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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

If \(A\) and \(B\) are symmetric matrices \(\left(A^{1}=A\right)\) and \(C=A B-B A\), show that \(C\) is antisymmetric \(\left(C^{\mathrm{T}}=-C\right)\)

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

From (3.4), show that if \(M\) is orthogonal, then det \(M=1\) or \(-1 .\) (When det \(M=1\), the transformation is called a proper rotation; when det \(M=-1\), one or all three axes have, been reflected, in addition to rotation.) Hint: Find det \(\left(M M^{T}\right) ;\) how is a determinant affected by interchanging rows and columns?

The characteristic equation for a second-order matrix \(M\) is a quadratic equation. We have, considered in detail the case in which \(M\) is a real symmetric matrix and the roots of thecharacteristic equation (eigenvalues) are real, positive, and unequal. Discuss some other possibilities as follows: (a) \(M\) real and symmetric, eigenvalues real, one positive and one negative. Show that the plane is reflected in one of the cigenvector lines (as well as stretched or shrunk). Consider as a simple special case $$ M=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right) $$ (b) \(M\) real and symmetric, eigenvalues equal (and therefore real). Show that \(M\) must he a multiple of the unit matrix. Thus show that the deformation consists of dilation or shrinkage in the radial direction (the same in all directions) with no rotation (and reflection in the origin if the root is negative). (c) \(M\) real, not symmetric, eigenvalues real and not equal. Show that in this case the eigenvectors are not orthogonal. Hint: Find their dot product. (d) \(M\) real, not symmetric, eigenvalues complex, Show that all vectors are rotated, that is, there are no (real) eigenvectors which are unchanged in direction by the transformation. Consider the characteristic equation of a rotation matrix as a special case.

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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