if \(\underline{u}\) and \(\underline{\mathrm{v}}\) are unit vectors and \(\theta\) is acute angle between them find \(\theta\) such that \(2 \underline{\mathrm{u}} \times 3 \underline{\mathrm{v}}\) becomes unit vectors. (a) \(\sin ^{-1}(1 / 3)\) (b) \(\sin ^{-1}(1 / 6)\) (c) \(\cos ^{-1}(1 / 6)\) (d) none of these

Unit vectors play a fundamental role in vector operations. A unit vector has a magnitude of 1, making it a standard for specifying directions without a scale. Imagine an arrow pointing in a specific direction; if this arrow has a length of 1, no matter the units, it is called a unit vector. In mathematics, unit vectors are often represented with a hat, such as \( \hat{i} \) for the unit vector in the direction of the x-axis in three-dimensional space.

When dealing with unit vectors, the operations become simpler. In the exercise, \( \underline{u} \) and \( \underline{v} \) are given as unit vectors, which means their magnitudes are each one. Multiplying a vector by a scalar value, such as 2 or 3, changes its magnitude but not its direction. So, the cross product of these scaled unit vectors \( 2\underline{u} \times 3\underline{v} \) is still in a direction perpendicular to both \( \underline{u} \) and \( \underline{v} \) but has a magnitude that is based on the scalars and the sine of the angle between them.

Inverse trigonometric functions allow us to determine the angles from the ratios of the sides of a right triangle. Specifically, the arcsine function, denoted as \( \sin^{-1} \) or \( \arcsin \), is used to find the angle when the opposite side and hypotenuse are known. In the context of vectors, we use inverse trigonometric functions to find the angle between vectors when we know their magnitudes and the magnitude of their cross product.

In this problem, we have found that \( \sin \theta = 1/6 \) through the step by step solution. To find the angle \( \theta \), we apply the inverse sine function to 1/6, which yields \( \sin^{-1}(1/6) \). This step is crucial as it translates the ratio obtained from the cross product magnitude back into an angle, \( \theta \)—the acute angle between the two unit vectors.

The cross product magnitude reveals not just the size but also the geometrical interpretation of the result when two vectors are multiplied using the cross product. The formula \( |\underline{u} \times \underline{v}| = |\underline{u}||\underline{v}|\sin \theta \) encapsulates the relationship between the vectors' lengths, the angle between them, and their cross product's magnitude.

For unit vectors, the formula simplifies to \( \sin \theta \) because the magnitudes of the vectors are one. By understanding the magnitudes' scalar multiples, in this case, 2 and 3, you can find the resulting vector's magnitude, which is required to be a unit vector. Setting this magnitude to 1 provides a straightforward equation that can be solved for \( \sin \theta \), and consequently, the angle \( \theta \) itself. This ties into the concept of vector cross product, showing how cross product magnitude directly informs us about the relationship between the involved vectors and their resultant vector.