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The direction vector is crucial when dealing with line segments in a three-dimensional space. It points in the direction from one point to another. To find it, you simply subtract the coordinates of the starting point from the coordinates of the endpoint. For example, given points \(P = (-4, 7, 5)\) and \(Q = (2, -2, -3)\), you subtract \(P\) from \(Q\):\[\textbf{d} = Q - P = (2 - (-4), -2 - 7, -3 - 5) = (6, -9, -8).\] This direction vector \((6, -9, -8)\) tells you how to move from point \(P\) to point \(Q\). It's essential for constructing equations and finding points along the line segment.

A fractional distance parameter \( t \) determines how far you are along the line segment from one point to another. For instance, if \( t = \frac{1}{7} \), it means you're one-seventh of the way from point \(P\) to point \(Q\). This parameter can be used to find new points on the line segment. With the direction vector \(\textbf{d} \), the formula is: \[ A = P + t \times \textbf{d}. \] This equation helps you find any point on the segment at a specified fraction of the total distance.

To find a specific point on a line segment defined by two endpoints \(P\) and \(Q\), you need both the direction vector \(\textbf{d}\) and the fractional distance \( t \). For instance, to find the point \(A\) which is \(\frac{1}{7}\) of the way from \(P\) to \(Q\), follow these steps:

• Compute the direction vector: \(\textbf{d} = (6, -9, -8)\).
• Use the formula \(A = P + \frac{1}{7} \times \textbf{d}\).

Substituting the values:
\[ A = (-4, 7, 5) + \frac{1}{7} \times (6, -9, -8) = \bigg(-4 + \frac{6}{7}, 7 - \frac{9}{7}, 5 - \frac{8}{7}\bigg) \] Simplify to get: \[ A = \bigg(-\frac{28}{7} + \frac{6}{7}, \frac{49}{7} - \frac{9}{7}, \frac{35}{7} - \frac{8}{7}\bigg) = \bigg(-\frac{22}{7}, \frac{40}{7}, \frac{27}{7}\bigg) \] Now, you've found the coordinates \(\left(-\frac{22}{7}, \frac{40}{7}, \frac{27}{7} \right)\), which is the precise point one-seventh of the way from \(P\) to \(Q\).