Question

Let \(\mathbf{v}_1=\begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix}\) and \(\mathbf{v}_2=\begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}\) be two vectors. The value of the coefficient \(\alpha\) in the expression \(\mathbf{v}_1=\alpha \mathbf{v}_2+\mathbf{e}\), which minimizes the length of the error vector \(\mathbf{e}\), is

• A. $\dfrac{7}{2}$
• B. $-\dfrac{2}{7}$
• C. $\dfrac{2}{7}$
• D. $-\dfrac{7}{2}$

Hint:

For $\mathbf{v}_1\approx \alpha\mathbf{v}_2$, the best $\alpha$ in least squares is always the projection coefficient: $\alpha=\dfrac{\mathbf{v}_2.\mathbf{v}_1}{\mathbf{v}_2.\mathbf{v}_2}$.

Answer 1

The Correct Option is C

Step 1: Least–squares condition (orthogonal projection).
To minimize $\|\mathbf{e}\|$ in $\mathbf{v}_1=\alpha\mathbf{v}_2+\mathbf{e}$, we need $\mathbf{e}\perp \mathbf{v}_2$. Hence $\mathbf{v}_2^{\top}(\mathbf{v}_1-\alpha\mathbf{v}_2)=0 \Rightarrow \alpha=\dfrac{\mathbf{v}_2^{\top}\mathbf{v}_1}{\mathbf{v}_2^{\top}\mathbf{v}_2}$. 
Step 2: Compute the dot products.
$\mathbf{v}_2^{\top}\mathbf{v}_1 = [2,1,3].[1,2,0] = 2.1 + 1.2 + 3.0 = 4$.
$\mathbf{v}_2^{\top}\mathbf{v}_2 = 2^2 + 1^2 + 3^2 = 4+1+9=14$. 
Step 3: Evaluate $\alpha$.
\[ \alpha=\frac{4}{14}=\frac{2}{7}. \] \[ \boxed{\dfrac{2}{7}} \]