Vectors 𝐀 and 𝐁 have equal magnitudes of 5.00 . The sum of 𝐀...

Vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ have equal magnitudes of 5.00 . The sum of $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ is the vector $6.00 \hat{\mathbf{j}}$. Determine the angle between $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$.

Key Concepts

Vector Addition

Vector addition involves combining two or more vectors to obtain a resultant vector. It is fundamental in physics and engineering to determine the net effect when multiple vector quantities, such as forces or velocities, act simultaneously. The process often uses the head-to-tail method or the parallelogram law, and it requires considering both the magnitudes and directions of the vectors being added.

Law of Cosines in Vector Problems

When two vectors are added, the law of cosines is a powerful tool for relating the magnitudes of the individual vectors to the magnitude of their resultant and the angle between them. This law generalizes the Pythagorean theorem and is particularly useful when the vectors have known magnitudes and the angle between them is the unknown that needs to be determined.

Dot Product and Angle Between Vectors

The dot product of two vectors provides a scalar quantity that encapsulates both the magnitudes of the vectors and the cosine of the angle between them. This relationship is pivotal in calculating the angle between two vectors when their magnitudes and dot product are known, making it an essential concept in vector analysis.

Transcript

00:01 Okay, for this problem, we have two vectors, a and b, that both have the same length of five units.

00:07 And when we add them together, we end up with the vector that has a length of six in the j -hat direction.

00:17 So six upward.

00:19 And we're asked what the angle is between the two vectors.

00:22 Now, the only way that we can do this is if our two vectors, a and b, come out either here to the right or here to the left.

00:35 And hopefully you realize that if i went to the left, it would be the same angle between them.

00:41 And there's two different ways to solve this problem.

00:43 There's a little bit harder and a little bit easier depending on how you view it.

00:48 I'll start with the harder one, and then i'll finish up with the easy one.

00:52 Because i'm dealing with an angle and the opposite length, i can use the law of cosines.

01:00 So i have d squared is equal to a squared plus b squared.

01:08 That's not a plus, that's a minus 2, ab times the cosine of the angle.

01:16 But remember that a and b are the same length, and so i can use just one of the lengths.

01:22 So that's a squared plus a squared minus 2, a times a, which is a squared over, cosine theta or times cosine theta and then these also turn into a 2 a squared.

01:40 Now i'm solving for cosine.

01:42 So this is going to be d squared minus 2a squared because i move that over the left side equals minus 2a squared cosine theta.

01:55 So cosine theta is equal to d squared minus 2a squared over minus 2 a squared.

02:06 We can plug in numbers now.

02:09 We know that d is equal to 6, so that's 6 squared times 2.

02:14 A squared, a is 5...

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