Question

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

Short answer

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

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 = \( \vec{b} - \vec{a} \).

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 θ, where θ is the angle between the vectors. If the vectors are perpendicular to each other, then θ is 90° and cos θ is 0. As a result, the dot product will be 0 if the vectors are perpendicular to each other.

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 = \( (\vec{a} + \vec{b}) \) . \( (\vec{b} - \vec{a}) \)).

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.