В тетраэдре ABCD, ребра которого равны 1, найдите угол между плоскостями АВС и BCD.

Let's analyze the problem carefully: **Given:** - Tetrahedron \( ABCD \) with all edges equal to 1. - Find the angle between the planes \( ABC \) and \( BCD \). --- ### Step 1: Understand the structure Since all edges are equal to 1, the tetrahedron is **regular**. - All edges are length 1. - So, faces \( ABC \) and \( BCD \) are equilateral triangles. --- ### Step 2: Coordinates setup To find the angle between the planes, we can assign coordinates to the points. **Choose coordinate axes:** - Place \( B \) at the origin: \( B = (0, 0, 0) \). - Place \( A \) at \( (1, 0, 0) \)? - Place \( C \)? Since \( ABC \) is an equilateral triangle with side 1: The points \( A \) and \( C \) are at a distance 1 from \( B \). To make calculations easier, let's set: \[ A = (1, 0, 0), \quad B = (0, 0, 0) \] - Find \( C \): Because \( AC = 1 \), and \( BC = 1 \), point \( C \) must satisfy: \[ |C - A| = 1, \quad |C - B| = 1 \] - Let's pick \( C \) in the \( xy \)-plane: \[ C = (x, y, 0) \] From \( |C - B| = 1 \): \[ \sqrt{x^2 + y^2} = 1 \Rightarrow x^2 + y^2 = 1 \] From \( |C - A| = 1 \): \[ \sqrt{(x - 1)^2 + y^2} = 1 \] Square both sides: \[ (x - 1)^2 + y^2 = 1 \] Subtract \( x^2 + y^2 = 1 \): \[ (x - 1)^2 + y^2 - (x^2 + y^2) = 0 \Rightarrow (x^2 - 2x + 1) + y^2 - x^2 - y^2 = 0 \] Simplify: \[ -2x + 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} \] Now, from \( x^2 + y^2 = 1 \): \[ \left(\frac{1}{2}\right)^2 + y^2 = 1 \Rightarrow \frac{1}{4} + y^2 = 1 \Rightarrow y^2 = \frac{3}{4} \] So: \[ y = \pm \frac{\sqrt{3}}{2} \] Choose \( y > 0 \): \[ C = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right) \] --- ### Step 3: Coordinates of point \( D \) Similarly, \( D \) is such that all edges are length 1: - \( D \) is at distance 1 from \( A \), \( B \), and \( C \). Find coordinates for \( D = (x, y, z) \): \[ |D - A|^2 = 1 \] \[ |D - B|^2 = 1 \] \[ |D - C|^2 = 1 \] - From \( B = (0, 0, 0) \): \[ x^2 + y^2 + z^2 = 1 \] - From \( A = (1, 0, 0) \): \[ (x - 1)^2 + y^2 + z^2 = 1 \] Subtract the first from the second: \[ (x - 1)^2 + y^2 + z^2 - (x^2 + y^2 + z^2) = 0 \] \[ x^2 - 2x + 1 + y^2 + z^2 - x^2 - y^2 - z^2 = 0 \] \[ -2x + 1 = 0 \Rightarrow x = \frac{1}{2} \] - From \( C = (\frac{1}{2}, \frac{\sqrt{3}}{2}, 0) \): \[ \left( \frac{1}{2} - \frac{1}{2} \right)^2 + \left( y - \frac{\sqrt{3}}{2} \right)^2 + z^2 = 1 \] \[ 0 + \left( y - \frac{\sqrt{3}}{2} \right)^2 + z^2 = 1 \] Recall from the first that: \[ x^2 + y^2 + z^2 = 1 \Rightarrow \left(\frac{1}{2}\right)^2 + y^2 + z^2 = 1 \] \[ \frac{1}{4} + y^2 + z^2 = 1 \] Similarly, substitute \( y = y \), and consider the earlier equation: \[ \left( y - \frac{\sqrt{3}}{2} \right)^2 + z^2 = 1 \] Express \( z^2 \): \[ z^2 = 1 - \left( y - \frac{\sqrt{3}}{2} \right)^2 \] From the previous: \[ \frac{1}{4} + y^2 + z^2 = 1 \Rightarrow z^2 = 1 - \frac{1}{4} - y^2 = \frac{3}{4} - y^2 \] Set equal these two expressions for \( z^2 \): \[ \frac{3}{4} - y^2 = 1 - \left( y - \frac{\sqrt{3}}{2} \right)^2 \] Expanding: \[ \left( y - \frac{\sqrt{3}}{2} \right)^2 = y^2 - 2 y \frac{\sqrt{3}}{2} + \frac{3}{4} = y^2 - y \sqrt{3} + \frac{3}{4} \] Now: \[ \frac{3}{4} - y^2 = 1 - ( y^2 - y \sqrt{3} + \frac{3}{4} ) \] \[ \frac{3}{4} - y^2 = 1 - y^2 + y \sqrt{3} - \frac{3}{4} \] Simplify RHS: \[ 1 - y^2 + y \sqrt{3} - \frac{3}{4} = \left( 1 - \frac{3}{4} \right) - y^2 + y \sqrt{3} = \frac{1}{4} - y^2 + y \sqrt{3} \] LHS: \[ \frac{3}{4} - y^2 \] So: \[ \frac{3}{4} - y^2 = \frac{1}{4} - y^2 + y \sqrt{3} \] Subtract \( \frac{1}{4} - y^2 \) from both sides: \[ \left( \frac{3}{4} - y^2 \right) - \left( \frac{1}{4} - y^2 \right) = y \sqrt{3} \] \[ \frac{3}{4} - y^2 - \frac{1}{4} + y^2 = y \sqrt{3} \] \[ \frac{2}{4} = y \sqrt{3} \] \[ \frac{1}{2} = y \sqrt{3} \] Therefore: \[ y = \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{6} \] Now, find \( z^2 \): \[ z^2 = \frac{3}{4} - y^2 = \frac{3}{4} - \left( \frac{\sqrt{3}}{6} \right)^2 \] \[ z^2 = \frac{3}{4} - \frac{3}{36} = \frac{3}{4} - \frac{1}{12} = \frac{9}{12} - \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \] Thus: \[ z = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3} \] Take \( z > 0 \): \[ D = \left( \frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3} \right) \] --- ### **Step 4: Find normal vectors of \( \triangle ABC \) and \( \triangle BCD \)** - **Plane \( ABC \):** Vectors: \[ \vec{AB} = B - A = (0 - 1, 0 - 0, 0 - 0) = (-1, 0, 0) \] \[ \vec{AC} = C - A = \left(\frac{1}{2} - 1, \frac{\sqrt{3}}{2} - 0, 0 - 0 \right) = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right) \] Normal vector: \[ \vec{n}_{ABC} = \vec{AB} \times \vec{AC} \] Cross product: \[ \vec{n}_{ABC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 0 & 0 \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \end{vmatrix} = \mathbf{i} (0 \cdot 0 - 0 \cdot \frac{\sqrt{3}}{2}) - \mathbf{j} ( -1 \cdot 0 - 0 \cdot (-\frac{1}{2}) ) + \mathbf{k} ( -1 \cdot \frac{\sqrt{3}}{2} - 0 \cdot (-\frac{1}{2}) ) \] Simplify: \[ \vec{n}_{ABC} = \mathbf{i}(0 - 0) - \mathbf{j}(0 - 0) + \mathbf{k} \left( - \frac{\sqrt{3}}{2} - 0 \right) = (0, 0, - \frac{\sqrt{3}}{2}) \] The normal vector: \[ \boxed{\vec{n}_{ABC} = \left( 0, 0, - \frac{\sqrt{3}}{2} \right)} \] - **Plane \( BCD \):** Vectors: \[ \vec{BC} = C - B = \left(\frac{1}{2} - 0, \frac{\sqrt{3}}{2} - 0, 0 - 0 \right) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right) \] \[ \vec{BD} = D - B = \left(\frac{1}{2} - 0, \frac{\sqrt{3}}{6} - 0, \frac{\sqrt{6}}{3} - 0 \right) = \left(\frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3}\right) \] Compute: \[ \vec{n}_{BCD} = \vec{BC} \times \vec{BD} \] Cross product: \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{6} & \frac{\sqrt{6}}{3} \end{vmatrix} \] Calculate