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The line from a vertex of a triangle to the midpoint of the opposite side is called a median of the triangle. If the vertices of a triangle have vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w},\) show that the point on each median that is \(\frac{1}{3}\) the way from the midpoint to the vertex has vector \(\frac{1}{3}(\mathbf{u}+\mathbf{v}+\mathbf{w})\). Conclude that the point \(C\) with vector \(\frac{1}{3}(\mathbf{u}+\mathbf{v}+\mathbf{w})\) lies on all three medians. This point \(C\) is called the centroid of the triangle.

Short Answer

Expert verified
The centroid is \(\frac{1}{3}(\mathbf{u} + \mathbf{v} + \mathbf{w})\), and it lies on all medians.

Step by step solution

01

Identify the Midpoint of a Side

Consider the side opposite the vertex with vector \(\mathbf{u}\). The midpoint of the side between vertices \(\mathbf{v}\) and \(\mathbf{w}\) is given by the average of the position vectors of these two vertices. This is calculated as: \[ \text{Midpoint} = \frac{\mathbf{v} + \mathbf{w}}{2} \].
02

Determine the Vector of the Median

The median from vertex \(\mathbf{u}\) to the midpoint of the opposite side \(\frac{\mathbf{v} + \mathbf{w}}{2}\) can be represented parametrically. The vector equation for this median is: \[ \mathbf{r}(t) = t \mathbf{u} + (1-t) \left( \frac{\mathbf{v} + \mathbf{w}}{2} \right) \].
03

Find the Point One-Third the Way Along the Median

We need the point that is \(\frac{1}{3}\) the way from the midpoint to the vertex \(\mathbf{u}\). Substituting \(t = \frac{2}{3}\) into the vector equation, we have: \[ \mathbf{r}\left(\frac{2}{3}\right) = \frac{2}{3} \mathbf{u} + \frac{1}{3} \left(\mathbf{v} + \mathbf{w}\right) \].
04

Show It's the Centroid

Simplify the expression \(\frac{2}{3} \mathbf{u} + \frac{1}{3} \mathbf{v} + \frac{1}{3} \mathbf{w}\) to \(\frac{1}{3}(\mathbf{u} + \mathbf{v} + \mathbf{w})\). This shows that the point \(C\), with vector \(\frac{1}{3}(\mathbf{u} + \mathbf{v} + \mathbf{w})\), is one-third of the way from the midpoint to the vertex for all three medians.
05

Conclusion

Since the derivation holds for all three medians, the point \(C\) lies on all three medians of the triangle. Therefore, \(C\) is the centroid of the triangle, where each median intersects.

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Key Concepts

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

Median of a Triangle
The concept of the median is a central theme in geometry. A median is a line segment that connects a vertex of a triangle to the midpoint of the opposite side. This does not only highlight the symmetry and balance within the triangle but also divides it into two smaller triangles of equal area.
  • Each triangle has three medians, one from each vertex.
  • The medians help in locating the centroid, which is the center of mass or balance point of the triangle.
  • The point where all three medians intersect is known as the centroid, denoted usually as point C.
Understanding the median enriches your comprehension of a triangle's properties and aids in various geometrical and physical applications.
Vector Representation
Vectors are essential mathematical tools that are used to describe directions and magnitudes in space. In the context of triangles, vectors can represent the position of vertices.
  • For example, if the vertices of a triangle are represented as vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\), you can describe each vertex's position in mathematical terms.
  • The representation of a median using vectors aids in calculations involving points along the median, the midpoint, and vector-based measurements.
  • Vectors simplify computations and make it easier to manipulate equations for finding centroids or other geometric properties.
Thus, vector representation is crucial for translating geometrical problems into algebraic forms, which are easier to solve and analyze.
Parametric Equation
Parametric equations are a powerful tool in geometry. They allow the description of a path, such as a line or curve, using a parameter to express each coordinate.
  • For the median in a triangle, the parametric equation represents every point along the median as a function of a parameter \( t \).
  • The median from vertex \( \mathbf{u} \) to the midpoint \( \frac{\mathbf{v} + \mathbf{w}}{2} \) is expressed using \( t \) as: \( \mathbf{r}(t) = t \mathbf{u} + (1-t) \frac{\mathbf{v} + \mathbf{w}}{2} \).
  • By manipulating \( t \), you can determine specific points along the median, like the point that is \( \frac{1}{3} \) the way from the midpoint to the vertex.
This kind of equation provides a dynamic way to understand movement along paths in the triangle, which is particularly useful in calculus and physics.
Geometric Proof
A geometric proof is a logical series of statements, drawing upon definitions, axioms, and previously established theorems to arrive at a certain conclusion. When proving a point lies on all three medians of a triangle, as is the case with the centroid, you follow a methodical approach.
  • To show that the point \( C \) with vector \( \frac{1}{3}(\mathbf{u} + \mathbf{v} + \mathbf{w}) \) is the centroid, we use step-by-step algebraic manipulation.
  • First, we calculate midpoints and use the vector forms of each median.
  • Then, verify that this point fulfills the conditions for a median across all three sides, thus proving the point is the intersection of medians.
This detailed logical framework convinces mathematically that the calculations are correct, allowing you to understand and prove properties within triangles with full confidence.

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Most popular questions from this chapter

Find the distance between the following pairs of points. a. \(\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right]\) and \(\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]\) b. \(\left[\begin{array}{r}2 \\ -1 \\ 2\end{array}\right]\) and \(\left[\begin{array}{l}2 \\ 0 \\ 1\end{array}\right]\) c. \(\left[\begin{array}{r}-3 \\ 5 \\ 2\end{array}\right]\) and \(\left[\begin{array}{l}1 \\ 3 \\ 3\end{array}\right]\) d. \(\left[\begin{array}{r}4 \\ 0 \\ -2\end{array}\right]\) and \(\left[\begin{array}{l}3 \\ 2 \\ 0\end{array}\right]\)

Let \(\mathbf{u}=\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{l}4 \\ 0 \\ 1\end{array}\right],\) and \(\mathbf{w}=\left[\begin{array}{c}-1 \\ 1 \\ 5\end{array}\right] .\) In each case, find \(\mathbf{x}\) such that: a. \(3(2 \mathbf{u}+\mathbf{x})+\mathbf{w}=2 \mathbf{x}-\mathbf{v}\) b. \(2(3 \mathbf{v}-\mathbf{x})=5 \mathbf{w}+\mathbf{u}-3 \mathbf{x}\)

In each case, find all points of intersection of the given plane and the line \(\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}1 \\\ -2 \\ 3\end{array}\right]+t\left[\begin{array}{r}2 \\ 5 \\\ -1\end{array}\right]\) a. \(x-3 y+2 z=4\) b. \(2 x-y-z=5\) c. \(3 x-y+z=8\) d. \(-x-4 y-3 z=6\)

Find two orthogonal vectors that are both orthogonal to \(\mathbf{v}=\left[\begin{array}{l}1 \\ 2 \\ 0\end{array}\right]\).

Given a rectangular solid with sides of lengths \(1,1,\) and \(\sqrt{2}\), find the angle between a diagonal and one of the longest sides.

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