What should be the angle between two vectors of magnitudes 3.20 and 5.70 units, so that their resultant has a magnitude of 6.10 units?
Cos x = (b^2 + c^2 - a^2) / 2bc
Cos x = (3.2^2 + 5.7^2 - 6.1^2) / (2 * 3.2 * 5.7)
Cos x = 5.52/36.48
Cos x = 0.15
x = 81.3 degrees
Looks good to me.
To find the angle between two vectors in this scenario, we can use the law of cosines.
The law of cosines states that for a triangle with sides of lengths a, b, and c, and an angle opposite side c, the equation is:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we have two vectors with magnitudes 3.20 units and 5.70 units, and we want to find the angle between them so that their resultant has a magnitude of 6.10 units.
Let's say the angle between the two vectors is x. The magnitudes of the vectors would be the sides of the triangle, and the resultant magnitude (6.10 units) would be side c.
Using the law of cosines, we have:
6.10^2 = 3.20^2 + 5.70^2 - 2 * 3.20 * 5.70 * cos(x)
Simplifying this equation, we get:
37.21 = 10.24 + 32.49 - 36.48 * cos(x)
Rearranging the equation, we have:
0 = 5.52 - 36.48 * cos(x)
Now, we can solve for cos(x):
cos(x) = 5.52 / 36.48
Using a calculator or a trigonometric table, we find that cos(x) is approximately 0.15.
To find the value of x, we take the inverse cosine (cos^(-1)) of 0.15:
x = cos^(-1)(0.15)
Using a calculator or a trigonometric table, we find that x is approximately 81.3 degrees.
Therefore, the angle between the two vectors should be approximately 81.3 degrees.