已知 alpha_1 = ((1)/(sqrt(3)), (1)/(sqrt(3)), (1)/(sqrt(3)) ), alpha_2 = (-(1)/(sqrt(2)), (1)/(sqrt(2)), 0 ), alpha_3 = (-(1)/(sqrt(6)), -(1)/(sqrt(6)), (2)/(sqrt(6)) ) 是 mathbb(R)^3 的一个正交规范基(即标准正交基),若用这个基来线性表示 mathbb(R)^3 中的向量 alpha = (1, -1, -1),则 alpha = ( )。

已知

$\alpha_1 = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$, $\alpha_2 = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)$, $\alpha_3 = \left(-\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right)$

是 $\mathbb{R}^3$ 的一个正交规范基(即标准正交基),若用这个基来线性表示 $\mathbb{R}^3$ 中的向量 $\alpha = (1, -1, -1)$,则 $\alpha = (\ )$。

  • A. $\frac{1}{\sqrt{3}}\alpha_1 - \sqrt{2}\alpha_2 - \frac{2}{\sqrt{6}}\alpha_3$
  • B. $\frac{1}{\sqrt{3}}\alpha_1 + \sqrt{2}\alpha_2 - \frac{2}{\sqrt{6}}\alpha_3$
  • C. $-\frac{1}{\sqrt{3}}\alpha_1 - \sqrt{2}\alpha_2 - \frac{2}{\sqrt{6}}\alpha_3$
  • D. $-\frac{1}{\sqrt{3}}\alpha_1 + \sqrt{2}\alpha_2 + \frac{2}{\sqrt{6}}\alpha_3$

参考答案与解析:

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