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Parametric equations are an effective way to represent lines in three-dimensional space. These equations express the coordinates of points on a line as functions of a parameter, typically labeled as \(t\). For example, the parametric equations \(x = 5 + 3t\), \(y = 5 + 4t\), and \(z = -3 - 5t\) describe a line where:

• The initial point, often called an anchor point, is \((5, 5, -3)\).
• The direction the line extends is captured by the coefficients of \(t\): \(3\) for \(x\), \(4\) for \(y\), and \(-5\) for \(z\).

This characteristic makes parametric equations particularly useful in engineering and physics. They allow for the modeling of motion and paths by varying \(t\) to represent different points in time. When working with these equations, note how each term contributes to the position of the line in space.

The cross product is a critical tool when working with vectors, especially in three-dimensional space. It results in a vector that is perpendicular to the plane formed by two vectors. For two vectors \( \overrightarrow{PQ} = \langle -5, -5, 3 \rangle \) and \( \overrightarrow{d} = \langle 3, 4, -5 \rangle \), the cross product helps in determining the shortest distance between a point and a line.

The cross product formula involves determinants and produces a vector denoted by:\[\overrightarrow{PQ} \times \overrightarrow{d} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-5 & -5 & 3 \3 & 4 & -5 \end{vmatrix}\]After calculating, as shown in the solution, the result is \(\langle 13, -16, -5 \rangle\). This vector can be further used to find the perpendicular distance from a point to a line by examining its magnitude. The cross product is foundational in geometry and physics, especially in cases involving rotational movement or torque.

A direction vector is a vector that indicates the direction of a line in space. In our example, it's derived from the parametric equations as \( \overrightarrow{d} = \langle 3, 4, -5 \rangle \). The components of this vector come from the coefficients of \(t\) in the parametric form of the line:

• \(3\) in the \(x\)-direction
• \(4\) in the \(y\)-direction
• \(-5\) in the \(z\)-direction

This vector not only shows the path of the line but also helps determine distances from points perpendicular to this path. The direction vector is an essential tool in vector calculus and physics for trajectory prediction and line-of-sight analysis.

The magnitude of a vector gives its length or size and is calculated using the Pythagorean theorem in three dimensions. To find the magnitude of a vector like the cross product \( \overrightarrow{PQ} \times \overrightarrow{d} = \langle 13, -16, -5 \rangle \), use the formula:\[\| \overrightarrow{v} \| = \sqrt{x^2 + y^2 + z^2}\]Applying this:\[\| \overrightarrow{PQ} \times \overrightarrow{d} \| = \sqrt{13^2 + (-16)^2 + (-5)^2} = \sqrt{625} = 25\]Similarly, the magnitude of the direction vector \( \overrightarrow{d} = \langle 3, 4, -5 \rangle \) is:\[\| \overrightarrow{d} \| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2}\]The calculated magnitudes are crucial when using formulas that involve distances or orthogonal projections, making them indispensable in both mathematics and physical analysis.