91Ó°ÊÓ

Understanding the magnitude of a vector is crucial when working with vector operations. The magnitude of a vector, often represented as \(|\mathbf{a}|\) or \(|\mathbf{b}|\), is a measure of its length.

• For a vector \(\mathbf{a}\) in two-dimensional space, with components \(a_x\) and \(a_y\), the magnitude is given by \(|\mathbf{a}| = \sqrt{a_x^2 + a_y^2}\).
• In three-dimensional space, if a vector \(\mathbf{a}\) has components \(a_x\), \(a_y\), and \(a_z\), the magnitude becomes \(|\mathbf{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\).

The magnitude tells us how long the vector is in the coordinate system. It does not give us information about direction. In the given problem, the magnitudes of vectors \(\mathbf{a}\) and \(\mathbf{b}\) are 6 and \(\frac{1}{2}\) respectively. This sets the stage for further calculations.

Finding the angle between two vectors helps in understanding their spatial relationship. This angle, often denoted as \(\theta\), can significantly influence the result of vector operations such as the cross product.The angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\) is used in the cross product formula: \(| \mathbf{a} \times \mathbf{b} | = | \mathbf{a} | | \mathbf{b} | \sin(\theta)\). The angle impacts how much of one vector moves perpendicularly towards another.

• If the angle is \(0^{\circ}\), the vectors are pointing in the same direction and their cross product is zero because there is no perpendicular component.
• If the angle is \(90^{\circ}\), the vectors are at a perfect right angle to each other, maximizing the cross product.

In our problem, the angle \(\theta = 60^{\circ}\) contributes a factor of \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\) in the cross product calculation.

Trigonometric functions like sine, cosine, and tangent are pivotal in vector mathematics.Sine Function:The sine function relates an angle in a right triangle to the ratio of the opposite side over the hypotenuse. Within the context of vectors, the sine function helps determine the magnitude of the perpendicular component.For an angle \(\theta\):- \(\sin(\theta) = \text{opposite side} / \text{hypotenuse}\) - For a \(60^{\circ}\) angle, this results in \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\).This trigonometric function plays a critical role in our vector cross product formula. It scales the product of the magnitudes of the two vectors based on how "twisted" they are relative to each other. Since vectors \(\mathbf{a}\) and \(\mathbf{b}\) form a \(60^{\circ}\) angle, the sine of this angle crucially defines the proportion of their interaction perpendicular to each other.