ps1sol - 18.02 Fall 2008 – Problem Set 1, Part B...

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Unformatted text preview: 18.02 Fall 2008 – Problem Set 1, Part B Solutions 1. a) P 1 = (1 , 1 , 1), P 2 = (1 , − 1 , − 1), P 3 = ( − 1 , − 1 , 1), P 4 = ( − 1 , 1 , − 1). Each pair of points differs by two sign changes from the others. All six edges of the tetrahedron are diagonals of faces of the cube and hence have the same length. For instance, | −−−→ P 1 P 2 | = |( 1 − 1 , − 1 − 1 , − 1 − 1 )| = |( , − 2 , − 2 )| = 2 √ 2 . P 3 P 1 P 4 P 2 b) cos θ = A · B / | A || B | = ( 1 , 1 , 1 )·( 1 , − 1 , − 1 ) / ( √ 3) 2 = − 1 / 3. θ ≈ 1 . 91 radians (109 . 5 ◦ ). c) Adjacent edges: −−−→ P 1 P 2 = ( , − 2 , − 2 ) , −−−→ P 1 P 3 = (− 2 , − 2 , ) . cos α = −−−→ P 1 P 2 · −−−→ P 1 P 3 | −−−→ P 1 P 2 || −−−→ P 1 P 3 | = 4 (2 √ 2) 2 = 1 2 ; α = π/ 3 = 60 ◦ The faces are equilateral triangles, so the angles are 60 ◦ . Opposite edges: −−−→ P 1 P 2 = ( , − 2 , − 2 ) , −−−→ P 3 P 4 = ( , 2 , − 2 ) . cos β = −−−→ P 1 P 2 · −−−→ P 3 P 4 | −−−→ P 1 P 2 || −−−→ P 3 P 4 | = (2 √ 2) 2 = 0; β = π/ 2 = 90 ◦ By symmetry the perpendicular bisector to an edge contains the opposite edge, so these two edges are perpendicular to each other. d) −−−→ P 1 P 2 = ( , − 2 , − 2 ) , −−−→ P 1 P 3 = (− 2 , − 2 , ) . −−−→ P 1 P 2 × −−−→ P 1 P 3 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ˆ ı ˆ ˆ k − 2 − 2 − 2 − 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = −...
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This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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ps1sol - 18.02 Fall 2008 – Problem Set 1, Part B...

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