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Find the "distance"' between the " points "? \(\begin{array}{ll}\text { (a) }(4,-1,2,7) & \text { and }(2,3,1,9) ; \\ \text { (b) }(-1,5,-3,2,4) & \text { and }(2,6,2,7,6)\end{array}\)

Short Answer

Expert verified
Part (a): 6. Part (b): 8.

Step by step solution

01

- Understanding the Distance Formula in Higher Dimensions

The distance between two points in a space of dimension can be found using the generalization of the Euclidean distance formula. The formula for the distance between points is: -----
02

- Applying the Formula to Part (a)

Given points The distance
03

- Simplifying the Expression

Subtract the respective coordinates and square the results: The difference in x-coordinates is -2 = 2. Squaring this gives 4.The difference in y-coordinates is -3-1 = -4. Squaring this gives 16.The difference in z-coordinates is -1 = 1. Squaring this gives 1.The difference in w-coordinates is -7-9 = -2. Squaring this gives 4. Adding these up: , then taking the square root gives us .
04

- Applying the Formula to Part (b)

Apply the distance formula to the points and : :Simplifying this, we get:Adding these, we have .Taking the square root of this sum, we get the final distance.

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Key Concepts

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

distance formula
The distance formula is crucial for determining the space between two points in any dimension. It's a generalization of the 2D Euclidean distance formula you've probably encountered: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
In higher dimensions, we extend this concept by adding more terms for each additional dimension. For example, the distance \(d\) between two points \(P_1(x_1, y_1, z_1, w_1, \.\.)\) and \(P_2(x_2, y_2, z_2, w_2, \.\.)\) in a four-dimensional space would be:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 + (w_2 - w_1)^2}\]
Let's walk through the application of this formula step-by-step to solidify your understanding.
higher-dimensional geometry
Higher-dimensional geometry extends the concepts of shapes and distances into more than three dimensions. While it may seem abstract, the mathematical principles remain consistent.
We use coordinates to describe locations in 4D, 5D, etc., just like we do in 3D space using \(x, y, z\)-coordinates. For our problem, we need to calculate distances in a 4-dimensional space for part (a) and a 5-dimensional space for part (b). Here’s how you approach it:
- Identify the coordinates of your points.
- Substitute the coordinates into the distance formula extended into the required number of dimensions.
These steps simplify seemingly complex distances into a straightforward calculation, helping you tackle higher-dimensional problems effectively.
coordinate differences
Calculating coordinate differences is the first step in finding the distance between points in any dimension. This involves subtracting corresponding coordinates of the two points, squaring each result, and then summing them all up:
- For part (a), given points \( (4,-1,2,7)\) and \( (2,3,1,9)\), we find the differences:
\[ (4-2) = 2, (-1-3) = -4, (2-1) = 1, (7-9) = -2.\]
Squaring these differences, we get:
\(2^2 = 4, (-4)^2 = 16, 1^2 = 1, (-2)^2 = 4\).
- Summing these: \( 4 + 16 + 1 + 4 = 25\).
- The distance is the square root of this sum: \( \sqrt{25} = 5\).
For part (b), repeat the process for points \( (-1,5,-3,2,4)\) and \( (2,6,2,7,6)\) in 5-dimensional space. This breakdown shows the power of coordinate differences in higher-dimensional spaces.

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Most popular questions from this chapter

(a) If \(C\) is orthogonal and \(M\) is symmetric, show that \(C^{-1} M C\) is symmetric. (b) If \(C\) is orthogonal and \(M\) is antisymmetric, show that \(C^{-1} M C\) is antisymmetric. In the following problems, assume that \(M\) is symmetric, \(C\) is orthogonal, and \(D=C^{-1} M C^{*}\) is diagonal.

(a) If a body is rotating about a fixed axis, then its angular momentum \(L\) and its angular velocity \(\boldsymbol{o}\) are parallel vectors and \(\mathrm{L}=\mathrm{le} \mathrm{s}\), where \(I\) is the (scalar) moment of inertia of the body about the axis. However, in general, \(L\) and \(\omega\) are not parallel and \(I\) in the equation must be a second-order tensor; let us call it I. Find I in dyadic form in the following way: For simplicity, first consider a point mass \(m\) at the point \(r\). The angular momentum of \(m\) about the origin is by definition \(m r \times v\), where \(v\) is the linear velocity.From Chapter \(6, v=\omega \times r .\) Write out the triple vector product for \(L\) and from it write each component of \(\mathrm{L}\) in terms of the three components of 6 . Write your results in matrix form and in dyadic form $$ \mathbf{L}=\mathbf{I} \cdot \mathbf{\omega}=\left(\mathrm{ii} l_{x x}+\mathrm{ij} l_{x y}+\cdots\right)+\omega $$ You should have $$ I_{x z}=m\left(y^{2}+z^{2}\right), \quad I_{x y}=-m x y, \quad \text { etc. } $$ For a set of masses \(m_{i}\) or an extended body, replace the cxpressions for \(I_{z x+}, \operatorname{ctc}_{4}\), by the corresponding sums or integrals: $$ \begin{aligned} &I_{x x}=\sum m_{i}\left(y_{i}^{2}+z_{i}^{2}\right) \quad \text { or } \quad \int\left(y^{2}+z^{2}\right) d m_{4} \\ &I_{x y}=-\sum m_{j} x_{i} y_{i} \quad \text { or }-\int x y d m_{4} \quad \text { etc. } \end{aligned} $$ (b) Show that 1 is a second-order (Cartesian) tensor by expressing its components relative to a rotated system \(\left[I_{s^{\prime} x^{\prime}}=m\left(y^{2}+z^{\prime 2}\right)\right.\), etc. \(]\) in terms of \(x, y, z\) using \((11.7)\) or \((11.11)\), and hence in terms of \(I_{x \pi}\), etc., to show that I obeys the transformation equations \((11,13)\). (c) Show that if \(\mathrm{n}\) is a unit vector, the expression \(\mathbf{n}+\mathbf{1} \cdot \mathbf{n}\) gives the moment of inertia about an axis through the origin parallel to n. Hint: Consider I rotated to a system in which one of the axes is along \(n\). (d) Observe that the I matrix is symmetric and recall that a symmetric matrix may be diagonalized by a rotation of axes. The eigenvalues of the I matrix are called the principal moments of inertia. Show by part (c) that they are moments of inertia abour the new axes \(\left(x^{\prime}, y^{\prime}, z^{\prime}\right)\) relative to which \(\mathbf{I}\) is diagonal. These new axes are called the principal axes of incrtia. For the mass distribution consisting of point masses \(m\) at \((1,1,1)\) and \((1,1,-1)\), find the nine components of 1, and find the principal moments of inertia and the principal axes.

The trace of a matrix is the sum of the elements on the main diagonal. Show that the trace is not changed by cyclic permutation of the matrices, that is, \(\operatorname{Tr}(A B C)=\operatorname{Tr}(C A B)=\) Tr \((B C A)\). [Caution: \(\operatorname{Tr}(A B C) \neq \operatorname{Tr}(A C B)\) in general. ]

Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{rrr}-1 & 1 & 3 \\ 1 & 2 & 0 \\ 3 & 0 & 2\end{array}\right)\)

Show that if \(C\) is a matrix whose columns are the components \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) of tuo perpendicular vectors each of unit length, then \(C\) is an orthogonal matrix. Hint: Find \(C^{1} C\).

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