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The cross product is a vector operation in three-dimensional space that involves two vectors. Its main goal is to produce another vector that is perpendicular to the original two vectors \( \mathbf{u} \) and \( \mathbf{v} \). In simpler terms, the cross product takes vectors pointing in different directions and generates a new vector that stands at a right angle to both.
To calculate the cross product of vectors \( \mathbf{u} = \langle 2, 6, 1 \rangle \) and \( \mathbf{v} = \langle 3, 0, 1 \rangle \), we use the formula:

• First component: \( u_2v_3 - u_3v_2 = 6 \cdot 1 - 1 \cdot 0 = 6 \)
• Second component: \( u_3v_1 - u_1v_3 = 1 \cdot 3 - 2 \cdot 1 = 1 \)
• Third component: \( u_1v_2 - u_2v_1 = 2 \cdot 0 - 6 \cdot 3 = -18 \)

Therefore, the cross product \( \mathbf{u} \times \mathbf{v} \) results in the vector \( \langle 6, 1, -18 \rangle \). This new vector is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \), illustrating the geometric property of cross products.

The magnitude of a vector is essentially its length. It is calculated using the Pythagorean theorem extended into three dimensions. For a vector \( \mathbf{w} = \langle a, b, c \rangle \), the magnitude \( \| \mathbf{w} \| \) is given by:
\[\| \mathbf{w} \| = \sqrt{a^2 + b^2 + c^2}\]
For our cross product vector \( \langle 6, 1, -18 \rangle \), we substitute the components into the formula to find:
\[\| \mathbf{w} \| = \sqrt{6^2 + 1^2 + (-18)^2} = \sqrt{36 + 1 + 324} = \sqrt{361} = 19\]
This calculation shows us that the vector \( \langle 6, 1, -18 \rangle \) has a length of 19. Knowing the magnitude is crucial when we want to normalize the vector, turning it into a unit vector.

Unit vectors are vectors with a magnitude of exactly 1. They point in a specific direction but have standard length. To convert any vector into a unit vector, we divide each component of the vector by its magnitude.
Standard unit vectors often serve as building blocks in physics and engineering because they allow easy description of vector quantities in terms of direction.
For the cross product vector \( \mathbf{w} = \langle 6, 1, -18 \rangle \), we find the corresponding unit vector by dividing each component by the magnitude (19):
\[\mathbf{w}_{unit} = \frac{1}{19} \langle 6, 1, -18 \rangle = \langle \frac{6}{19}, \frac{1}{19}, \frac{-18}{19} \rangle\]
Now, \( \mathbf{w}_{unit} \) represents a vector pointing in the same direction as \( \mathbf{w} \) but with a length of 1. Such unit vectors maintain directional properties while standardizing the vector's length, which is vital for many applications in vector calculus and 3D modelling.