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Let $v_1=\left[\begin{array}{l}1 \\ 2 \\ 0\end{array}\right]$ and $v_2=\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]$ be two vectors. The value of the coefficient $\alpha$ in the expression $v_1=\alpha v_2+e$, which minimizes the length of the error vector $e$, is

  1. $\frac{7}{2}$
  2. $\frac{-2}{7}$
  3. $\frac{2}{7}$
  4. $\frac{-7}{2}$

1 Answer

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answer c .
expl:

e = v1-alpha(v2)
e = [1-2a,2-a,0-3a]

minimizes length means orthogonal so using dot product with v2
2(1-2a)+1(2-a)+3(-3a) = 0
a = 2/7
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