Question

In Exercises 19-22, the magnitudes of vectors \(\vec{u}\) and \(\vec{v}\) in \(\mathbb{R}^{3}\) are given, along with the angle \(\theta\) between them. Use this information to find the magnitude of \(\vec{u} \times \vec{v}\). \(\|\vec{u}\|=2, \quad\|\vec{v}\|=5, \quad \theta=30^{\circ}\)

Short answer

The magnitude of \(\vec{u} \times \vec{v}\) is 5.

Step-by-step solution

Understanding the Cross Product Magnitude Formula

The magnitude of the cross product of two vectors \(\vec{u}\) and \(\vec{v}\) can be found using the formula: \(\|\vec{u} \times \vec{v}\| = \|\vec{u}\| \cdot \|\vec{v}\| \cdot \sin(\theta)\). This formula requires the magnitudes of the vectors and the sine of the angle between them.

Substituting the Known Values

Given \(\|\vec{u}\| = 2\), \(\|\vec{v}\| = 5\), and \(\theta = 30^{\circ}\), substitute these values into the formula: \[\|\vec{u} \times \vec{v}\| = 2 \cdot 5 \cdot \sin(30^{\circ})\].

Calculating \(\sin(30^{\circ})\)

Recognize that \(\sin(30^{\circ}) = \frac{1}{2}\). This value is useful as it simplifies the computation of the cross product's magnitude.

Solving the Expression

Substitute \(\sin(30^{\circ})\) into the formula to find: \[\|\vec{u} \times \vec{v}\| = 2 \cdot 5 \cdot \frac{1}{2} = 5\]. Therefore, the magnitude of the cross product is 5.

Key concepts

Vector Magnitude

Vector magnitude is a way of describing how long a vector is. Think of it as the length of an arrow. The magnitude of a vector \( \vec{u} \) in three-dimensional space \( \mathbb{R}^3 \) is calculated using the following formula: \[ \|\vec{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2} \] where \( u_1 \), \( u_2 \), and \( u_3 \) are the components of the vector \( \vec{u} \).
When solving problems, knowing the magnitude of a vector helps us understand its strength or size, without worrying about its direction. This is particularly useful in physics and engineering.
In our exercise example, each given vector has a magnitude: \( \|\vec{u}\| = 2 \) and \( \|\vec{v}\| = 5 \). By knowing these values, you can calculate the magnitude of their cross product.

Angle Between Vectors

The angle between two vectors is crucial for understanding their relationship in space. Specifically, it tells us how two vectors are oriented relative to each other.
The angle \( \theta \) can vary from \( 0^{\circ} \) to \( 180^{\circ} \), where \( 0^{\circ} \) indicates they are in the same direction, and \( 180^{\circ} \) indicates they are pointing in exactly opposite directions. For our exercise, though, the angle given is \( 30^{\circ} \).
Calculating any mathematical expressions involving vectors, often requires knowing the angle between them. This angle plays a crucial role, especially when using the cross product formula.

Cross Product Formula

The cross product is a vector operation that returns another vector, which is perpendicular to the original two vectors. The magnitude of the cross product is determined using the following formula:
\[ \|\vec{u} \times \vec{v}\| = \|\vec{u}\| \cdot \|\vec{v}\| \cdot \sin(\theta) \]
This formula multiplies the magnitudes of \( \vec{u} \) and \( \vec{v} \), and then multiplies the result by the sine of the angle between \( \theta \). In our example, by plugging in the values, the magnitude is calculated as 5.
A cross product is useful for determining an area spanned by two vectors or for finding a direction normal to a surface created by those vectors.

Sine Function in Trigonometry

The sine function helps determine relationships between the angles and sides of a triangle. In the context of vectors, sine is used to calculate how much of one vector is perpendicular to another.
For any angle \( \theta \), \( \sin(\theta) \) ranges from \(-1\) to \(1\). It measures the ratio of the length of the side opposite the angle to the hypotenuse in a right-angled triangle.
In our exercise, \( \sin(30^{\circ}) = \frac{1}{2} \). Knowing the sine value for specific angles simplifies calculations, making it easier to solve expressions like those found in vector cross product problems. The sine function is a fundamental part of trigonometry and it's indispensable in physics and engineering calculations.