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The cross product is a mathematical operation that takes two vectors and produces a third vector. This new vector is perpendicular to the plane formed by the original vectors, making it orthogonal to each of them.
To find the cross-product in 3-dimensional space, we arrange the vectors in a specific way using a determinant:

• Given two vectors, \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \mathbf{a} \times \mathbf{b} \) is found using: \( \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \).
• The resulting vector is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).

In the given exercise, the cross product of vectors \( \mathbf{a} = \langle 1,0,4 \rangle \) and \( \mathbf{b} = \langle 1,-4,2 \rangle \) is calculated to be \( \langle 16,2,-4 \rangle \). This vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).
Cross product technique is crucial in physics and engineering to determine a normal vector to a plane.

A unit vector is a vector with a magnitude (length) of exactly one. It's mainly used to represent a direction without scaling. Converting any vector into a unit vector involves dividing the vector by its magnitude.

• The unit vector \( \mathbf{u} \) of a vector \( \mathbf{v} \) is \( \mathbf{u} = \mathbf{v} / ||\mathbf{v}|| \), where \( ||\mathbf{v}|| \) is the magnitude of \( \mathbf{v} \).
• This transformation keeps the vector's direction intact while adjusting its length to 1.

In the problem at hand, after calculating the vector \( \mathbf{c} = \langle 16, 2, -4 \rangle \), it was divided by its magnitude, 18, to yield the unit vector \( \langle 8/9, 1/9, -2/9 \rangle \). Unit vectors are useful because they allow you to describe a direction in a pure form, helpful in various applications such as normalizing directional vectors.

Vector magnitude is a measure of a vector's length in space. You calculate it using the Pythagorean theorem by taking the square root of the sum of the squares of its components.
For a vector \( \mathbf{v} = \langle a, b, c \rangle \), its magnitude is given by:

• \( ||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2} \)

The magnitude helps in determining how large or small a vector is and is crucial for various operations like finding unit vectors. In the exercise, for vector \( \mathbf{c} = \langle 16, 2, -4 \rangle \), the magnitude was found to be 18, which is used to generate a corresponding unit vector. This concept is widely applicable in physics, where it can represent speed or force intensity, among other things.

3-dimensional vectors consist of three components, usually aligned with the x, y, and z axes. They are used to represent entities or values that require three dimensions to be fully described, such as force, velocity, or position in space.

• Vectors are denoted as \( \langle x, y, z \rangle \), where each component represents a portion along the respective axis.
• A valuable property of 3D vectors is their ability to describe both magnitude and direction simultaneously.

In the context of this exercise, vectors \( \mathbf{a} = \langle 1,0,4 \rangle \) and \( \mathbf{b} = \langle 1,-4,2 \rangle \) are examples of 3-dimensional vectors. They encompass quantities along each axis, allowing for precise and comprehensive mathematical operations like the cross product. Understanding 3D vectors is essential for applications in physics simulations, computer graphics, and engineering design.