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Chapter 9: Problem 51

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

Short Answer

Expert verified
The diagonals of a rhombus are perpendicular to each other as their dot product comes out to be zero, which signifies perpendicularity in terms of vectors.

Step by step solution

01

Construct the Vectors

Consider a rhombus ABCD. Suppose AB=AD=a and BC=CD=b. Let's assign vectors to the sides of the rhombus such that the vector AB is \( \vec{a} \) and the vector AD is \( \vec{b} \). Now the diagonals can be represented using these vectors as AC = AB + BC = \( \vec{a} + \vec{b} \) and BD = AD -AB = \( \vec{b} - \vec{a} \).
02

Find the Dot Product

The next step is to find the dot product of vectors AC and BD. The dot product between two vectors is given by the formula A.B = |A| |B| cos 胃, where 胃 is the angle between the vectors. If the vectors are perpendicular to each other, then 胃 is 90掳 and cos 胃 is 0. As a result, the dot product will be 0 if the vectors are perpendicular to each other.
03

Plug in the Vectors

Here, when plugging in our vectors AC and BD into the dot product equation, it should look like this: (AC.BD = \( (\vec{a} + \vec{b}) \) . \( (\vec{b} - \vec{a}) \)).
04

Simplify the Dot Product

By expanding and simplifying this, we get \( \vec{a}.\vec{b} + \vec{b}.\vec{b} - \vec{a}.\vec{a} - \vec{b}.\vec{a} \) which will simplify to 0, as \( \vec{a}.\vec{b} = \vec{b}.\vec{a} \) and \( \vec{a}.\vec{a} = \vec{b}.\vec{b} \). So, the dot product is 0, which means the vectors are perpendicular.

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