3d: Return to basic: other side: one of considers 0 < t < 1s = t - 1 L1 and L2 are lines. t * L1 + s * L2 (formula1) the above doesn't have parabra.And, from L1 to L2 are lined if you display fomula1.The start always has been here.And, we'…

Basically, a, b, c, d : constantx, y, z, d : arbitaryax + by + cz + d = 0 If we has 3sete of ( x, y, z ), we could the above formula. And, a + bx + c + d = ax + 2c if we found the above, we solved it lke the following: a = ba + c + d = 2c …

so, we forcus on the bold below formula. 1/(a+b) = c/a + d/b c*b/a*b + a*d/a*b = ( c*b + a*d ) / a*b c*b + a*d = C a+b=C*a*b a + b = (c*b + a*d)*a* b the above looks also interesting. abc = 1 ...(1) abd = 1 ...(2) we could replace, bc = 1/…

1/(a+b) = c/a + d/b a/(a+b) = c + d*a/b a= bc + ad ... (1) a+b = b ...(2) b or c equals 0 I hope c = 0 ... and don't know that. The above formula doesn't succeed. Because 1/2 = 2/4 But, it looks interesting. At least, 1/(a+b) = C*(c/a + d/…