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Use vectors to show that the line joining the midpoints of two sides of a triangle is parallel to the third side and half as long.

Short Answer

Expert verified
The line joining midpoints of two sides in a triangle is parallel to the third side and half as long.

Step by step solution

01

Define Triangle Vertices

Consider a triangle with vertices labeled as \(A\), \(B\), and \(C\). Let's define the position vectors of these vertices relative to a common origin as \(\vec{a}\), \(\vec{b}\), and \(\vec{\), respectively.
02

Calculate Midpoints of Sides

The midpoint \(M\) of the side \(AB\) can be calculated using the formula for the midpoint of a line segment: \(\vec{m} = \frac{\vec{a} + \vec{b}}{2}\). Similarly, the midpoint \(N\) of the side \(AC\) is \(\vec{n} = \frac{\vec{a} + \vec{c}}{2}\).
03

Derive Vector for Midpoint Line

The vector connecting the midpoint \(M\) to midpoint \(N\) is given by: \(\vec{MN} = \vec{n} - \vec{m} = \left(\frac{\vec{a} + \vec{c}}{2}\right) - \left(\frac{\vec{a} + \vec{b}}{2}\right) = \frac{\vec{c} - \vec{b}}{2}\).
04

Establish Parallelism to Third Side

The vector form of the third side \(BC\) is \(\vec{BC} = \vec{c} - \vec{b}\). Comparing this with the vector for line \(MN\), we can see \(\vec{MN} = \frac{1}{2} \vec{BC}\), thus proving that \(MN\) is parallel to \(BC\).
05

Verify Length Relationship

Since \(\vec{MN} = \frac{1}{2} \vec{BC}\), it is evident that the magnitude of \(MN\) is half the magnitude of \(BC\), i.e., \(|\vec{MN}| = \frac{1}{2} |\vec{BC}|\). This confirms that \(MN\) is half as long as \(BC\).

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Key Concepts

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

Midpoint Theorem
The midpoint theorem is a fundamental concept in geometry that relates the midpoints of the sides of a triangle. This theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is exactly half its length. This theorem can serve as a useful tool for proving properties related to similar triangles and parallel line segments.

To understand how this theorem applies using vector geometry, consider triangle ABC. First, we find the midpoints of sides AB and AC. The midpoint of line segment AB, labeled as M, has the position vector \( \vec{m} = \frac{\vec{a} + \vec{b}}{2} \). Similarly, the midpoint of AC, labeled as N, has the position vector \( \vec{n} = \frac{\vec{a} + \vec{c}}{2} \).
  • The line segment MN connects these midpoints.
  • This line segment is crucial in applying the midpoint theorem.
  • Using vectors allows for clear and simple derivation of relationships.
Vectors in Triangles
Vectors are a powerful tool for analyzing triangles in geometry. They allow for precise calculations and prove geometric properties such as parallelism and proportional lengths. In the scenario of the triangle ABC, each vertex is assigned a position vector relative to an origin.

For this exercise, the vertices A, B, and C have vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \). Calculating midpoints uses these vectors to create expressions like \( \vec{m} = \frac{\vec{a} + \vec{b}}{2} \) and \( \vec{n} = \frac{\vec{a} + \vec{c}}{2} \).
By using these midpoint vectors, you can define the vector for the line segment MN, which is \( \vec{MN} = \vec{n} - \vec{m} = \frac{\vec{c} - \vec{b}}{2} \). When vectors represent these components:
  • They help visualize and verify parallelism and proportions in geometry.
  • They ensure accuracy in the geometric proofs and properties.
  • The simplicity of vector operations simplifies complex geometric problems.
Line Segments and Parallelism
In triangle geometry, recognizing parallelism between line segments is key to understanding the relationships between various parts of the triangle. By calculating the vector from one midpoint to another, you can demonstrate this parallelism easily.

For the triangle ABC, the segment MN connecting midpoints M and N has the vector \( \vec{MN} = \frac{\vec{c} - \vec{b}}{2} \). On the other hand, the vector \( \vec{BC} \) for the side BC is given as \( \vec{c} - \vec{b} \). By comparing these vectors:
  • The line segment MN is parallel to BC because \( \vec{MN} \) is a scalar multiple (\( \frac{1}{2} \)) of \( \vec{BC} \).
  • The scalar factor signifies that the length of MN is half the length of BC.
  • This parallelism confirms the midpoint theorem using a vector approach.
Parallel line segments and their relationships assist in understanding and proving geometric properties within triangles.

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Most popular questions from this chapter

Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be pairwise orthogonal vectors. a. Show that \(\|\mathbf{u}+\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{u}\|^{2}+\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}\). b. If \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are all the same length, show that they all make the same angle with \(\mathbf{u}+\mathbf{v}+\mathbf{w}\).

Let \(P_{0}\) be a point with vector \(\mathbf{p}_{0},\) and let \(a x+b y+c z=d\) be the equation of a plane with normal \(\mathbf{n}=\left[\begin{array}{l}a \\\ b \\ c\end{array}\right]\) a. Show that the point on the plane closest to \(P_{0}\) has vector \(\mathbf{p}\) given by $$ \mathbf{p}=\mathbf{p}_{0}+\frac{d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)}{\|\mathbf{n}\|^{2}} \mathbf{n} $$ \(\left[\right.\) Hint \(: \mathbf{p}=\mathbf{p}_{0}+t \mathbf{n}\) for some \(t,\) and \(\left.\mathbf{p} \cdot \mathbf{n}=\mathbf{d} .\right]\) b. Show that the shortest distance from \(P_{0}\) to the plane is \(\frac{\left|d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)\right|}{\|\mathbf{n}\|}\). c. Let \(P_{0}^{\prime}\) denote the reflection of \(P_{0}\) in the planethat is, the point on the opposite side of the plane such that the line through \(P_{0}\) and \(P_{0}^{\prime}\) is perpendicular to the plane. Show that \(\mathbf{p}_{0}+2 \frac{d-\left(\mathbf{p}_{0} \cdot \mathbf{n}\right)}{\|\mathbf{n}\|^{2}} \mathbf{n}\) is the vector of \(P_{0}^{\prime}\)

Show that \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})\) need not equal \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) by calculating both when $$ \mathbf{u}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathbf{v}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right], \text { and } \mathbf{w}=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] $$

In each case, verify that the points \(P\) and \(Q\) lie on the line. $$ \begin{array}{ll} \text { a. } & x=3-4 t \quad P(-1,3,0), Q(11,0,3) \\ & y=2+t \\ & z=1-t \\ \text { b. } & x=4-t \quad P(2,3,-3), Q(-1,3,-9) \\ & y=3 \\ & z=1-2 t \end{array} $$

a. Does the line through \(P(1,2,-3)\) with direction vector \(\mathbf{d}=\left[\begin{array}{r}1 \\ 2 \\ -3\end{array}\right]\) lie in the plane \(2 x-y-z=3 ?\) Explain. b. Does the plane through \(P(4,0,5), Q(2,2,1),\) and \(R(1,-1,2)\) pass through the origin? Explain.

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