If a=b+c and the magnitude of a b and c are 5,4 and 3 units respectively the angle between a and c is:
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Given that |A⃗ |=5 , |B⃗ |=4 , |C⃗ |=3 &
A⃗ =B⃗ +C⃗
A⃗ −B⃗ =C⃗
Taking self scalar product on both the sides as follows
(A⃗ −B⃗ )⋅(A⃗ −B⃗ )=C⃗ ⋅C⃗
A⃗ ⋅A⃗ −B⃗ ⋅A⃗ −A⃗ ⋅B⃗ +B⃗ ⋅B⃗ =|C⃗ |2
|A⃗ |2−2A⃗ ⋅B⃗ +|B⃗ |2=|C⃗ |2
52−2|A⃗ ||B⃗ |cosθ+42=32
41−2⋅5⋅4cosθ=9
32−40cosθ=0
cosθ=45
θ=cos−1(45)
=36.86989764584401∘
Hence, the angle ( θ ) between given vectors A⃗ & B⃗ is 36.86989764584401∘
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