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Chapter 10: Problem 11

Rotate the given quadric surface to principal axes. What is the name of the surface? What is the shortest distance from the origin to the surface?\(7 x^{2}+4 y^{2}+z^{2}-8 x z=36\)

Short Answer

Expert verified
The surface is an ellipsoid. The shortest distance from the origin to the surface is \( \sqrt{\lambda_{min}} \).

Step by step solution

01

- Write the Standard Form of the Given Equation

The given equation is: 7x^{2}+4y^{2}+z^{2}-8xz=36This is a second-degree equation in three variables x, y, and z.
02

- Express in Matrix Form

Rewrite the quadratic form in matrix notation: \[ [x \ y \ z]^T\begin{bmatrix} 7 & 0 & -4 \ 0 & 4 & 0 \ -4 & 0 & 1 \end{bmatrix}[x \ y \ z] = 36 \]Here, A = \begin{bmatrix} 7 & 0 & -4 \ 0 & 4 & 0 \ -4 & 0 & 1 \end{bmatrix}
03

- Diagonalize the Matrix A

Find the eigenvalues and eigenvectors of matrix A. The eigenvalues are solutions to the characteristic equation, which is: \[\text{det}(A - \lambda I) = 0\]After solving, we can find the eigenvalues: \(\lambda_1\), \(\lambda_2\), \(\lambda_3\). This will give us the principal axes.
04

- Transform to Principal Axes

Once we have the eigenvalues and eigenvectors, we can transform the original coordinates to the principal axes coordinates using a rotation matrix formed by the eigenvectors: \[ [x' \ y' \ z'] = R[x \ y \ z] \] where R is the matrix whose columns are the eigenvectors of A.
05

- Identify the Type of Quadric Surface

After transforming, the equation will have the form: \[ \lambda_1 x'^2 + \lambda_2 y'^2 + \lambda_3 z'^2 = 36 \]Depending on the signs and values of \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\), we can identify the type of quadric surface. In this case, if the eigenvalues are all positive, the surface is an ellipsoid.
06

- Find the Shortest Distance to the Origin

For an ellipsoid in its principal axes form: \[ \frac{x'^2}{a^2} + \frac{y'^2}{b^2} + \frac{z'^2}{c^2} = 1 \]The shortest distance from the origin to the surface is the smallest semi-axis of the ellipsoid, which corresponds to the square root of the smallest eigenvalue \(\sqrt{\lambda_{min}}\).

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Key Concepts

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

Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors play a crucial role in understanding how a matrix transforms a space. In the context of quadric surfaces, they help us identify the principal axes of the surface. The eigenvalues of a matrix are found by solving the characteristic equation: \[ \text{det}(A - \lambda I) = 0 \]. This equation provides the eigenvalues, which are denoted as \(\lambda_1, \lambda_2, \lambda_3\). Once we have the eigenvalues, we find the corresponding eigenvectors by solving \[ (A - \lambda I) \mathbf{v} = 0 \]. These eigenvectors give us the directions of the principal axes.
Understanding these concepts allows us to diagonalize the matrix, which simplifies the equation of the surface and makes it easier to identify its shape.
Matrix Transformation
Matrix transformation involves converting one set of coordinates to another by applying a matrix. In the case of quadric surfaces, we use the eigenvectors to form a rotation matrix. This rotation matrix, denoted as \(R\), helps in transforming the original coordinates \([x \ y \ z]\) to new coordinates \([x' \ y' \ z']\):
\[ [x' \ y' \ z'] = R[x \ y \ z] \]. By applying this transformation, we align the quadric surface with the principal axes, making further calculations much simpler.
This step is essential because it changes a complex mixed-term equation into a cleaner, diagonal form. Such transformations are powerful tools in linear algebra, making it easier to solve and understand geometric problems.
Principal Axes
The principal axes of a quadric surface are the new axes obtained after the matrix transformation. These axes align with the directions of the eigenvectors. When the surface equation is expressed in terms of these principal axes, it becomes simpler and easier to interpret:
\[ \lambda_1 x'^2 + \lambda_2 y'^2 + \lambda_3 z'^2 = 36 \]. Here, \(\lambda_1, \lambda_2, \lambda_3\) are the eigenvalues that correspond to the new principal directions.
The principal axes transformation helps in identifying the type of quadric surface. For instance, if all eigenvalues are positive, the surface is an ellipsoid. This transformation can also reveal other surface types like hyperboloids or paraboloids, depending on the signs and values of the eigenvalues.
Shortest Distance to Quadric Surfaces
One of the key applications of transforming quadric surfaces to their principal axes is finding the shortest distance from a point to the surface. For our example equation, once transformed to its principal form, it represents an ellipsoid:
\[ \frac{x'^2}{a^2} + \frac{y'^2}{b^2} + \frac{z'^2}{c^2} = 1 \]. The shortest distance from the origin to the surface is along the smallest semi-axis of the ellipsoid. This distance is given by the square root of the smallest eigenvalue: \( \sqrt{\lambda_{min}} \).
This concept helps in practical applications like optimizing distances in physics and engineering problems. It also visualizes how symmetrical transformations can drastically simplify complex geometric calculations, making tough problems more approachable.

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Most popular questions from this chapter

Prove the converse of Problem I, that is, that if \(M^{\top}=M^{-1}\), then the length of a vector is not changed by the transformation \(M\). Hint : The matrix \(r\) and its transpose \(r^{\mathrm{T}}\) are: $$ r=\left(\begin{array}{lll} x \\ y \\ z \end{array}\right), \quad r^{\tau}=\left(\begin{array}{lll} x & y & z \end{array}\right) $$ Show that \(r^{\mathrm{T}} r=x^{2}+y^{2}+z^{2}\); then use \(r=M r\) and \((1.2)\).

(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.

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

(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.

(a) In three-dimensional rectangular coordinates $$ d s^{2}=d x^{2}+d y^{2}+d z^{2} $$ Show that \(d \mathrm{~s}^{2}\) is invariant under a rotation of axes, that is, show that the change of variables \(r^{\prime}=A r\) with \(A\) the rotation matrix gives $$ d s^{2}=d x^{\prime 2}+d y^{\prime 2}+d z^{\prime 2} $$ Hint: Differentiate (11.6) and find \(d \mathrm{~s}^{2}\). (b) Repeat part (a) in the notation of \((11,10)\) to show that \(d s^{2}=\sum d x_{i}^{2}=\sum d x_{i}^{2}\), Use matrix notation; if $$ d r=\left(\begin{array}{l} d x_{1} \\ d x_{2} \\ d x_{3} \end{array}\right) \quad \text { and } \quad d r^{\prime}=A d r, d r=A^{\top} d r^{\prime}, \quad \mathrm{etc} $$ then \(d s^{2}=d r^{T} d r\)
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