91Ó°ÊÓ

The cross product of two vectors, often symbolized by the "×" sign, is a technique to find a vector that is perpendicular to the plane formed by the two original vectors in a 3D space. It is especially useful when dealing with geometric problems that involve right angles or finding an orthogonal vector.
In this exercise, to determine a direction vector for the line, which is perpendicular to both provided vectors \(\mathbf{u}\) and \(\mathbf{v}\), we perform a cross product operation. Here's how the process works:

• Arrange the components of \(\mathbf{u}\) and \(\mathbf{v}\) in a 3x3 matrix.
• The first row consists of the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).
• The second row lists the components of \(\mathbf{u} = \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}\).
• The third row contains \(\mathbf{v} = \begin{pmatrix} 3 & 4 & 5 \end{pmatrix}\).

The computation involves calculating the determinant of the matrix, which involves multiplying and subtracting the components as seen in the set equations for \(\mathbf{i}, \mathbf{j}, \mathbf{k}\):
\[ \mathbf{n} = \mathbf{i}(2 \cdot 5 - 3 \cdot 4) - \mathbf{j}(1 \cdot 5 - 3 \cdot 3) + \mathbf{k}(1 \cdot 4 - 2 \cdot 3) \]
This results in the vector \( \begin{pmatrix} -2 & -4 & -2 \end{pmatrix} \), which serves as our necessary perpendicular direction vector.

A direction vector is crucial in creating parametric equations for a line in space, as it gives us the precise direction in which the line travels. With a specific point given, the direction vector helps establish the line's progression through that point.
For our example, after calculating the cross product of \(\mathbf{u}\) and \(\mathbf{v}\), we determined the direction vector \(\begin{pmatrix} -2 & -4 & -2 \end{pmatrix}\).
This direction vector tells us the rate of change along the line with respect to the parameter \(t\).
To construct the parametric equations, we use the form:

• \(x = x_0 + at\)
• \(y = y_0 + bt\)
• \(z = z_0 + ct\)

Where \((x_0, y_0, z_0)\) is a known point on the line, here \((2, 3, 0)\), and \(a, b, c\) are the respective components of the direction vector. Each parameter describes how far from the starting point \((x_0, y_0, z_0)\) the line extends as \(t\) varies.

Perpendicular lines, also described as orthogonal, are lines that intersect at a right angle (90 degrees). In vector mathematics, showing that two lines are perpendicular implies discovering a vector normal to the surfaces they describe.
In this exercise, to find a line perpendicular to vectors \(\mathbf{u}\) and \(\mathbf{v}\), we utilize the cross product. The resulting vector, \(\begin{pmatrix} -2 & -4 & -2 \end{pmatrix}\), is inherently perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\).
Here's why it works:

• The cross product of two vectors gives a third vector orthogonal to the original two.
• This is because the dot product between the resulting vector from the cross product and the original vectors equal zero, confirming perpendicularity.
• For a line to be perpendicular to given vectors, its direction vector must maintain an orthogonal relationship with them, as accomplished through the cross product.

This property aids in defining the direction vector for the line, ensuring the newly formed line will intersect both vectors at right angles, leading to the accurate formation of the intended line path across the coordinate plane.