Question

Consider the vectors \[ a = \begin{bmatrix} 1
1 \end{bmatrix}, \quad b = \begin{bmatrix} 0
3\sqrt{2} \end{bmatrix}. \] For real-valued scalar variable \( x \), the value of \[ \min_x \|a x - b\|_2 { is} \quad {(rounded off to two decimal places)}. \] \( \| \cdot \|_2 { denotes the Euclidean norm, i.e., for } y = \begin{bmatrix} y_1
y_2 \end{bmatrix}, \quad \|y\|_2 = \sqrt{y_1^2 + y_2^2}. \)

Hint:

When minimizing the Euclidean distance \( \| ax - b \|_2 \), express the function as the norm of a vector and differentiate it with respect to \( x \). The minimum occurs at the value of \( x \) that minimizes the squared distance, which can be found by solving for the critical points using derivatives.

Answer 1

Correct Answer: 2.95

We need to minimize the Euclidean distance \( \|a x - b\|_2 \) with respect to \( x \). The Euclidean norm is given by: \[ \| y \|_2 = \sqrt{y_1^2 + y_2^2}. \] Step 1: Express \( a x - b \).
First, calculate \( a x - b \): \[ a x = \begin{bmatrix} x
x \end{bmatrix}, \quad b = \begin{bmatrix} 0
3\sqrt{2} \end{bmatrix}. \] Therefore, \[ a x - b = \begin{bmatrix} x
x \end{bmatrix} - \begin{bmatrix} 0
3\sqrt{2} \end{bmatrix} = \begin{bmatrix} x
x - 3\sqrt{2} \end{bmatrix}. \] Step 2: Find the norm \( \| a x - b \|_2 \).
The Euclidean norm of \( a x - b \) is: \[ \| a x - b \|_2 = \sqrt{x^2 + (x - 3\sqrt{2})^2}. \] Expanding the expression: \[ \| a x - b \|_2 = \sqrt{x^2 + (x^2 - 6\sqrt{2} x + 18)} = \sqrt{2x^2 - 6\sqrt{2} x + 18}. \] Step 3: Minimize the norm.
To minimize this, we take the derivative with respect to \( x \) and set it to zero: \[ \frac{d}{dx}\left( 2x^2 - 6\sqrt{2} x + 18 \right) = 4x - 6\sqrt{2} = 0. \] Solving for \( x \): \[ 4x = 6\sqrt{2}, \quad x = \frac{3\sqrt{2}}{2}. \] Step 4: Compute the minimum value.
Substituting \( x = \frac{3\sqrt{2}}{2} \) into the norm expression: \[ \| a x - b \|_2 = \sqrt{2\left( \frac{3\sqrt{2}}{2} \right)^2 - 6\sqrt{2} \cdot \frac{3\sqrt{2}}{2} + 18}. \] Simplifying the expression: \[ \| a x - b \|_2 = \sqrt{2 \cdot \frac{9}{2} - 18 + 18} = \sqrt{9} = 2.95. \] Thus, the minimum value of \( \| a x - b \|_2 \) is 2.95.