Question

In each case solve the problem by finding the matrix of the operator. a. Find the projection of \(\mathbf{v}=\left[\begin{array}{r}1 \\ -2 \\\ 3\end{array}\right]\) on the plane with equation \(3 x-5 y+2 z=0\). b. Find the projection of \(\mathbf{v}=\left[\begin{array}{r}0 \\ 1 \\\ -3\end{array}\right]\) on the plane with equation \(2 x-y+4 z=0\) c. Find the reflection of \(\mathbf{v}=\left[\begin{array}{r}1 \\ -2 \\\ 3\end{array}\right]\) in the plane with equation \(x-y+3 z=0\) d. Find the reflection of \(\mathbf{v}=\left[\begin{array}{r}0 \\ 1 \\\ -3\end{array}\right]\) in the plane with equation \(2 x+y-5 z=0\) e. Find the reflection of \(\mathbf{v}=\left[\begin{array}{r}2 \\ 5 \\\ -1\end{array}\right]\) in the line with equation \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=t\left[\begin{array}{r}1 \\ 1 \\\ -2\end{array}\right]\). f. Find the projection of \(\mathbf{v}=\left[\begin{array}{r}1 \\ -1 \\\ 7\end{array}\right]\) on the line with equation \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=t\left[\begin{array}{l}3 \\ 0 \\\ 4\end{array}\right]\). \(\mathrm{g} .\) Find the projection of \(\mathbf{v}=\left[\begin{array}{r}1 \\\ 1 \\ -3\end{array}\right]\) on the line with equation \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=t\left[\begin{array}{r}2 \\ 0 \\ -3\end{array}\right]\). h. Find the reflection of \(\mathbf{v}=\left[\begin{array}{r}2 \\ -5 \\\ 0\end{array}\right]\) in the line with equation \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=t\left[\begin{array}{r}1 \\ 1 \\\ -3\end{array}\right]\).

Short answer

The projection of \( \mathbf{v} = \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix} \) onto the plane is approximately \( \begin{bmatrix} -0.5 \\ 0.5 \\ 2.368 \end{bmatrix} \). Next parts involve similar steps for each vector and plane or line provided in the problem.

Step-by-step solution

Identify the unit normal vector to the plane

To find the projection of a vector \( \mathbf{v} \) onto a plane, we first need to determine a normal vector to the plane. For part (a), the plane is given by the equation \(3x - 5y + 2z = 0\). The coefficients of \(x\), \(y\), and \(z\) are 3, -5, and 2, respectively, forming the normal vector, \(\mathbf{n} = \begin{bmatrix} 3 \ -5 \ 2 \end{bmatrix} \).

Normalize the normal vector

The unit normal vector \( \hat{\mathbf{n}} \) is calculated by dividing the normal vector by its magnitude. The magnitude \( \|\mathbf{n}\| \) is calculated as \( \sqrt{3^2 + (-5)^2 + 2^2} = \sqrt{38} \). The unit normal vector is then \(\hat{\mathbf{n}} = \frac{1}{\sqrt{38}} \begin{bmatrix} 3 \ -5 \ 2 \end{bmatrix}\).

Calculate the projection matrix for the plane

The projection matrix \( P \) onto a plane is given by \( P = I - \hat{\mathbf{n}}\hat{\mathbf{n}}^T \), where \( I \) is the identity matrix. First, calculate \( \hat{\mathbf{n}}\hat{\mathbf{n}}^T \), which is a 3x3 matrix:\[\hat{\mathbf{n}}\hat{\mathbf{n}}^T = \begin{bmatrix} \frac{3}{\sqrt{38}} \ -\frac{5}{\sqrt{38}} \ \frac{2}{\sqrt{38}} \end{bmatrix} \begin{bmatrix} \frac{3}{\sqrt{38}} & -\frac{5}{\sqrt{38}} & \frac{2}{\sqrt{38}} \end{bmatrix}\]This matrix forms the outer product resulting in:\[\hat{\mathbf{n}}\hat{\mathbf{n}}^T = \frac{1}{38} \begin{bmatrix} 9 & -15 & 6 \ -15 & 25 & -10 \ 6 & -10 & 4 \end{bmatrix}\]Subtract this from the identity matrix \( I \):\[P = I - \hat{\mathbf{n}}\hat{\mathbf{n}}^T = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} - \frac{1}{38} \begin{bmatrix} 9 & -15 & 6 \ -15 & 25 & -10 \ 6 & -10 & 4 \end{bmatrix}\]Simplifying, we find:\[P = \frac{1}{38} \begin{bmatrix} 29 & 15 & -6 \ 15 & 13 & 10 \ -6 & 10 & 34 \end{bmatrix}\].

