Question

大一高等数学

大一高等数学空间曲面的切平面与法线。
希望有步骤。

Answer 1

令 F = x^2 + y^2 + z^2 - 1
F'<x> = 2x, F'<y> = 2y, F'<z> = 2z
2x /1 = 2y/(-1) = 2z/2 = t
x = t/2, y = -t/2, z = t
代入 x^2 + y^2 + z^2 = 1, (3/2)t^2 = 1, t = ± √(2/3) = ± √6/3,
切点 P(√6/6, -√6/6, √6/3 ), Q(-√6/6, √6/6, -√6/3 ),
切平面方程 (x-√6/6) - (y+√6/6) + 2(z-√6/3) = 0, 即 x - y + 2z - √6 = 0;
或 (x+√6/6) - (y-√6/6) + 2(z+√6/3) = 0, 即 x - y + 2z + √6 = 0。