Question

if the cube shown above is sliced by a plane to create a triangle, which sets of vertices could the plane pass through?
a. 1, 3, and 4
2, 3, and 8
1, 3, and 7
b. 4, 5, and 7
1, 2, and 7
2, 3, and 7
c. 4, 5, and 7
1, 3, and 8
2, 4, and 7
d. 3, 5, and 8
1, 2, and 6
2, 4, and 7

Explanation:

Step1: Analyze each option

• For option A:
• Consider the vertices \(1\), \(3\), and \(7\).
• In a cube, if we try to visualize the plane passing through these three non - collinear vertices.
• A plane passing through three non - collinear points forms a triangle.
• For other options in A (e.g., \(1\), \(3\), \(4\) are in the same face (a rectangle, not a triangle - forming set as the intersection would be a rectangle - like shape in 2D when sliced)); \(2\), \(3\), \(8\) won't form a triangle (by cube's structure analysis).
• For option B:
• \(4\), \(5\), \(7\) are in the same face (a rectangle, not a triangle - forming set as the intersection would be a rectangle - like shape in 2D when sliced); \(1\), \(2\), \(7\) won't form a triangle (by cube's structure analysis); \(2\), \(3\), \(7\) won't form a triangle (by cube's structure analysis).
• For option C:
• \(4\), \(5\), \(7\) are in the same face (a rectangle, not a triangle - forming set as the intersection would be a rectangle - like shape in 2D when sliced); \(1\), \(3\), \(8\) won't form a triangle (by cube's structure analysis); \(2\), \(4\), \(7\) won't form a triangle (by cube's structure analysis).
• For option D:
• \(3\), \(5\), \(8\) won't form a triangle (by cube's structure analysis); \(1\), \(2\), \(6\) won't form a triangle (by cube's structure analysis); \(2\), \(4\), \(7\) won't form a triangle (by cube's structure analysis).

Answer:

A. \(1\), \(3\), and \(7\)