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Understanding how to calculate a displacement vector is crucial in mapping the direct path between two points in space. Displacement is a vector quantity, which means it has both magnitude and direction. It represents the straight-line distance from the starting point to the final position, including the direction.

To calculate the displacement vector, you subtract the coordinates of the starting point from the end point for each dimension. In the context of the given exercise, we have points C and D; we calculate the displacement vector \( \overrightarrow{DC} \) by subtracting each coordinate of point D from point C, resulting in the vector \( \overrightarrow{DC} = \langle 1, 5, -8 \rangle \) for the two points C(2,9,0) and D(1,4,8). This simplification gives you not just the distance but also the direction from D to C.

When we talk about the distance in a spatial context, we refer to the length of the shortest path between two points. Unlike displacement, distance does not consider direction; it’s a scalar quantity meaning it is only concerned with magnitude.

To find the distance between two points in three-dimensional space, you use the distance formula derived from the Pythagorean theorem, applicable when dealing with right triangles. In the context of the points C and D from our exercise, we use the displacement vector \( \overrightarrow{DC} = \langle 1, 5, -8 \rangle \) to find the distance with the formula \( \sqrt{x_{DC}^2 + y_{DC}^2 + z_{DC}^2} \), ultimately finding the straight-line distance to be \( \sqrt{90} \) units.

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This method combines algebra and geometry by graphing shapes and curves to represent them with equations.

In coordinate geometry, the position of a point in space is described with respect to an ordered pair (in two dimensions) or triplet (in three dimensions), which corresponds to the coordinates on the axes. For instance, point C in our problem is denoted as (2,9,0) which tells us its position relative to the coordinate axes in a three-dimensional space. It's this structured system that allows us to make precise calculations about distances and vectors, serving as the backbone for the earlier discussed computations.

A vector's magnitude corresponds to its 'length' or 'size,' and can be understood as the distance from the initial point, often the origin, to the terminal point of the vector.

For finding the magnitude of a displacement vector in three-dimensional space, one would employ the same distance formula used to calculate the straight-line distance between two points. The magnitude of the displacement vector \( \overrightarrow{DC} \) from our example would be calculated as \( \sqrt{1^2 + 5^2 + (-8)^2} \), or \( \sqrt{90} \), which quantifies how 'far away' and in what direction point C is from point D, regardless of the path taken to get there.