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The cross product of two vectors is a powerful tool in geometry and physics. It helps us find a vector that is perpendicular (orthogonal) to two given vectors in three-dimensional space. The cross product of two vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \) is symbolized as \( \mathbf{a} \times \mathbf{b} \). To compute it, you need to form a determinant with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row and the components of the vectors \( \mathbf{a} \) and \( \mathbf{b} \) in the subsequent rows.

The result is another vector \( \mathbf{c} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k} \). This vector \( \mathbf{c} \) will be orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).

• A key property of the cross product is that reversing the order of the vectors results in a vector pointing in the opposite direction. \( \mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}) \).
• The magnitude of the cross product \( |\mathbf{a} \times \mathbf{b}| \) represents the area of a parallelogram formed by the vectors \( \mathbf{a} \) and \( \mathbf{b} \).

By finding the cross product \( \mathbf{u} \times \mathbf{v} \) as in the original problem, \( \mathbf{w} = -2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k} \) becomes the direction vector for the line that is orthogonal to vectors \( \mathbf{u} \) and \( \mathbf{v} \).

In the context of parametric equations for lines, the direction vector defines the line's direction and is crucial in determining its equations. A direction vector \( \mathbf{d} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) indicates how the line moves in three-dimensional space. The components \( a, b, \) and \( c \) show the change in the \( x, y, \) and \( z \) directions per unit change in the parameter \( t \), respectively.

For a line described by parametric equations, the vector function can be expressed as \( \mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{d} \), where \( \mathbf{r}_0 \) is a point on the line and \( \mathbf{d} \) is the direction vector. The given point provides a starting location, while the direction vector defines the path of the line as \( t \) varies.

• If the direction vector is multiplied by a scalar, it does not change the direction of the line but adjusts its pace by expanding or contracting the vector's magnitude.

This concept was applied in the original solution to create the parametric equations using the point \((2,3,0)\) and direction vector \(-2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\). These equations describe a line moving in the specified direction.

Orthogonal vectors are vectors that meet at a right angle, meaning they are perpendicular to each other in geometric space. This property can be checked using the dot product: two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal if and only if their dot product is zero, \( \mathbf{a} \cdot \mathbf{b} = 0 \).

Orthogonality is a critical concept for ensuring that the line in the original problem is properly oriented. Given vectors \( \mathbf{u} \) and \( \mathbf{v} \), finding a direction vector that is orthogonal to both involves calculating the cross product, as it yields a vector that is perpendicular to both original vectors.

• This concept underpins much of three-dimensional space analysis and has practical applications in fields such as physics, engineering, and computer graphics.
• Orthogonality simplifies problems by reducing the dimensionality, such as converting a multi-dimensional problem into a series of single-dimensional problems.

The calculated vector \( \mathbf{w} = -2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k} \) from the cross product is orthogonal to both \( \mathbf{u} \) and \( \mathbf{v} \), ensuring the line's direction is correct relative to these vectors.