(a²+b²+a.b.cos Ф)
1 = √(3² + 4² + cos Ф)
Ф = 180°
(b) If resultant , R = 5 unit then,
R = √(a²+b²+a.b.cos Ф)
5 = √(3² + 4² + cos Ф)
Ф = 90°
(c) If resultant , R = 7 unit then,
R = √(a²+b²+a.b.cos Ф)
7 = √(3² + 4² + cos Ф)
Ф = 0°
Hope it helps!
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