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Chapter 11: Problem 83

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

Short Answer

Expert verified
Yes, the diagonals of a rhombus are perpendicular to each other which is proven by using vectors and their dot product.

Step by step solution

01

Define the Rhombus Vectors

Let's assume we have a rhombus defined by the vectors \( \mathbf{a} \) and \( \mathbf{b} \). The diagonal vectors \( \mathbf{d1} \) and \( \mathbf{d2} \) of the rhombus can be determined by vector addition. \( \mathbf{d1} = \mathbf{a} + \mathbf{b} \) and \( \mathbf{d2} = \mathbf{a} - \mathbf{b} \)
02

Prove That the Diagonals Are Perpendicular

For the diagonals to be perpendicular, the dot product of \( \mathbf{d1} \) and \( \mathbf{d2} \) should be zero as for any two vectors \( \mathbf{d1} \) and \( \mathbf{d2} \), \( \mathbf{d1} \cdot \mathbf{d2} = 0 \) implies \( \mathbf{d1} \) is perpendicular to \( \mathbf{d2} \). Calculate \( \mathbf{d1} \cdot \mathbf{d2} = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} = |\mathbf{a}|^2 - |\mathbf{b}|^2 \). But since it's a rhombus, all sides are equal thus \( |\mathbf{a}| = |\mathbf{b}| \) which gives \( \mathbf{d1} \cdot \mathbf{d2} = 0 \). Thus the diagonals are perpendicular.
03

Conclude

So, we have proved that the diagonals of a rhombus are perpendicular by using vectors. This uses the principle that if the dot product of two vectors is zero, then the vectors are perpendicular.

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