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Chapter 6: Problem 2

If \(\mathbf{A}\) and \(\mathbf{B}\) are the diagonals of a parallelogram, find a vector formula for the area of the parallelogram.

Short Answer

Expert verified
\(\frac{1}{2}\|\mathbf{A} \times \mathbf{B}\|\)

Step by step solution

01

Understand the relationship between diagonals and area

In a parallelogram, the area can be found using the cross product of two adjacent sides. However, given the diagonals \(\mathbf{A}\) and \(\mathbf{B}\), another approach involves using these diagonals.
02

Express sides in terms of diagonals

Let the vertices of the parallelogram be \(\mathbf{O}\), \(\mathbf{P}\), \(\mathbf{Q}\), and \(\mathbf{R}\) such that \(\mathbf{A} = \mathbf{OP} + \mathbf{OR}\) and \(\mathbf{B} = \mathbf{OQ} + \mathbf{OR}\). The sides of the parallelogram formed by \(\mathbf{P}\) and \(\mathbf{Q}\), for example, can be expressed as \(\mathbf{a}\) and \(\mathbf{b}\).
03

Use vector properties to find area

The area of the parallelogram is mathematically \(\|\mathbf{a} \times \mathbf{b}\|\). Since \(\mathbf{a}\) and \(\mathbf{b}\) can be represented as a half-sum of diagonals, we modify it to \(\|\frac{1}{2}(\mathbf{A} \times \mathbf{B})\|\).
04

Simplify the expression

Finally, the area of the parallelogram can be simplified to \(\frac{1}{2}\|\mathbf{A} \times \mathbf{B}\|\). This is derived from the fact that diagonals of a parallelogram bisect each other and split it into two congruent triangles.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product
The cross product is a powerful tool in vector calculus, used to find a vector perpendicular to two given vectors. If you have vectors \(\textbf{u}\) and \(\textbf{v}\), their cross product, denoted \(\textbf{u} \times \textbf{v}\), is a vector perpendicular to both. Mathematically, the cross product is given by the determinant formula:

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Most popular questions from this chapter

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint(\operatorname{curl} \mathbf{V}) \cdot \mathbf{n} d \sigma\) over the part of the surface \(z=9-x^{2}-9 y^{2}\) above the \((x, y)\) plane if \(\mathbf{V}=2 x y \mathbf{i}+\left(x^{2}-2 x\right) \mathbf{j}-x^{2} z^{2} \mathbf{k}\)

The force acting on a moving charged particle in a magnetic field \(\mathbf{B}\) is \(\mathbf{F}=q(\mathbf{v} \times \mathbf{B})\) where \(q\) is the electric charge of the particle, and \(\mathbf{v}\) is its velocity. Suppose that a particle moves in the \((x, y)\) plane with a uniform \(\mathbf{B}\) in the \(z\) direction. Assuming Newton's second law, \(m d \mathbf{v} / d t=\mathbf{F},\) show that the force and velocity are perpendicular and that both have constant magnitude. Hint: Find \((d / d t)(\mathbf{v} \cdot \mathbf{v}).\)

Evaluate each integral in the simplest way possible. $$\iint \mathbf{P} \cdot \mathbf{n} d \sigma \text { over the upper half of the sphere } r=1 \text { if } \mathbf{P}=\operatorname{curl}(\mathbf{j} x-\mathbf{k} z)$$

Evaluate each integral in the simplest way possible. $$\oint \mathbf{F} \cdot d \mathbf{r} \text { around the circle } x^{2}+y^{2}+2 x=0, \text { where } \mathbf{F}=y \mathbf{i}-x \mathbf{j}$$

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iiint \nabla \cdot \mathbf{V} d \tau\) over the unit cube in the first octant, where $$\mathbf{V}=\left(x^{3}-x^{2}\right) y \mathbf{i}+\left(y^{3}-2 y^{2}+y\right) x \mathbf{j}+\left(z^{2}-1\right) \mathbf{k}$$.
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