题目
已知两点 M_(1)(4,sqrt(2),1) 和 M_(2)(3,0,2)，则 overrightarrow(M_{1)M_(2)} 的方向余弦为A. cos alpha = -(1)/(2), cos beta = -(sqrt(2))/(2), cos gamma = (1)/(2)B. cos alpha = (1)/(2), cos beta = -(sqrt(2))/(2), cos gamma = (1)/(2)C. cos alpha = -(1)/(2), cos beta = (sqrt(2))/(2), cos gamma = (1)/(2)D. cos alpha = -(1)/(2), cos beta = -(sqrt(2))/(2), cos gamma = -(1)/(2)
已知两点 $M_{1}(4,\sqrt{2},1)$ 和 $M_{2}(3,0,2)$，则 $\overrightarrow{M_{1}M_{2}}$ 的方向余弦为
A. $\cos \alpha = -\frac{1}{2}, \cos \beta = -\frac{\sqrt{2}}{2}, \cos \gamma = \frac{1}{2}$
B. $\cos \alpha = \frac{1}{2}, \cos \beta = -\frac{\sqrt{2}}{2}, \cos \gamma = \frac{1}{2}$
C. $\cos \alpha = -\frac{1}{2}, \cos \beta = \frac{\sqrt{2}}{2}, \cos \gamma = \frac{1}{2}$
D. $\cos \alpha = -\frac{1}{2}, \cos \beta = -\frac{\sqrt{2}}{2}, \cos \gamma = -\frac{1}{2}$
题目解答
答案
A. $\cos \alpha = -\frac{1}{2}, \cos \beta = -\frac{\sqrt{2}}{2}, \cos \gamma = \frac{1}{2}$
解析
步骤 1：计算向量 $\overrightarrow{M_{1}M_{2}}$
向量 $\overrightarrow{M_{1}M_{2}}$ 的坐标为： \[ \overrightarrow{M_{1}M_{2}} = (3-4, 0-\sqrt{2}, 2-1) = (-1, -\sqrt{2}, 1) \]
步骤 2：计算向量 $\overrightarrow{M_{1}M_{2}}$ 的模长
向量 $\overrightarrow{M_{1}M_{2}}$ 的模长为： \[ |\overrightarrow{M_{1}M_{2}}| = \sqrt{(-1)^2 + (-\sqrt{2})^2 + 1^2} = \sqrt{1 + 2 + 1} = \sqrt{4} = 2 \]
步骤 3：计算方向余弦
方向余弦为： \[ \cos \alpha = \frac{-1}{2} = -\frac{1}{2}, \quad \cos \beta = \frac{-\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}, \quad \cos \gamma = \frac{1}{2} \]
向量 $\overrightarrow{M_{1}M_{2}}$ 的坐标为： \[ \overrightarrow{M_{1}M_{2}} = (3-4, 0-\sqrt{2}, 2-1) = (-1, -\sqrt{2}, 1) \]
步骤 2：计算向量 $\overrightarrow{M_{1}M_{2}}$ 的模长
向量 $\overrightarrow{M_{1}M_{2}}$ 的模长为： \[ |\overrightarrow{M_{1}M_{2}}| = \sqrt{(-1)^2 + (-\sqrt{2})^2 + 1^2} = \sqrt{1 + 2 + 1} = \sqrt{4} = 2 \]
步骤 3：计算方向余弦
方向余弦为： \[ \cos \alpha = \frac{-1}{2} = -\frac{1}{2}, \quad \cos \beta = \frac{-\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}, \quad \cos \gamma = \frac{1}{2} \]