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By applying Pythagoras's theorem (the usual two-dimensional version) twice over, prove that the length \(r\) of a three-dimensional vector \(\mathbf{r}=(x, y, z)\) satisfies \(r^{2}=x^{2}+y^{2}+z^{2}\)

Short Answer

Expert verified
\( r^2 = x^2 + y^2 + z^2 \) for a vector \( \mathbf{r} = (x, y, z) \).

Step by step solution

01

Consider the Base Case for Two Dimensions

In two dimensions, the position vector is \( \mathbf{r} = (x, y) \). According to Pythagoras's theorem, the length of this vector \( r \) is calculated as \( r_1 = \sqrt{x^2 + y^2} \). Therefore, \( r_1^2 = x^2 + y^2 \).
02

Incorporate the Third Dimension

Extend the two-dimensional space to three dimensions by adding the third component \( z \). The position vector then becomes \( \mathbf{r} = (x, y, z) \). Initially, ignore the \( z \) component and consider only the 2D calculation: \( r_1 = \sqrt{x^2 + y^2} \).
03

Apply Pythagoras's Theorem in Three Dimensions

Now, treat \( z \) as forming a right triangle with the hypotenuse \( r_1 \) and the length \( r \). Applying Pythagoras's theorem to this new triangle, we have: \( r = \sqrt{r_1^2 + z^2} \).
04

Substitute and Simplify

Substitute \( r_1^2 = x^2 + y^2 \) into the formula from Step 3: \( r = \sqrt{(x^2 + y^2) + z^2} \). Simplifying this expression, we get \( r = \sqrt{x^2 + y^2 + z^2} \). Consequently, \( r^2 = x^2 + y^2 + z^2 \).

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Key Concepts

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

Pythagorean theorem
In mathematics, the Pythagorean theorem is a fundamental principle used to determine the relationship between the sides of a right triangle. The theorem is best known for its application in two dimensions, where it states that the square of the hypotenuse's length is equal to the sum of the squares of the lengths of the other two sides. It is expressed as follows: \[ c^2 = a^2 + b^2 \] Here, \(c\) represents the hypotenuse, while \(a\) and \(b\) are the other two sides. This principle helps in calculating the distance in the Euclidean space, which can be extended to the third dimension.
To apply the Pythagorean theorem in three dimensions, we first handle two of the dimensions, calculating the hypotenuse for them, and then further apply the theorem including the third dimension. This process involves using the theorem twice, which is beautifully demonstrated through the exercise question. By applying the theorem twice, we derive the magnitude of a three-dimensional vector, connecting this ancient principle to modern-day vector mathematics.
three-dimensional vectors
A three-dimensional vector expands beyond the usual x-y plane to include depth, represented by the z-coordinate. Unlike two-dimensional vectors which lie flat, three-dimensional vectors give a more comprehensive picture of spatial positioning.
Mathematically, a three-dimensional vector is written as \( \mathbf{r} = (x, y, z) \), where \(x\), \(y\), and \(z\) are the magnitudes in their respective axes. Understanding these vectors is crucial in fields ranging from physics to computer graphics, where position, velocity, or forces are analyzed in three-dimensional space.
The exercise mentioned utilizes a three-dimensional vector, calculating its magnitude through a step-by-step enhancement of a two-dimensional approach. Using the concept of the Pythagorean theorem, we find the length of this vector, which incorporates all three spatial coordinates.
coordinate systems
Coordinate systems are essential in representing the particular location of points in space. In our daily applications, the most common coordinate system is the Cartesian system. This system uses perpendicular axes (x, y, and z) to define the position of any point in a two-dimensional or three-dimensional space.
The Cartesian coordinate system allows us to break down geometric problems, like computing the magnitude of vectors, with the help of coordinate measurements. Each coordinate represents an axis, which can be handled step-by-step as shown in the exercise solution.
Having a clear understanding of coordinate systems is essential, as it allows for consistent mathematical analysis and is pivotal in translating real-world objects to mathematical models. When dealing with three-dimensional spaces, knowing how these coordinate systems interact enables accurate calculations and designs. Consequently, strengthening our grasp on coordinate systems invariably improves our skill in tackling complex multi-dimensional problems.

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Most popular questions from this chapter

A parallelepiped (a six-faced solid with opposite faces parallel) has one corner at the origin \(O\) and the three edges that emanate from \(O\) defined by vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) Show that the volume of the parallelepiped is \(|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|\)

By evaluating their dot product, find the values of the scalar \(s\) for which the two vectors \(\mathbf{b}=\hat{\mathbf{x}}+s \hat{\mathbf{y}}\) and \(\mathbf{c}=\hat{\mathbf{x}}-s \hat{\mathbf{y}}\) are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.

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In case you haven't studied any differential equations before, I shall be introducing the necessary ideas as needed. Here is a simple excercise to get you started: Find the general solution of the firstorder equation \(d f / d t=f\) for an unknown function \(f(t) .\) [There are several ways to do this. One is to rewrite the equation as \(d f / f=d t\) and then integrate both sides.] How many arbitrary constants does the general solution contain? [Your answer should illustrate the important general theorem that the solution to any \(n\) th-order differential equation (in a very large class of "reasonable" equations) contains \(n\) arbitrary constants.]

[Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from \(\phi_{\mathrm{o}}=20\) degrees, using the values \(R=5 \mathrm{m}\) and \(g=9.8 \mathrm{m} / \mathrm{s}^{2} .\) Make a plot of \(\phi\) against time for two or three periods. ( \(\mathbf{b}\) ) On the same picture, plot the approximate solution (1.57) with the same \(\phi_{\mathrm{o}}=20^{\circ}\) Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for "NDSolve”and how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

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