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?
Well, if I were to put it in a humorous way, the angle between two vectors is like trying to find the perfect angle for a selfie. You want to make sure you capture all the "magnitude" of your fabulousness, but at the same time, not too close that it distorts the true beauty.
In this case, let's call the angle between the vectors "Mr. Selfie Angle." So, Mr. Selfie Angle should allow us to create a resultant vector with a magnitude of 6.10 units.
Now, imagine you have two friends, each holding a giant arrow representing the magnitude of their vectors. One friend has a 3.20-unit arrow, and the other friend has a 5.70-unit arrow. To achieve a resultant of 6.10 units, you want your friends to hold their arrows at an angle that perfectly aligns with the "Mr. Selfie Angle."
I apologize if I got a little carried away with this analogy, but in terms of math, you can use the law of cosines to find the angle between the vectors. The formula is:
c^2 = a^2 + b^2 - 2ab * cos(C)
Where:
c is the magnitude of the resultant vector (6.10 units),
a is the magnitude of one vector (3.20 units),
b is the magnitude of the other vector (5.70 units), and
C is the angle between the two vectors.
By rearranging the formula and solving for C, you can calculate the angle between the vectors. But remember, finding the perfect selfie angle is a serious task!
To find the angle between two vectors, we can use the cosine formula. Let's denote the angle between the vectors as θ.
The cosine formula states that the dot product of two vectors, A and B, is equal to the product of their magnitudes multiplied by the cosine of the angle between them.
So, the dot product of vectors A and B is given by:
A · B = |A| * |B| * cos(θ)
In this case, the magnitudes of the two vectors are given as 3.20 and 5.70 units, and the magnitude of their resultant vector is given as 6.10 units.
Let's substitute these values into the equation:
3.20 * 5.70 * cos(θ) = 6.10
Now, let's solve for cos(θ):
cos(θ) = 6.10 / (3.20 * 5.70)
cos(θ) ≈ 0.349
To find the value of θ, we need to take the inverse cosine (or arccos) of 0.349:
θ ≈ arccos(0.349)
Using a calculator, we find that θ ≈ 69.3 degrees (rounded to one decimal place).
Therefore, the angle between the two vectors should be approximately 69.3 degrees.
To find the angle between two vectors, you can use the dot product formula:
A · B = |A| |B| cos(theta)
Where A and B are the magnitudes of the two vectors, and theta is the angle between them.
In this case, the magnitudes of the two vectors are 3.20 and 5.70 units, and the magnitude of the resultant vector is 6.10 units.
Let's assume the angle between the two vectors is theta.
Using the given information, we can rewrite the dot product formula as:
3.20 * 5.70 * cos(theta) = 6.10
Now, we can solve for cos(theta):
cos(theta) = 6.10 / (3.20 * 5.70)
cos(theta) ≈ 0.3485
To find the angle theta, we can take the inverse cosine (also called arccosine) of cos(theta):
theta ≈ arccos(0.3485)
Using a calculator, we find theta ≈ 69.02 degrees.
Therefore, the angle between the two vectors should be approximately 69.02 degrees.
Use the law of cosines.
c^2 = a^2 + b^2 - 2ab cos C
6.1^2 = 3.2^2 + 5.7^2 - 2*3.2*5.7 cos C
cos C = (10.24 + 32.49 - 37.21)/36.84
= 0.1498
Solve for C and verify my numbers.