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Chapter 2: Problem 18

Give the \(4 \times 4\) matrix that rotates points in \(\mathbb{R}^{3}\) about the \(z\) -axis through an angle of \(-30^{\circ},\) and then translates by \(\mathbf{p}=(5,-2,1)\)

Short Answer

Expert verified
The matrix is: \( \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 & 5 \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 & -2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \).

Step by step solution

01

Understand Rotation Matrix

To rotate a point around the z-axis by an angle \( \theta \), the rotation matrix can be represented as follows:\[R_z(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta & 0 & 0 \ \sin \theta & \cos \theta & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}\]
02

Calculate the Rotation Components

Given \( \theta = -30^{\circ} \), convert degrees to radians: \(-30^{\circ} = -\frac{\pi}{6}\). Calculate the trigonometric values: \( \cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \), \( \sin(-\frac{\pi}{6}) = -\frac{1}{2}\). Substitute these into the rotation matrix:\[R_z(-\frac{\pi}{6}) = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 & 0 \ -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}\]
03

Understand Translation Matrix

In a 4x4 homogeneous coordinate system, a translation by \( \mathbf{p}=(5,-2,1) \) is expressed as:\[T(\mathbf{p}) = \begin{bmatrix} 1 & 0 & 0 & 5 \ 0 & 1 & 0 & -2 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \end{bmatrix}\]
04

Combine Rotation and Translation

The transformation matrix that first rotates, then translates, is obtained by combining both matrices. This involves placing the rotation matrix in the 3x3 top-left portion and the translation vector in the last column:\[M = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 & 5 \ -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 & -2 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \end{bmatrix}\]

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Key Concepts

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

Rotation Matrix
A rotation matrix is a critical concept in 3D transformations that helps rotate points or objects around a specified axis in 3D space. In the case of rotating around the z-axis, the rotation matrix is given by:
  • Rows and columns forming a square matrix (usually 3x3 or 4x4)
  • Containing trigonometric functions that represent the angle of rotation
  • Keeping axis of rotation aligned
To rotate about the z-axis, use:\[R_z(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta & 0 & 0 \ \sin \theta & \cos \theta & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}\]By substituting the angle in radians into this matrix, you achieve the desired rotation. Here, an angle of \,\(-30^{\circ}\,\) corresponds to \,\(-\frac{\pi}{6}\,\) radians. Implementing this into the matrix, gives the rotated transformation for points in 3D space.
Translation Matrix
The translation matrix is another core component in 3D transformations, used to move, or 鈥榯ranslate鈥�, an object to a different position in space. It's expressed in a 4x4 matrix form, which can easily integrate with rotation matrices using homogeneous coordinates.

Translate using vector \( \mathbf{p} = (5, -2, 1) \), represented as:\[T(\mathbf{p}) = \begin{bmatrix} 1 & 0 & 0 & 5 \ 0 & 1 & 0 & -2 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \end{bmatrix}\]The translation matrix shifts the position of each point by adding a translation vector to its original position.
  • The last column represents the translation values
  • Maintains size as a 4x4 matrix for consistency with rotation matrix dimensions
This allows a seamless combination of transformations in a single matrix.
Homogeneous Coordinates
Homogeneous coordinates are crucial for combining transformations like rotation and translation in 3D space. By using an extra dimension, they allow us to represent transformations in a unified matrix form. A key feature is:
  • All transformations can be expressed as a multiplication of 4x4 matrices
  • The additional dimension enables translation and rotation to coexist
  • Provides a systematic approach to complex transformations
In homogeneous coordinates, a point in 3D space \( (x, y, z) \) is represented as \( (x, y, z, 1) \). This extra dimension (typically set to 1) simplifies computation and allows for more flexible manipulation of points in space.
Linear Algebra
Linear algebra forms the basis of all transformation matrices in 3D graphics. It involves the study and manipulation of vectors and matrices, which are fundamental to representing and transforming geometric shapes in 3D.
  • Describes linear transformations through matrix multiplications
  • Utilizes concepts like vector spaces, matrix operations
  • Essential for understanding rotations, translations, and scaling
In practical applications, linear algebra helps seamlessly integrate multiple transformations, such as rotation followed by translation, using comprehensive matrices like the one combining both a rotation matrix:\[\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 & 5 \ -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 & -2 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \end{bmatrix}\]This matrix succinctly encapsulates both the rotation and translation, demonstrating the power and flexibility of linear algebra in geometric modeling.

