What is angle between 3i^-4j^ and 2i^+3j^
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Dot product of vector A and B is
A · B = |A||B| cosθ …………(1)
Here θ = angle between vectors A and B
A = 3i - 4j
B = 2i + 3j
|A| = √(3² + 4²) = 5
|B| = √(2² + 3²) = 3.60
Dot product of A and B
A · B = (3i - 4j) · (2i + 3j)
= 6 - 12
= -6
Using equation (1)
-6 = 5 × 3.60 × cosθ
cosθ = -1/3
θ = 109.2 degrees
Angle between them is 109.2 degrees
A · B = |A||B| cosθ …………(1)
Here θ = angle between vectors A and B
A = 3i - 4j
B = 2i + 3j
|A| = √(3² + 4²) = 5
|B| = √(2² + 3²) = 3.60
Dot product of A and B
A · B = (3i - 4j) · (2i + 3j)
= 6 - 12
= -6
Using equation (1)
-6 = 5 × 3.60 × cosθ
cosθ = -1/3
θ = 109.2 degrees
Angle between them is 109.2 degrees
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