the components: - \( x \)-component: \[ \left(\frac{\sqrt{3}}{2}\right) \cdot \frac{\sqrt{6}}{3} - 0 \cdot \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2} \times \frac{\sqrt{6}}{3} \] Simplify: \[ = \frac{\sqrt{3} \times \sqrt{6}}{2 \times 3} = \frac{\sqrt{3} \times \sqrt{6}}{6} \] Note: \[ \sqrt{3} \times \sqrt{6} = \sqrt{3 \times 6} = \sqrt{18} = 3 \sqrt{2} \] So: \[ \frac{3 \sqrt{2}}{6} = \frac{\sqrt{2}}{2} \] - \( y \)-component: \[ - \left( \frac{1}{2} \times \frac{\sqrt{6}}{3} - 0 \times \frac{1}{2} \right) = - \left( \frac{\sqrt{6}}{6} \right) = - \frac{\sqrt{6}}{6} \] - \( z \)-component: \[ \frac{1}{2} \cdot \frac{\sqrt{3}}{6} - \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{12} - \frac{\sqrt{3}}{4} \] Rewrite: \[ \frac{\sqrt{3}}{12} - \frac{3 \sqrt{3}}{12} = - \frac{2 \sqrt{3}}{12} = - \frac{\sqrt{3}}{6} \] **Normal vector:** \[ \boxed{ \vec{n}_{BCD} = \left( \frac{\sqrt{2}}{2}, - \frac{\sqrt{6}}{6}, - \frac{\sqrt{3}}{6} \right) } \] --- ### **Step 5: Find the angle between the planes** The angle \( \theta \) between the planes is related to the normal vectors: \[ \cos \theta = \frac{\left| \vec{n}_{ABC} \cdot \vec{n}_{BCD} \right|}{\|\vec{n}_{ABC}\| \, \|\vec{n}_{BCD}\|} \] - Dot product: \[ \vec{n}_{ABC} \cdot \vec{n}_{BCD} = 0 \times \frac{\sqrt{2}}{2} + 0 \times \left(- \frac{\sqrt{6}}{6}\right) + \left(- \frac{\sqrt{3}}{2}\right) \times \left(- \frac{\sqrt{3}}{6}\right) \] Calculate: \[ \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{6} = \frac{\sqrt{3} \times \sqrt{3}}{2 \times 6} = \frac{3}{12} = \frac{1}{4} \] Since both are negative signs: \[ \text{dot product} = \frac{1}{4} \] - Norms: \[ \|\vec{n}_{ABC}\| = \sqrt{0^2 + 0^2 + \left(- \frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] \[ \|\vec{n}_{BCD}\| = \sqrt{ \left( \frac{\sqrt{2}}{2} \right)^2 + \left( - \frac{\sqrt{6}}{6} \right)^2 + \left( - \frac{\sqrt{3}}{6} \right)^2 } \] Calculate each component: \[ \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{2}{4} = \frac{1}{2} \] \[ \left( - \frac{\sqrt{6}}{6} \right)^2 = \frac{6}{36} = \frac{1}{6} \] \[ \left( - \frac{\sqrt{3}}{6} \right)^2 = \frac{3}{36} = \frac{1}{12} \] Sum: \[ \frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{6}{12} + \frac{2}{12} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4} \] Norm: \[ \|\vec{n}_{BCD}\| = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] --- ### **Step 6: Final calculation** \[ \cos \theta = \frac{\left| \frac{1}{4} \right|}{\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}} = \frac{\frac{1}{4}}{\frac{\sqrt{3} \times \sqrt{3}}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3} \] Thus, \[ \boxed{ \theta = \arccos \left(\frac{1}{3}\right) } \] --- ## **Answer:** **The angle between the planes \( ABC \) and \( BCD \) in the regular tetrahedron with side length 1 is** \[ \boxed{\theta = \arccos\left(\frac{1}{3}\right)} \]