题目
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 = ( )。 A. (1)/(sqrt(3))alpha_1 - sqrt(2)alpha_2 - (2)/(sqrt(6))alpha_3B. (1)/(sqrt(3))alpha_1 + sqrt(2)alpha_2 - (2)/(sqrt(6))alpha_3C. -(1)/(sqrt(3))alpha_1 - sqrt(2)alpha_2 - (2)/(sqrt(6))alpha_3D. -(1)/(sqrt(3))alpha_1 + sqrt(2)alpha_2 + (2)/(sqrt(6))alpha_3
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. $-\frac{1}{\sqrt{3}}\alpha_1 + \sqrt{2}\alpha_2 + \frac{2}{\sqrt{6}}\alpha_3$
题目解答
答案
已知正交规范基 $\alpha_1, \alpha_2, \alpha_3$，向量 $\alpha$ 可表示为：
\[
\alpha = x\alpha_1 + y\alpha_2 + z\alpha_3
\]
其中系数 $x = \alpha \cdot \alpha_1$，$y = \alpha \cdot \alpha_2$，$z = \alpha \cdot \alpha_3$。
计算得：
\[
x = \alpha \cdot \alpha_1 = -\frac{1}{\sqrt{3}}, \quad y = \alpha \cdot \alpha_2 = -\sqrt{2}, \quad z = \alpha \cdot \alpha_3 = -\frac{2}{\sqrt{6}}
\]
因此，$\alpha = -\frac{1}{\sqrt{3}}\alpha_1 - \sqrt{2}\alpha_2 - \frac{2}{\sqrt{6}}\alpha_3$，对应选项 **C**。
\[
\boxed{C}
\]