Explanation:
1a. The dark line through the middle of the figure is intended to show where the planes intersect. Points A, D, and B are shown on the line. The line can be referred to using any two of those points:
The line of intersection is line AB.
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1b. The only three points shown on any line in the figure are the ones shown on the line of intersection:
points A, B, D
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1c. A plane can be named using any three non-collinear points on the plane. Possible names are ...
plane ABF
plane BDG
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1d. The points shown to be in plane P are {A, B, C, D, E}. The points shown to be in plane R are {A, B, D, F, G}. To answer this question, you can choose any four points from either list.
points A, B, F, G are coplanar
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1e. In parts (a) and (b) we identified that points A, B, and D are on the line that lies in both planes. Any of these points will do.
point A lies in both planes
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2. The distance formula can be used to find the distances between the points. Then those distances can be added to give the perimeter.
AB = √((-1-(-2))^2+(4-1)^2) = √(1+9) = √10 ≈ 3.16
BC = √((2-(-2))^2+(1-1)^2) = √16 = 4
CA = √((2-(-1))^2+(1-4)^2) = √(9+9) = 3√2 ≈ 4.24
So the perimeter is ...
P = AB +BC +CA = 3.16 +4.00 +4.24 = 11.40
The perimeter of ABC is about 11.4 units.
