题目
已M两点 _(1)(sqrt (2),4,1) 和M2(0,3,2),计算向量M1M2的方向余弦:-|||-A cos x=-dfrac (1)(2) cos beta =-dfrac (sqrt {2)}(2) cos y=dfrac (sqrt {2)}(2)-|||-B cos alpha =-dfrac (sqrt {2)}(2) cos beta =-dfrac (1)(2) , cos gamma =dfrac (1)(2)-|||-C cos alpha =-dfrac (1)(2) , cos beta =-dfrac (1)(2) , cos y=dfrac (1)(2)-|||-D cos alpha =-dfrac (sqrt {2)}(2) cos beta =-dfrac (sqrt {2)}(2) cos y=dfrac (sqrt {2)}(2)
题目解答
答案
B. $\cos \alpha =-\dfrac {\sqrt {2}}{2}$ $\cos \beta =-\dfrac {1}{2}$ , $\cos \gamma =\dfrac {1}{2}$
解析
步骤 1：计算向量M1M2
向量M1M2的坐标为M2坐标减去M1坐标，即：
$\overrightarrow{M_{1}M_{2}} = (0 - \sqrt{2}, 3 - 4, 2 - 1) = (-\sqrt{2}, -1, 1)$
步骤 2：计算向量M1M2的模
向量M1M2的模为：
$|\overrightarrow{M_{1}M_{2}}| = \sqrt{(-\sqrt{2})^2 + (-1)^2 + 1^2} = \sqrt{2 + 1 + 1} = \sqrt{4} = 2$
步骤 3：计算方向余弦
方向余弦为向量M1M2的各分量除以向量的模，即：
$\cos\alpha = \dfrac{-\sqrt{2}}{2}$
$\cos\beta = \dfrac{-1}{2}$
$\cos\gamma = \dfrac{1}{2}$
向量M1M2的坐标为M2坐标减去M1坐标，即：
$\overrightarrow{M_{1}M_{2}} = (0 - \sqrt{2}, 3 - 4, 2 - 1) = (-\sqrt{2}, -1, 1)$
步骤 2：计算向量M1M2的模
向量M1M2的模为：
$|\overrightarrow{M_{1}M_{2}}| = \sqrt{(-\sqrt{2})^2 + (-1)^2 + 1^2} = \sqrt{2 + 1 + 1} = \sqrt{4} = 2$
步骤 3：计算方向余弦
方向余弦为向量M1M2的各分量除以向量的模，即：
$\cos\alpha = \dfrac{-\sqrt{2}}{2}$
$\cos\beta = \dfrac{-1}{2}$
$\cos\gamma = \dfrac{1}{2}$