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Chapter 9: Problem 52

Use vectors to prove that a parallelogram is a rectangle if and only if its diagonals are equal in length.

Short Answer

Expert verified
To prove a parallelogram is a rectangle if its diagonals are equal, vectors can be used to represent the sides and diagonals. Equating the squared magnitudes gives a relation, which implies that the vectors representing sides are perpendicular to each other, thus proving that the parallelogram is a rectangle.

Step by step solution

01

Represent the Parallelogram

Represent the parallelogram ABCD using vectors. We will consider one point (for instance, A) as the origin. Now let's denote AB as vector \( \vec{a} \), AD as vector \( \vec{b} \) and AC and BD as vectors \( \vec{d1} \) and \( \vec{d2} \) respectively.
02

Expression for Diagonals

First, we aim to express the vectors for our diagonals AC and BD i.e., \( \vec{d1} \) and \( \vec{d2} \) in term of vectors \( \vec{a} \) and \( \vec{b} \). The diagonal AC can be given as the sum of vectors AB and BC. Remember that BC = AD = \( \vec{b} \). Hence, \( \vec{d1} = \vec{a} + \vec{b} \). Similarly the diagonal BD can be represented as the sum of vectors BA and AD. Since BA= - AB = -\( \vec{a} \), Hence, \( \vec{d2} = -\vec{a} + \vec{b} \).
03

Equalize the Diagonals

Now, if the parallelogram is a rectangle, then the lengths of diagonals will be equal. According to the property of vectors, the magnitude of vectors \( \vec{d1} \) and \( \vec{d2} \) will be equal, if these vectors represent equal lengths. So, we equate the magnitudes. We square the magnitudes to eliminate the square root for convenience: \( \vec{d1} \cdot \vec{d1} = \vec{d2} \cdot \vec{d2} \).
04

Simplify

We will simplify this equation by distributing and simplifying: \( (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{a} + \vec{b}) \cdot (-\vec{a} + \vec{b}) \). This will simplify to \( \vec{a} \cdot \vec{a} + 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \). This simplifies to \( 4\vec{a} \cdot \vec{b} = 0 \). If \( \vec{a} \) and \( \vec{b} \) are not equal to zero, then \( \vec{a} \cdot \vec{b} = 0 \). Their dot product is zero, which means they are orthogonal (perpendicular) vectors.
05

Final Proof

Given that AD is perpendicular to AB (as shown in previous step), and we know for a parallelogram, opposite sides are parallel and equal, hence the shape ABCD is a rectangle. Therefore we have shown in the problem that a parallelogram is indeed a rectangle if its diagonals are equal.

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Most popular questions from this chapter

Find the angle \(\theta\) between the vectors. $$ \begin{array}{l} \mathbf{u}=3 \mathbf{i}+4 \mathbf{j} \\ \mathbf{v}=-2 \mathbf{j}+3 \mathbf{k} \end{array} $$

Find the vector \(z,\) given that \(\mathbf{u}=\langle 1,2,3\rangle\) \(\mathbf{v}=\langle 2,2,-1\rangle,\) and \(\mathbf{w}=\langle 4,0,-4\rangle\) \(2 \mathbf{u}+\mathbf{v}-\mathbf{w}+3 \mathbf{z}=\mathbf{0}\)

Find the direction cosines of \(u\) and demonstrate that the sum of the squares of the direction cosines is 1. $$ \mathbf{u}=\langle 0,6,-4\rangle $$

In Exercises \(25-28,\) find the direction cosines of \(u\) and demonstrate that the sum of the squares of the direction cosines is 1. $$ \mathbf{u}=\mathbf{i}+2 \mathbf{j}+2 \mathbf{k} $$

Find \((\mathbf{a}) \mathbf{u} \cdot \mathbf{v},(\mathbf{b}) \mathbf{u} \cdot \mathbf{u},(\mathbf{c})\|\mathbf{u}\|^{2},(\mathbf{d})(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) and \((e) u \cdot(2 v)\). $$ \mathbf{u}=\langle 2,-3,4\rangle, \quad \mathbf{v}=\langle 0,6,5\rangle $$
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