If angle between \(\underline{a}\) and \(\underline{b}\) is \((\pi / 6)\) and \(|\underline{a}|=4,|\underline{b}|=2\), then \(|\underline{\mathrm{a}} \times \underline{\mathrm{b}}|=\) (A) 4 (B) 16 (C) 8 (D) 2

The magnitude of a vector is a measure of its length and is a crucial concept in physics and mathematics. In our exercise, magnitudes are represented as \( |\textbf{a}| \) for vector \( \textbf{a} \) and \( |\textbf{b}| \) for vector \( \textbf{b} \). These are absolute values, meaning they are always positive and denote the size of the vector in a space.

To clarify, consider a vector as an arrow pointing from one point to another. The length of this arrow is the magnitude of the vector. For example, if \( |\textbf{a}| = 4 \) and \( |\textbf{b}| = 2 \) from our textbook exercise, imagine two arrows where one is twice as long as the other. Understanding vector magnitude is essential before you can delve into vector operations such as vector cross product.

The sine function is one of the fundamental trigonometric functions and it relates a given angle to a specific ratio of sides in a right triangle. In the context of our problem, the angle in question is \( \frac{\pi}{6} \) or 30 degrees. The sine of an angle, represented as \( \sin(\theta) \) for angle \( \theta \), is particularly significant when dealing with vector cross products.

The value of \( \sin(\frac{\pi}{6}) \) is \( \frac{1}{2} \) as given in the solution. This ratio comes into play when calculating the strength of the perpendicularity between two vectors. Remember, the stronger the perpendicularity—or the closer to 90 degrees the angle is—the greater the result of the cross product will be.

Vector multiplication can be done in several ways, but one of the most interesting and useful forms is the cross product. The cross product of two vectors yields another vector that is perpendicular to both original vectors and thus, it is used to explore concepts of area and torque among many others.

Using the formula \( |\textbf{a} \times \textbf{b}| = |\textbf{a}| |\textbf{b}| \sin(\theta) \), where \( \theta \) is the angle between the vectors, we have a clear path to finding the magnitude of this new perpendicular vector. As noted in our solution, since we have all the necessary values, the calculation simplifies the understanding of how vector quantities interact in different dimensions through cross product.

The Joint Entrance Examination (JEE) in Mathematics encompasses a vast array of topics, including Algebra, Calculus, and, notably, Vectors and 3D Geometry. Vector Algebra plays a vital role as it forms the basis for many physics and engineering concepts.

In JEE Maths, understanding vectors, including their multiplication like dot and cross products, is imperative. Problems like our exercise are typical of JEE questions and are designed to test students' grasp of vector properties and operations. Preparing to solve such problems requires a strong foundational knowledge of both the arithmetic and the geometric interpretation of vectors.