Geometry in architecture and building by Hans Sterk

By Hans Sterk

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If the mirror is not a coordinate plane, like the y, z–plane, things usually become a lot harder to describe. In later chapters we will go into this in more detail. In the case at hand, the ‘normal vector approach’ runs as follows. Reflect the normal vector (2, 1, −3) to obtain (−2, 1, −3), so the equation of U ′ is of the form −2x + y − 3z = d for some d. To determine d, pick a point on U , say (2, 3, 0) and reflect it: (−2, 3, 0). Substitute in −2x + y − 3z = d and conclude that d = 7. The equation of U ′ is therefore −2x + y − 3z = 7.

For example, for λ = 2 and µ = −3 we find 2 · (2, −1, 0) − 3 · (3, 0, 1) = (−5, −2, −3). Next we turn to a general strategy for finding vector parametric descriptions. The way to find a vector parametric equation of a plane, starting from an equation, is to solve the equation (and don’t forget: there are infinitely many solutions; it’s a plane after all). Suppose V is the plane with equation x1 + 2x2 − 3x3 = 4; so V is parallel to U , since the vector (1, 2, −3) is perpendicular to both planes.

10 Exercises 55 a) Use two planes from each fan to describe the two fans using parameters. b) Find a plane U1 from the first fan that makes an angle of 45◦ with the x, y–plane. Then find a plane U2 from the second fan that make an angle of 90◦ with U1 . c) Find a plane V1 from the first fan that makes an angle of 30◦ with the x, y–plane. Then find a plane V2 from the second fan that makes an angle of 90◦ with V1 . 24 Constructing a house shaped like a cube A building is shaped like a cube, say with bounding planes x = 0, x = 1, y = 0, y = 1, z = 0 and z = 1.

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