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\(\mathrm{A}\) fisherman wants to know if his fly rod will fit in a rectangular \(2 \mathrm{ft} \times 3 \mathrm{ft} \times 4 \mathrm{ft}\) packing box. What is the longest rod that fits in this box?

Short Answer

Expert verified
Answer: The maximum length of a fishing rod that will fit inside the rectangular packing box is approximately 5.39 ft.

Step by step solution

01

Identify the dimensions of the box

The dimensions of the rectangular box are given as 2 ft x 3 ft x 4 ft. Let's denote these dimensions as a, b, and c respectively. So, we have a = 2 ft, b = 3 ft, and c = 4 ft.
02

Apply Pythagorean theorem in three dimensions

We need to find the length of the longest rod, which is the length of the box's diagonal. In three dimensions, the Pythagorean theorem is given by the formula: \(d^2 = a^2 + b^2 + c^2\), where d is the diagonal length and a, b, and c are the dimensions of the box.
03

Calculate the diagonal

Plug the values of a, b, and c into the formula: \(d^2 = (2)^2 + (3)^2 + (4)^2\). Simplifying, we have: \(d^2 = 4 + 9 + 16 = 29\). To find d, we need to take the square root of 29. Therefore, \(d = \sqrt{29}\).
04

Find the numerical value of the diagonal

Using a calculator, find the square root of 29: \(d \approx 5.39 \mathrm{ft}\).
05

Final answer

The longest rod that fits in this packing box is approximately 5.39 ft in length.

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Key Concepts

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

Pythagorean theorem
The Pythagorean theorem is a fundamental principle in geometry. It's originally applicable in two dimensions, specifically for right triangles. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Mathematically, it is expressed as:\[ c^2 = a^2 + b^2 \]Where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides. This principle is very useful, not just in theoretical problems, but in real-world applications as well. It allows us to calculate distances and lengths that aren't immediately apparent.

In a three-dimensional world, like when calculating the diagonal of a rectangular prism (box), we adjust the Pythagorean theorem to include three dimensions. It works by incorporating a third perpendicular dimension. This will help you find the longest distance across a box, which is crucial for determining if an object like a fishing rod can fit inside it.
Rectangular prism
A rectangular prism is a 3D object which is also known as a cuboid. It's a shape that has six faces, all of which are rectangles. Its edges are mutually perpendicular, making it a simple, and highly symmetrical shape.

The dimensions of a rectangular prism are defined by its length, width, and height, often denoted as \(a\), \(b\), and \(c\). For example, in the given problem, the dimensions are 2 ft x 3 ft x 4 ft, where 2 ft can be the length, 3 ft the width, and 4 ft the height.

Rectangular prisms are common in daily life; think of packing boxes, rooms, and bricks. Understanding how such shapes function, particularly when determining space and volume or fitting other objects like rods within them, is a helpful skill. The key in these calculations is to remember that such prisms extend the basic principles of two-dimensional rectangles into three-dimensional space.
Diagonal calculation
The concept of diagonal calculation in three-dimensional space is an extension of the two-dimensional diagonal. For a rectangle, the diagonal can be found using the simple Pythagorean theorem. However, for a rectangular prism, we incorporate all three dimensions to find the space diagonal.

The formula for calculating this length, \(d\), incorporates all dimensions of the prism: \[ d^2 = a^2 + b^2 + c^2 \]Using this, we can find \(d\), the longest diagonal that spans from one corner of the box to its opposite.

After you've set up the equation, simply substitute the dimensions into it, sum the squares, and take the square root to get the diagonal length. For example, substituting the dimensions 2 ft, 3 ft, and 4 ft into the formula provides:\[ d^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29\]Thus, \(d = \sqrt{29}\) and the diagonal approximately equals 5.39 ft. This method ensures that you find the maximum length that fits within the box, crucial when considering objects like fishing rods.

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Most popular questions from this chapter

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}\)

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Let \(\mathbf{v}=\langle a, b, c\rangle\) and let \(\alpha, \beta\) and \(\gamma\) be the angles between \(\mathbf{v}\) and the positive \(x\) -axis, the positive \(y\) -axis, and the positive \(z\) -axis, respectively (see figure). a. Prove that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) b. Find a vector that makes a \(45^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? c. Find a vector that makes a \(60^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? d. Is there a vector that makes a \(30^{\circ}\) angle with i and \(\mathbf{j}\) ? Explain. e. Find a vector \(\mathbf{v}\) such that \(\alpha=\beta=\gamma .\) What is the angle?

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Find a general expression for a nonzero vector orthogonal to the plane containing the curve. $$\begin{aligned} \mathbf{r}(t)=&(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j} \\\ &+(e \cos t+f \sin t) \mathbf{k} \end{aligned}$$ where \(\langle a, c, e\rangle \times\langle b, d, f\rangle \neq \mathbf{0}.\)

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