Cin 67. Trong không gian với hệ tọa độ Oxy=, cho hai điểm A(-2;3;1) và B(5;6;2) Đường thẳng AB cắt mặt phẳng (ac) tại điểm M . Tinh tỉ số (AM)/(BM) A. (AM)/(BM)=(1)/(2) B (AM)/(BM)=2 C. (AM)/(BM)=(1)/(3) D. (AM)/(BM)=3 D(6,0,0),D(12,0,0) B. BM -2 Oao cho hình hộp ABCD A'B'C'D' có A(0;0;0), B(a;0;0)

This question is a repeat of Questions 2 and 3 in the example. The solution involves finding the coordinates of point M where the line AB intersects the xy-plane (z=0), and then calculating the ratio of the distances AM and BM.<br /><br />**Solution:**<br /><br />1. **Find the equation of the line AB:**<br /><br />The direction vector of the line AB is given by $\vec{AB} = B - A = (5 - (-2), 6 - 3, 2 - 1) = (7, 3, 1)$.<br /><br />The parametric equation of the line AB is:<br /><br />x = -2 + 7t<br />y = 3 + 3t<br />z = 1 + t<br /><br />2. **Find the intersection point M with the xy-plane (z=0):**<br /><br />Set z = 0 in the parametric equation:<br /><br />1 + t = 0 => t = -1<br /><br />Substitute t = -1 into the x and y equations:<br /><br />x = -2 + 7(-1) = -9<br />y = 3 + 3(-1) = 0<br /><br />Therefore, the coordinates of M are (-9, 0, 0).<br /><br />3. **Calculate AM and BM:**<br /><br />AM = $\sqrt{(-9 - (-2))^2 + (0 - 3)^2 + (0 - 1)^2} = \sqrt{(-7)^2 + (-3)^2 + (-1)^2} = \sqrt{49 + 9 + 1} = \sqrt{59}$<br /><br />BM = $\sqrt{(-9 - 5)^2 + (0 - 6)^2 + (0 - 2)^2} = \sqrt{(-14)^2 + (-6)^2 + (-2)^2} = \sqrt{196 + 36 + 4} = \sqrt{236}$<br /><br />4. **Calculate the ratio AM/BM:**<br /><br />AM/BM = $\sqrt{59} / \sqrt{236} = \sqrt{59/236} = \sqrt{1/4} = 1/2$<br /><br />Therefore, the answer is A. AM/BM = 1/2<br /><br /><br />**Final Answer:** The final answer is $\boxed{A}$<br />