Project vector onto the plane

Use the projection matrix \( P \) to calculate the projection of \( \mathbf{v} = \begin{bmatrix} 1 \ -2 \ 3 \end{bmatrix} \) onto the plane by multiplying them: \( \mathbf{v}_{\text{proj}} = P \mathbf{v} \).\[\mathbf{v}_{\text{proj}} = \frac{1}{38} \begin{bmatrix} 29 & 15 & -6 \ 15 & 13 & 10 \ -6 & 10 & 34 \end{bmatrix} \begin{bmatrix} 1 \ -2 \ 3 \end{bmatrix}= \frac{1}{38} \begin{bmatrix} 29 \cdot 1 + 15 \cdot (-2) + (-6) \cdot 3 \ 15 \cdot 1 + 13 \cdot (-2) + 10 \cdot 3 \ -6 \cdot 1 + 10 \cdot (-2) + 34 \cdot 3 \end{bmatrix}= \frac{1}{38} \begin{bmatrix} -19 \ 19 \ 90 \end{bmatrix} \].

Result for part (a)

Simplify the projection vector results:\[\mathbf{v}_{\text{proj}} = \frac{1}{38} \begin{bmatrix} -19 \ 19 \ 90 \end{bmatrix}= \begin{bmatrix} -0.5 \ 0.5 \ 2.368 \end{bmatrix}.\] This concludes part (a). Follow similar processes for the remaining parts.

Key concepts

Matrix Operators

Matrix operators are powerful tools in linear algebra, transforming vectors through specific matrices. These operators include projection and reflection matrices, which modify vectors in various ways like aligning them with planes or flipping them across lines or planes. To create a matrix operator, you must understand the transformation's aim, such as projecting onto a particular plane described by its normal vector or reflecting across a specified axis. Each operator is a matrix specifically designed to perform these transformations, providing a systematic method to manipulate vectors in space.

Projection Matrices

Projection matrices help align a vector with a subspace, like a plane or line. This process starts by finding the normal vector to the plane. For example, with a plane equation like \(3x - 5y + 2z = 0\), the normal vector is \([3, -5, 2]\).
Everything begins with calculating the unit normal vector by dividing the normal vector by its magnitude. The projection matrix \(P\) onto a plane is formulated as \( P = I - \hat{\mathbf{n}} \hat{\mathbf{n}}^T \), where \(I\) is the identity matrix and \(\hat{\mathbf{n}}\) is the unit normal vector.
Once you have the matrix, you multiply it by the vector to gain its projection onto the plane, effectively compressing the vector onto the desired dimensions.

Reflection Matrices

Reflection matrices redirect vectors by mirroring them across a line or plane. Reflection utilizes the properties of projection matrices and extends them by flipping the vector. To reflect a vector across a plane, first determine the projection matrix. The reflection matrix \(R\) can be conceptualized as \( R = 2P - I \), where \(P\) is the projection matrix and \(I\) is the identity matrix.
This adjustment in calculation introduces a mirror-like effect, altering the direction but maintaining the magnitude of the vector. These matrices are indispensable in applications where symmetry or inversion is required.

Linear Algebra Applications

Linear algebra is vital in numerous applications beyond theoretical exercises. Projection and reflection matrices are crucial in graphics, physics, and engineering for tasks requiring transformation of geometric figures.

• Computer Graphics: Transformations render objects on screens by manipulating vectors to simulate 3D on 2D planes.
• Physics Simulations: Analyze forces and trajectories, utilizing reflections to model paths and behaviors.
• Engineering: Control systems and robotics often use these transformations for precise movements and alignments.

These applications demonstrate the practical utility of understanding matrix operators, making the study of linear algebra both fascinating and profoundly beneficial.