91Ó°ÊÓ

The magnitude of a vector represents its size or length in the vector space. If we think of a vector as an arrow, the magnitude is how long that arrow is. This is key to understanding how vectors relate to each other in space. Let's consider the vectors given in the exercise: \( \vec{A} \) and \( \vec{B} \). Both have magnitudes of 3, as mentioned. You can think of them having the same length or size.

To find the magnitude of any vector \( \vec{v} = (v_x, v_y, v_z) \), we use the formula:\[ \|\vec{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]

For instance, if a vector has components \(-5.00 \hat{k} + 2.00 \hat{i}\), the magnitude is calculated as \( \sqrt{(-5.00)^2 + (2.00)^2} \), resulting in \( \sqrt{29} \). Recognizing this magnitude is crucial for operations such as finding the angle between vectors.

The angle between two vectors represents how much one vector needs to rotate to align with the other. The cross product, a vector perpendicular to both originals, provides information about this angle through its magnitude.

Given \( \vec{A} \times \vec{B} = -5.00 \hat{k} + 2.00 \hat{i} \), the magnitude of this cross product is calculated as \( \sqrt{29} \). This magintude tells us about the perpendicularity of \( \vec{A} \) and \( \vec{B} \).

To find the angle \( \theta \) between \( \vec{A} \) and \( \vec{B} \), we use the formula from the cross product's magnitude: - \( \|\vec{A} \times \vec{B}\| = AB \sin \theta \), where \( A \) and \( B \) are the magnitudes of \( \vec{A} \) and \( \vec{B} \).- Plugging in known values: \( \sqrt{29} = 9 \sin \theta \).

This equation helps us find \( \sin \theta \) and consequently \( \theta \) using inverse trigonometric functions.

Inverse trigonometric functions are the tools used to determine angles when given a trigonometric ratio, like sine, cosine, or tangent. These functions effectively reverse the process of regular trigonometric calculations, allowing us to find angles from known values.

In this exercise, we aim to find the angle \( \theta \) between two vectors by using the sine inverse function. Once we have \( \sin \theta = \frac{\sqrt{29}}{9} \), we can employ the \( \arcsin \) function to retrieve the angle:
- \( \theta = \arcsin\left(\frac{\sqrt{29}}{9}\right) \)

Using a calculator or appropriate software helps in determining this angle in radians. It’s crucial to choose the range of the angle that matches the context of geometric constraints in vector problems. These inverse functions make it possible to go from a ratio back to an angle, completing the cycle of our understanding of vector relationships.