Problem 42
Question
Find the vector emanating from the origin whose terminal point is the midpoint of the segment joining \((3,2,-1)\) and \((5,-7,2)\).
Step-by-Step Solution
Verified Answer
The vector is \((4, -2.5, 0.5)\).
1Step 1: Determine Midpoint Coordinates
To find the midpoint of the segment joining two points \((3, 2, -1)\) and \((5, -7, 2)\), we use the midpoint formula: \(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)\). Substitute the given points: \(M = \left( \frac{3+5}{2}, \frac{2+(-7)}{2}, \frac{-1+2}{2} \right)\).
2Step 2: Calculate Midpoint
Compute the midpoint by performing the arithmetic: \(M = \left( \frac{8}{2}, \frac{-5}{2}, \frac{1}{2} \right)\). Simplify the fractions: \(M = (4, -2.5, 0.5)\).
3Step 3: Form the Vector from Origin
The vector from the origin \((0,0,0)\) to the midpoint \((4, -2.5, 0.5)\) is simply the coordinates of the midpoint. Therefore, the vector is \((4, -2.5, 0.5)\).
Key Concepts
Midpoint FormulaVector from OriginCoordinate Geometry
Midpoint Formula
When you have two points in space, the midpoint formula helps you find a point exactly halfway between them. This is useful in situations involving geometry and spatial analysis.
To find the midpoint of a segment joining two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), the formula is given by:
\[ M(x, y, z) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
This means you simply take the average of the x-coordinates, the y-coordinates, and the z-coordinates separately to find the midpoint.
In our exercise, plugging in the provided coordinates \( (3, 2, -1) \) and \( (5, -7, 2) \), we find that the midpoint calculation becomes:
1. Average of x-coordinates: \( \frac{3 + 5}{2} = 4 \)
2. Average of y-coordinates: \( \frac{2 + (-7)}{2} = -2.5 \)
3. Average of z-coordinates: \( \frac{-1 + 2}{2} = 0.5 \)
So, the midpoint is \( (4, -2.5, 0.5) \). This method ensures accuracy when working with points in three-dimensional space.
To find the midpoint of a segment joining two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), the formula is given by:
\[ M(x, y, z) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
This means you simply take the average of the x-coordinates, the y-coordinates, and the z-coordinates separately to find the midpoint.
In our exercise, plugging in the provided coordinates \( (3, 2, -1) \) and \( (5, -7, 2) \), we find that the midpoint calculation becomes:
1. Average of x-coordinates: \( \frac{3 + 5}{2} = 4 \)
2. Average of y-coordinates: \( \frac{2 + (-7)}{2} = -2.5 \)
3. Average of z-coordinates: \( \frac{-1 + 2}{2} = 0.5 \)
So, the midpoint is \( (4, -2.5, 0.5) \). This method ensures accuracy when working with points in three-dimensional space.
Vector from Origin
A vector is an entity with both a direction and a magnitude. In vector calculus, vectors originating from the origin are often discussed, as they simplify calculations and visualizations.
When a vector emanates from the origin \( (0, 0, 0) \) to another point, each coordinate of the other point describes the vector itself. For a vector pointing to \( (a, b, c) \), the vector is simply \( (a, b, c) \).
This is especially useful in our exercise because once we find the midpoint, this can be directly used to describe the vector. The coordinates of the midpoint become the components of the vector.
Hence, for the midpoint \( (4, -2.5, 0.5) \), the vector from the origin is
When a vector emanates from the origin \( (0, 0, 0) \) to another point, each coordinate of the other point describes the vector itself. For a vector pointing to \( (a, b, c) \), the vector is simply \( (a, b, c) \).
This is especially useful in our exercise because once we find the midpoint, this can be directly used to describe the vector. The coordinates of the midpoint become the components of the vector.
Hence, for the midpoint \( (4, -2.5, 0.5) \), the vector from the origin is
- x-component: 4
- y-component: -2.5
- z-component: 0.5
Coordinate Geometry
Coordinate geometry, or analytic geometry, involves describing the position of points and the analysis of geometric figures using a coordinate system. It is a critical component of vector calculus.
In coordinate geometry, points are expressed as coordinates, which can easily be used in calculations. The connection between algebra and geometry gives you the ability to precisely describe shapes, lines, and surfaces.
Through coordinate geometry, you can use formulae like the midpoint formula or calculate vectors that describe various paths in space.
For instance, in our example, we smoothly transition from determining midpoint coordinates using algebra to forming a vector, which provides a geometric representation.
Coordinate geometry allows you to easily and accurately calculate distances, slopes, and positions in either two-dimensional or three-dimensional spaces.
Therefore, mastering this concept is key to advancing your understanding of more complex geometric properties and vector calculus.
In coordinate geometry, points are expressed as coordinates, which can easily be used in calculations. The connection between algebra and geometry gives you the ability to precisely describe shapes, lines, and surfaces.
Through coordinate geometry, you can use formulae like the midpoint formula or calculate vectors that describe various paths in space.
For instance, in our example, we smoothly transition from determining midpoint coordinates using algebra to forming a vector, which provides a geometric representation.
Coordinate geometry allows you to easily and accurately calculate distances, slopes, and positions in either two-dimensional or three-dimensional spaces.
Therefore, mastering this concept is key to advancing your understanding of more complex geometric properties and vector calculus.
Other exercises in this chapter
Problem 42
Evaluate the integrals $$ \int_{-1}^{1}\left[(1+t)^{3 / 2} \mathbf{i}+(1-t)^{3 / 2} \mathbf{j}\right] d t $$
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find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1} .\)
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The sphere \((x-1)^{2}+(y+2)^{2}+(z+1)^{2}=10\) intersects the plane \(z=2\) in a circle. Find the circle's center and radius.
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A point moves around the circle \(x^{2}+y^{2}=25\) at constant angular speed of 6 radians per second starting at \((5,0)\). Find expressions for \(\mathbf{r}(t)
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