Vector Problem - Angle between two vectors

In summary, the problem involves two vectors A-> and B-> with equal magnitudes of 55.0. The sum of these vectors is a resultant vector of 17.7 j. To determine the angle between A-> and B->, we can use the formula R = [ A^2 + B^2 + 2ABcos(theta)]^1/2 and solve for theta.
  • #1
jaredm2012
5
0

Homework Statement



Vectors A-> and B-> have equal magnitudes of 55.0. If the sum of A-> and B-> is the vector 17.7 j, determine the angle between A-> and B->

Homework Equations



cos(theta) = {[A]^2 + ^2 - [A-B]^2}/2[A]

(sorry, not sure how to type out that formula!)

tan(theta) = (Ay + By)/(Ax + Bx)

R =[(Ax + Bx)^2 + (Ay + By)^2]^1/2

The Attempt at a Solution



This one I am not sure about how to start, at all. I plugged what I could into those formulas, but any answers I got were incorrect. I don't understand how I can find out any more information with what is given, namely the other components of the vector (the i for each, as well as discerning which values for j each vector has).
 
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  • #2
Since resultant is 17.7j, it is along y-axis. So its magnitude is 17.7. Magnitude of A ans B 55.0 each.
Use the formula R = [ A^2 + B^2 + 2ABcos(theta)]^1/2 and find theta.
 
  • #3


First, let's define the vectors A-> and B-> in terms of their components. Since we know that they have equal magnitudes of 55.0, we can represent them as:

A-> = 55.0cos(theta) i + 55.0sin(theta) j
B-> = 55.0cos(phi) i + 55.0sin(phi) j

Note that theta and phi represent the angles that each vector makes with the x-axis.

Next, we can use the given information that the sum of A-> and B-> is the vector 17.7 j, which can be represented as:

A-> + B-> = 0i + 17.7 j

Since the i components of A-> and B-> will cancel out, we are left with:

55.0sin(theta) j + 55.0sin(phi) j = 0i + 17.7 j

This means that:

55.0sin(theta) + 55.0sin(phi) = 17.7

We also know that the magnitude of the vector resulting from the sum of A-> and B-> is 17.7 units, so we can set up the following equation using the Pythagorean theorem:

(55.0cos(theta) + 55.0cos(phi))^2 + (55.0sin(theta) + 55.0sin(phi))^2 = 17.7^2

Simplifying this equation, we get:

3025(cos^2(theta) + cos^2(phi) + sin^2(theta) + sin^2(phi) + 2cos(theta)cos(phi) + 2sin(theta)sin(phi)) = 312.09

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can simplify this equation further to:

3025(1 + 1 + 2cos(theta)cos(phi) + 2sin(theta)sin(phi)) = 312.09

Dividing both sides by 3025, we get:

1 + cos(theta)cos(phi) + sin(theta)sin(phi) = 0.103

Now, we can use the trigonometric identities cos(theta - phi) = cos(theta)cos(phi) + sin(theta)sin(phi) and sin(theta - phi) = sin(theta)cos(phi) - cos(theta)sin(phi
 

Related to Vector Problem - Angle between two vectors

1. What is the angle between two vectors?

The angle between two vectors is the angle formed by the two vectors at their intersection point. It is measured in degrees or radians.

2. How do you find the angle between two vectors?

To find the angle between two vectors, you can use the dot product or cross product formula. Once you have the dot product or cross product value, you can use inverse trigonometric functions to find the angle.

3. Can the angle between two vectors be negative?

No, the angle between two vectors cannot be negative. It is always measured in the range of 0 to 180 degrees or 0 to π radians, depending on the unit of measurement.

4. What is the difference between acute, obtuse, and right angles in vector problems?

In vector problems, acute angles refer to angles less than 90 degrees, obtuse angles refer to angles greater than 90 degrees but less than 180 degrees, and right angles refer to angles equal to 90 degrees.

5. Can the angle between two vectors be greater than 180 degrees?

No, the angle between two vectors cannot be greater than 180 degrees. If the dot product or cross product of two vectors is negative, it means the angle between them is greater than 90 degrees and less than 180 degrees.

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