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Most popular questions from this chapter

Exercises \(42-44\) show how to use the condition number of a matrix \(A\) to estimate the accuracy of a computed solution of \(A \mathbf{x}=\mathbf{b}\) . If the entries of \(A\) and \(\mathbf{b}\) are accurate to about \(r\) significant digits and if the condition number of \(A\) is approximately \(10^{k}\) (with \(k\) a positive integer), then the computed solution of \(A \mathbf{x}=\mathbf{b}\) should usually be accurate to at least \(r-k\) significant digits. \([\mathbf{M}]\) Solve an equation \(A \mathbf{x}=\mathbf{b}\) for a suitable \(\mathbf{b}\) to find the last column of the inverse of the fifth-order Hilbert matrix $$ A=\left[\begin{array}{ccccc}{1} & {1 / 2} & {1 / 3} & {1 / 4} & {1 / 5} \\ {1 / 2} & {1 / 3} & {1 / 4} & {1 / 5} & {1 / 6} \\ {1 / 3} & {1 / 4} & {1 / 5} & {1 / 6} & {1 / 7} \\ {1 / 4} & {1 / 5} & {1 / 6} & {1 / 7} & {1 / 8} \\ {1 / 5} & {1 / 6} & {1 / 7} & {1 / 8} & {1 / 9}\end{array}\right] $$ How many digits in each entry of \(\mathbf{x}\) do you expect to be correct? Explain. [Note: The exact solution is \((630,-12600 \text { , }\) \(56700,-88200,44100 ) . ]\)

Let \(A\) be a lower triangular \(n \times n\) matrix with nonzero entries on the diagonal. Show that \(A\) is invertible and \(A^{-1}\) is lower triangular. I Hint: Explain why \(A\) can be changed into \(I\) using only row replacements and scaling. (Where are the pivots? Also, explain why the row operations that reduce \(A\) to \(I\) change \(I\) into a lower triangular matrix.]

Exercises \(23-26\) display a matrix \(A\) and an echelon form of \(A .\) Find a basis for \(\operatorname{Col} A\) and a basis for Nul \(A .\) $$ A=\left[\begin{array}{rrrr}{-3} & {9} & {-2} & {-7} \\ {2} & {-6} & {4} & {8} \\\ {3} & {-9} & {-2} & {2}\end{array}\right] \sim\left[\begin{array}{rrrr}{1} & {-3} & {6} & {9} \\ {0} & {0} & {4} & {5} \\ {0} & {0} & {0} & {0}\end{array}\right] $$

Exercises \(23-26\) display a matrix \(A\) and an echelon form of \(A .\) Find a basis for \(\operatorname{Col} A\) and a basis for Nul \(A .\) $$ \begin{aligned} A &=\left[\begin{array}{rrrrr}{1} & {4} & {8} & {-3} & {-7} \\\ {-1} & {2} & {7} & {3} & {4} \\ {-2} & {2} & {9} & {5} & {5} \\ {3} & {6} & {9} & {-5} & {-2}\end{array}\right] \\ \sim &\left[\begin{array}{ccccc}{1} & {4} & {8} & {0} & {5} \\ {0} & {2} & {5} & {0} & {-1} \\ {0} & {0} & {0} & {1} & {4} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right] \end{aligned} $$

In Exercises 37 and 38, construct bases for the column space and the null space of the given matrix A. Justify your work. $$ A=\left[\begin{array}{rrrrr}{5} & {2} & {0} & {-8} & {-8} \\ {4} & {1} & {2} & {-8} & {-9} \\ {5} & {1} & {3} & {5} & {19} \\ {-8} & {-5} & {6} & {8} & {5}\end{array}\right] $$
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