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 number binary operation +, * sq.root of number Result ------------------------ =

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 plane Equn Edge length (21):el1 area (124)=ABC 124 x +y +z +=0 Edge length (31):el2 area(456)=BCD 456 x +y +z +=0 Edge length (41):el3 area(235)=ACD 235 x +y +z +=0 Edge length (32):el4 area(136)=ABD 136 x +y +z +=0 Edge length (43):el5 ⊥from D to ABC Edge length (42):el6 ⊥ from A to BCD perimeter of tertahedron area of tertrahedron ⊥from B to ACD Vol. of tetrahedron (v124) Vol. of tetrahedron (v235) ⊥ from C to ABD Vol. of tetrahedron (v456) Vol. of tetrahedron (v136) inradius Vol. of tetrahedron (detA/3!) A: det tr: minor A tr: (x1) (y1) (z1) (t1) (xm1) (ym1) (zm1) (tm1) (x2) (y2) (z2) (t2) (xm2) (ym2) (zm2) (tm2) (x3) (y3) (z3) (t3) (xm3) (ym3) (zm3) (tm3) (x4) (y4) (z4) (t4) (xm4) (ym4) (zm4) (tm4) cofactor A adj.A tr: (xa1) (ya1) (za1) (ta1) (xa2) (ya2) (za2) (ta2) (xa3) (ya3) (za3) (ta3) (xa4) (ya4) (za4) (ta4) Inv A tr: det: trAm tr: (xi1) (yi1) (zi1) (ti1) (xi2) (yi2) (zi2) (ti2) (xi3) (yi3) (zi3) (ti3) (xi4) (yi4) (zi4) (ti4) minor invA tr: (xim1) (yim1) (zim1) (tim1) (xim2) (yim2) (zim2) (tim2) (xim3) (yim3) (zim3) (tim3) (xim4) (yim4) (zim4) (tim4) *Regular tetrahedron is one having 4 vertices, 4 faces with each face being equilateral triangle, 6 edges, having surface area √3(Edge length)2 and volume √2(edge length)3 /12 . Irregular tetrahedrons may have any or all sides having different lengths. Volume of a tetrahedron is (1/3) times the area of any face * perpendicular from the residual vertex on the face. * if there are 4 vertices with coordinates (x1,y1,z1),(x2,y2,z2),(x3,y3,z3) and (x4,y4,z4), then volume of tetrahedron is |x1  y1  z1 1| |x2  y2  z2 1|   *( 1/3!) |x3  y3  z3 1| |x4  y4  z4 1|* Regular tetrahedron with centroid at the origin (0,0,0)    D (0,0, 2√2a/(3√3) ) , A (0, a/√3, -√2a/(3√3) ), B (-a/2, -a/2√3,-√2a/(3√3) ), C (a/2, -a/2√3,-√2a/(3√3) ) * trAm : For each minor, there is a trace which are arranged in a matrix form corresponding to each minor.* Suppose a variable plane given by the equation lx+my+nz=p makes with the co-ordinate planes a tetrahedron, then the vertices are (p/l,0,0) , (0,p/m,0), (0,0,p/n) , (0,0,0)If (x1,y1,z1) is the centroid of the tetrahedron, then x1=p/4l, y1= p/4m, z1=p/4n. Vol. of tetrahedron is 1/6 |p/l 0   0                              0 p/m 0                              0  0    p/n| = 64x1y1z1/6; *tetrahedron with the following vertices (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)  Its dual tetrahedron will have vertices (-1,-1,-1), (-1,1,1), (1,-1,1), (1,1,-1)* Inradius of tetrahedron r= 3V / (A1+A2+A3+A4) where V is volume, and denominator is total area. * sum of the areas of any 3 faces > area of the 4th face * there exist tetrahedrons having inger valued edge length, face area,volume. Exa- 1edge length =896,opp. edge length=990, other 4 edge lengths each 1073.* consecutive integer as edges- 6,7,8,9,10,11. Volume=48.* Suppose 1 vertex at origin(0,0,0) and other 3 vertices are represented by matrix|x1  y1  z1 | |x2  y2  z2 |   |x3  y3  z3 | A rigid rotation of these points about the origin will leave the determinant of this matrix unchanged since determinant of rotation matrix is 1 by definition. In general, we can apply n-dimensional rotation that places n-1 of vertices into a subspace orthogonal to one of the axes.  Thus we can rotate the configuration so that first 2 vertices are in say x-y plane resulting in a transformed matrix with the same determinant but zeroes in all except the last element of last row. We have- |x1  y1  z1 |          |x'1  y'1    0 |           |x'1 y'1| |x2  y2  z2 |    =    |x'2  y'2   0 |    =h3 |x3  y3  z3 |          | x'3  y'3  z'3 |          |x'2  y'2| this 2x2 matrix contains the coordinates of 2 of the vertices. Now we can apply a rotation such that    |x'1  y'1 |   = |x''1     0 |    |x'2 y'2  |      |x"2   y"2| Here the y"2 coordinate is the height of the second vertex above the subspace (i.e., the line) containing the other two vertices (including the origin). Noting that x"1 is the height h1 of the first vertex over the origin, we have x1  y1  z1 |          |x2  y2  z2 |      =h3*h2*h1 |x3  y3  z3 |          * 6 face diagonals form a tetrahedron in a cube and volume of tetrahedron is 1/3 of that of the cube. * nth tetrahedral number Tn given by Tn = Σk=1 to n  k(k+1) /2 The numbers are 1,4,10,20,35,56,84,120,165,220, ......* Tn + Tn-1  = 12 + 22 +32 +... n2 * only 3 tetrahedral numbers are perfect squares. 1, 4, 19600. * The infinite sum of tetrahedral number reciprocals is 3/2.