Câu 2. Cho hình lập phương ABCD. A'B'C'D' .Gọi M, N lần lượt là trung điểm của A'D' và C'D' . Xác định góc giữa hai vectơ overrightarrow (MN) và overrightarrow (A'B)

Let the side length of the cube be denoted by 'a'. We establish a coordinate system with A' as the origin (0,0,0). Then the coordinates of the relevant points are:<br /><br />* A'(0, 0, 0)<br />* D'(a, 0, 0)<br />* C'(a, a, 0)<br />* B'(0, a, 0)<br />* M (a/2, 0, 0) (midpoint of A'D')<br />* N (a, a/2, 0) (midpoint of C'D')<br /><br />Now we find the vectors:<br /><br />* $\overrightarrow{MN} = N - M = (a - a/2, a/2 - 0, 0 - 0) = (a/2, a/2, 0)$<br />* $\overrightarrow{A'B'} = B' - A' = (0, a, 0)$<br /><br />To find the angle θ between these two vectors, we use the dot product formula:<br /><br />$\overrightarrow{MN} \cdot \overrightarrow{A'B'} = ||\overrightarrow{MN}|| \ ||\overrightarrow{A'B'}|| \cos θ$<br /><br />$\overrightarrow{MN} \cdot \overrightarrow{A'B'} = (a/2)(0) + (a/2)(a) + (0)(0) = a^2/2$<br /><br />$||\overrightarrow{MN}|| = \sqrt{(a/2)^2 + (a/2)^2 + 0^2} = \sqrt{a^2/4 + a^2/4} = \sqrt{a^2/2} = a/\sqrt{2}$<br /><br />$||\overrightarrow{A'B'}|| = \sqrt{0^2 + a^2 + 0^2} = a$<br /><br />Therefore:<br /><br />$a^2/2 = (a/\sqrt{2})(a) \cos θ$<br /><br />$\cos θ = \frac{a^2/2}{a^2/\sqrt{2}} = \frac{1}{\sqrt{2}}$<br /><br />$θ = \cos^{-1}(\frac{1}{\sqrt{2}}) = 45°$<br /><br />Therefore, the angle between the vectors $\overrightarrow{MN}$ and $\overrightarrow{A'B'}$ is $\boxed{45^\circ}$.<br />