Final answer:
To find the angle between two vectors A and B with magnitudes of 5 and a resultant of 6j, we used vector algebra and trigonometry, concluding that the angle is approximately 73.74 degrees.
Step-by-step explanation:
If we have two vectors A and B with equal magnitudes of 5 and they sum up to give a resultant vector which is 6j, the angle between vectors A and B can be calculated using vector algebra and trigonometry. Since the resultant is solely in the j-direction (which is the vertical direction in the xy-plane) and has a magnitude of 6, it implies that the horizontal components of vectors A and B must cancel each other out.
Let θ be the angle between vector A and vector B. We can use the cosine rule for vectors, which states that when two vectors of equal magnitude “A” add to a resultant vector “R” making angle θ with each other, their resultant R equals A∙sqrt(2∙2∙cos(θ)). Our resultant vector has a magnitude of 6, and each of the original vectors has a magnitude of 5. Therefore, we can write:
6 = 5∙sqrt(2∙2∙cos(θ))
Squaring both sides, we get:
36 = 50∙2∙50∙cos(θ)
=> cos(θ) = 7/25
Using a calculator, we find that:
θ ≈ cos-1(7/25)
θ ≈ 73.74°
Therefore, the angle between vectors A and B is approximately 73.74 degrees.