Calculus III

Linear Algebra

To solve this problem, we will follow these steps: 1. Find the projection matrix \( P \) onto the plane spanned by \( a_1 \) and \( a_2 \). 2. Use the projection matrix \( P \) to find the projection \( p \) of \( b \) onto the plane. 3. Calculate the error vector \( e \) as the difference between \( b \) and \( p \). 4. Verify if the error vector \( e \) is orthogonal to the plane by checking its dot product with \( a_1 \) and \( a_2 \). ### Step 1: Find the Projection Matrix \( P \) The projection matrix onto a plane spanned by two vectors \( a_1 \) and \( a_2 \) can be found using the formula: \[ P = A(A^TA)^{-1}A^T \] where \( A \) is the matrix whose columns are the spanning vectors \( a_1 \) and \( a_2 \). First, we need to construct matrix \( A \): \[ A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ 1 & 1 \end{bmatrix} \] Now, compute \( A^TA \): \[ A^TA = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 3 \end{bmatrix} \] Next, find the inverse of \( A^TA \): \[ (A^TA)^{-1} = \begin{bmatrix} 2 & 2 \\ 2 & 3 \end{bmatrix}^{-1} = \frac{1}{2\cdot3 - 2\cdot2} \begin{bmatrix} 3 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -2 & 2 \end{bmatrix} \] Now, compute \( P \): \[ P = A(A^TA)^{-1}A^T = \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \] \[ P = \begin{bmatrix} -2 & 2 \\ 1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \] \[ P = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \] ### Step 2: Find the Projection \( p \) of \( b \) To find the projection \( p \) of \( b \) onto the plane, we multiply the projection matrix \( P \) by vector \( b \): \[ p = P \cdot b = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} \] \[ p = \begin{bmatrix} -2 \\ 3 \\ 3 \end{bmatrix} \] ### Step 3: Find the Error Vector \( e \) The error vector \( e \) is the difference between the original vector \( b \) and its projection \( p \): \[ e = b - p = \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} - \begin{bmatrix} -2 \\ 3 \\ 3 \end{bmatrix} \] \[ e = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix} \] ### Step 4: Verify if the Error Vector \( e \) is Orthogonal to the Plane To verify if \( e \) is orthogonal to the plane, we check if it is orthogonal to both \( a_1 \) and \( a_2 \). This is done by computing the dot product of \( e \) with \( a_1 \) and \( a_2 \): \[ e \cdot a_1 = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = (1)(0) + (-2)(1) + (-1)(1) = 0 \] \[ e \cdot a_2 = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = (1)(1) + (-2)(1) + (-1)(1) = 0 \] Since the dot products are zero, the error vector \( e \) is orthogonal to both \( a_1 \) and \( a_2 \), and therefore, it is orthogonal to the plane spanned by these vectors.

Math

This question is all about orthogonal projection. Suppose we are given the following two vectors \( a_1 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \) and \( a_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \). 1. Find the projection matrix \( P \) onto the plane (subspace) in \( \mathbb{R}^3 \) through \( a_1 \) and \( a_2 \). 2. Find the projection (vector) \( p \) of \( b = \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} \) onto the same plane. 3. Find the error vector \( e \) of the projection in the previous part. 4. Is the error vector \( e \) orthogonal to the plane? Verify your yes/no answer.

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