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Chapter 7: Problem 86

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

Short Answer

Expert verified
The diagonals of a rhombus are always perpendicular because their dot product is equal to 0.

Step by step solution

01

Define the Vectors

Select any rhombus ABCD. Assign vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \) to the two adjacent sides of the rhombus. Represent the diagonals \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \) as the sum or difference of \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \), respectively. In other words, \( \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD} \) and \( \overrightarrow{BD} = \overrightarrow{AB} - \overrightarrow{AD} \).
02

Calculate the Dot Product

The dot product of \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \) can be computed as \( \overrightarrow{AC} \cdot \overrightarrow{BD} = (\overrightarrow{AB} + \overrightarrow{AD}) \cdot (\overrightarrow{AB} - \overrightarrow{AD}) \).
03

Simplify the Dot Product

Expanding the dot product using the distributive property, we get \( \overrightarrow{AC} \cdot \overrightarrow{BD} = (\overrightarrow{AB} \cdot \overrightarrow{AB}) - (\overrightarrow{AD} \cdot \overrightarrow{AD}) \). Taking dot product of a vector with itself gives the square of its magnitude. Since, in a rhombus, the side lengths \(\|\overrightarrow{AB}\|\) and \(\|\overrightarrow{AD}\|\) are equal, the difference is 0.
04

Conclude

Because the dot product \( \overrightarrow{AC} \cdot \overrightarrow{BD} = 0 \), we can conclude that diagonals \( AC \) and \( BD \) are perpendicular. So, for any rhombus, its diagonals are always perpendicular.

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