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Chapter 12: Problem 35

Find a. the direction of \(\overrightarrow{P_{1} P_{2}}\) and b. the midpoint of line segment \(P_{1} P_{2}\) $$P_{1}(-1,1,5) \quad P_{2}(2,5,0)$$

Short Answer

Expert verified
Direction as a unit vector is \((3/\sqrt{50}, 4/\sqrt{50}, -5/\sqrt{50})\); midpoint is (0.5, 3, 2.5).

Step by step solution

01

Identifying Components of Vector

To find the direction of the vector \(\overrightarrow{P_{1}P_{2}}\), we need to determine its components. Given \(P_{1}(-1,1,5)\) and \(P_{2}(2,5,0)\), calculate the changes in each coordinate: \(x\), \(y\), and \(z\). The components are \((2 - (-1), 5 - 1, 0 - 5)\), which simplifies to \((3,4,-5)\).
02

Calculating Unit Vector for Direction

The direction of \(\overrightarrow{P_{1}P_{2}}\) is expressed as a unit vector. Calculate the magnitude (length) of \(\overrightarrow{P_{1}P_{2}}\) using the formula: \(\sqrt{(3)^2 + (4)^2 + (-5)^2}\). This yields \(\sqrt{50}\). The unit vector is \((3/\sqrt{50}, 4/\sqrt{50}, -5/\sqrt{50})\).
03

Calculating Midpoint

To find the midpoint of the line segment \(P_{1} P_{2}\), use the midpoint formula: \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)\). For the given points, this computes to \(\left(\frac{-1 + 2}{2}, \frac{1 + 5}{2}, \frac{5 + 0}{2}\right)\), which results in \((0.5, 3, 2.5)\).
04

Final Result Assembly

The direction of \(\overrightarrow{P_{1}P_{2}}\) is a unit vector \((3/\sqrt{50}, 4/\sqrt{50}, -5/\sqrt{50})\) and the midpoint of \(P_{1} P_{2}\) is \((0.5, 3, 2.5)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coordinate Geometry
Coordinate Geometry is the study of geometric figures using a coordinate system. In three-dimensional space, each point is represented by an ordered triplet: \((x, y, z)\). This helps us describe geometric shapes and properties numerically and algebraically.
  • Points are plotted using coordinates, indicating their location in space.
  • Lines and planes can be expressed as linear equations that define their position.
  • Distance, angle, and area calculations become straightforward using coordinates.

In this context, vectors, like \(\overrightarrow{P_{1}P_{2}}\), are used to describe the direction and distance between two points. For points \(P_{1}(-1,1,5)\) and \(P_{2}(2,5,0)\), their vector connects these two specific locations in three-dimensional space.
Vector Components
A vector is defined by its components, which show the vector's direction and magnitude in a coordinate system. For a vector connecting two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), its components are the differences between the corresponding coordinates:
  • Change in x (horizontal): \((x_2 - x_1)\)
  • Change in y (vertical): \((y_2 - y_1)\)
  • Change in z (depth): \((z_2 - z_1)\)

For \(\overrightarrow{P_{1}P_{2}}\), with \(P_{1}(-1,1,5)\) and \(P_{2}(2,5,0)\), the components are calculated as:
  • \(\Delta x = 2 - (-1) = 3\)
  • \(\Delta y = 5 - 1 = 4\)
  • \(\Delta z = 0 - 5 = -5\)

This yields the vector components \(\overrightarrow{P_{1}P_{2}} = (3,4,-5)\). These components help in further calculations like finding magnitude and direction.
Unit Vectors
Unit vectors are vectors with a magnitude of one, used to indicate direction without magnitude. They are fundamental in representing direction in vector spaces.
  • Given a vector's components, its magnitude is found using: \(\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}\).
  • To get a unit vector, divide each component of the vector by its magnitude.

For \(\overrightarrow{P_{1}P_{2}} = (3,4,-5)\), the magnitude is \(\sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{50}\).
  • The unit vector then is \(\left(\frac{3}{\sqrt{50}}, \frac{4}{\sqrt{50}}, \frac{-5}{\sqrt{50}}\right)\).

Unit vectors simplify the understanding of vector direction, stripping away magnitude details while preserving directionality.
Midpoint Formula
The midpoint formula calculates the center point of a line segment defined by two endpoints. This formula provides the average of the coordinates of the endpoints, yielding a point directly between them.
  • The midpoint \(M\) of a segment with endpoints \(x_1, y_1, z_1\) and \(x_2, y_2, z_2\) is computed as:
\[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)\]
For our points \(P_{1}(-1,1,5)\) and \(P_{2}(2,5,0)\), the midpoint is
  • \(\left(\frac{-1+2}{2}, \frac{1+5}{2}, \frac{5+0}{2}\right) = (0.5, 3, 2.5)\).

This result provides a balanced point between \(P_{1}\) and \(P_{2}\), illustrating an exact middle position on the line segment.

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Most popular questions from this chapter

Location A bird flies from its nest \(5 \mathrm{km}\) in the direction \(60^{\circ}\) north of east, where it stops to rest on a tree. It then flies \(10 \mathrm{km}\) in the direction due southeast and lands atop a telephone pole. Place an \(x y\) -coordinate system so that the origin is the bird's nest, the \(x\) -axis points east, and the \(y\) -axis points north. a. At what point is the tree located? b. At what point is the telephone pole?

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Find the point equidistant from the points (0,0,0),(0,4,0) \((3,0,0),\) and (2,2,-3).
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