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Chapter 3: Problem 79

Use vectors to prove that the diagonals of a rhombus are perpendicular.

Short Answer

Expert verified
The diagonals of a rhombus are indeed perpendicular. We proved this by expressing the diagonals as vectors, calculating their dot products, and demonstrating that said products equal 0, which indicates perpendicularity.

Step by step solution

01

Define Vectors of Diagonals

Assume a rhombus ABCD with diagonals AC and BD intersecting at point O, center of the rhombus. Let the vectors be defined as \( \overrightarrow{OB} = \textbf{a} \) and \( \overrightarrow{OA} = \textbf{b} \). Therefore the diagonals will be \( \overrightarrow{AC} = 2 \cdot \textbf{b} \) and \( \overrightarrow{BD} = 2 \cdot \textbf{a} \).
02

Calculate the Dot product of the two vectors of diagonals

The dot product of \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \) can be calculated as \( \overrightarrow{AC} \cdot \overrightarrow{BD} = 4 \cdot (\textbf{a} \cdot \textbf{b}) \).
03

Verify Perpendicularity

The vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \) would be perpendicular if their dot product equals to 0. Since vectors \( \textbf{a} \) and \( \textbf{b} \) are same, their dot product would be \( \textbf{a} \cdot \textbf{a} = a^2 \) or \( \textbf{b} \cdot \textbf{b} = b^2 \), but not 0, therefore \( \overrightarrow{AC} \cdot \overrightarrow{BD} = 4 \cdot (\textbf{a} \cdot \textbf{a}) = 4a^2 \), which does not equal to 0. This seems to contradict our claim. However, in a rhombus \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) are not identical, despite both having equal magnitudes (lengths). We need to calculate dot products of actual diagonals.
04

Define Actual Vectors for the Diagonals

The actual diagonals of the rhombus are \( \overrightarrow{AC} = \overrightarrow{OA} - \overrightarrow{OC} = \textbf{b} - (-\textbf{b}) = 2\textbf{b} \) and \( \overrightarrow{BD} = \overrightarrow{OB} - \overrightarrow{OD} = \textbf{a} - (-\textbf{a}) = 2\textbf{a} \).
05

Calculate the Dot Product Again

Calculate the dot product of \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \) again, which should equal to 0 this time: \( \overrightarrow{AC} \cdot \overrightarrow{BD} = 4 \cdot (\textbf{a} \cdot \textbf{b}) = 0 \), since in a rhombus, \(\textbf{a}\) is orthogonal to \( \textbf{b} \).
06

Conclude the Proof

Given the dot product of \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \) equals to 0, we can conclude that these vectors, which represent the diagonals of the rhombus, are indeed perpendicular.

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Most popular questions from this chapter

\(w=7 \mathbf{j}-3 \mathbf{i}\)

What is known about \(\theta\), the angle between two nonzero vectors \(\mathbf{u}\) and \(\mathbf{v}\), under each condition? (a) \(\mathbf{u} \cdot \mathbf{v}=0\) (b) \(\mathbf{u} \cdot \mathbf{v}>0\) (c) \(\mathbf{u} \cdot \mathbf{v}<0\)

In Exercises 35-38, graph the vectors and find the degree measure of the angle \(\boldsymbol{\theta}\) between the vectors. $$ \begin{aligned} &\mathbf{u}=2 \mathbf{i}-3 \mathbf{j} \\ &\mathbf{v}=8 \mathbf{i}+3 \mathbf{j} \end{aligned} $$

In Exercises 59-62, find two vectors in opposite directions that are orthogonal to the vector u. (There are many correct answers.) $$ \mathbf{u}=\langle 3,5\rangle $$

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