GATE 2025 EC Q.59: solution and answer

Official answer: 2.95 to 3.05Source: Official answer key, GATE 2025 EC (IIT Roorkee)

1. Minimising $\Vert \mathbf{a}x-\mathbf{b}{\Vert }_2$ over the scalar $x$ is a least-squares problem: the minimiser makes $\mathbf{a}x$ the orthogonal projection of $\mathbf{b}$ onto the line spanned by $\mathbf{a}$, so the residual $\mathbf{a}x-\mathbf{b}$ must be perpendicular to $\mathbf{a}$.

2. Normal equation ${\mathbf{a}}^T(\mathbf{a}x-\mathbf{b})=0$ gives $$x^{\ast }=\frac{{\mathbf{a}}^T\mathbf{b}}{{\mathbf{a}}^T\mathbf{a}}=\frac{1\cdot 0+1\cdot 3\sqrt{2}}{1^2+1^2}=\frac{3\sqrt{2}}{2}.$$ (Equivalently, $\Vert \mathbf{a}x-\mathbf{b}{\Vert }_2^2=x^2+(x-3\sqrt{2}{)}^2=2x^2-6\sqrt{2}x+18$, a parabola opening upward with vertex at $x=\frac{6\sqrt{2}}{4}=\frac{3\sqrt{2}}{2}$.)

3. Residual at the minimiser: $$\mathbf{a}{x}^{\ast }-\mathbf{b}=\left[\begin{array}{c}\frac{3\sqrt{2}}{2}\\ \frac{3\sqrt{2}}{2}\end{array}\right]-\left[\begin{array}{c}0\\ 3\sqrt{2}\end{array}\right]=\left[\begin{array}{c}\frac{3\sqrt{2}}{2}\\ -\frac{3\sqrt{2}}{2}\end{array}\right].$$ Check: ${\mathbf{a}}^T(\mathbf{a}{x}^{\ast }-\mathbf{b})=\frac{3\sqrt{2}}{2}-\frac{3\sqrt{2}}{2}=0$, so the residual is indeed perpendicular to $\mathbf{a}$.

4. Norm of the residual: $$\sqrt{{\left(\frac{3\sqrt{2}}{2}\right)}^2+{\left(\frac{3\sqrt{2}}{2}\right)}^2}=\sqrt{\frac{9}{2}+\frac{9}{2}}=\sqrt{9}=3.$$

5. Check by Pythagoras: $\Vert \mathbf{b}{\Vert }_2^2-\frac{({\mathbf{a}}^T\mathbf{b}{)}^2}{{\mathbf{a}}^T\mathbf{a}}=18-\frac{18}{2}=9$, so the minimum distance is $\sqrt{9}=3.00$.