Finding the tangent equation of curve X ^ 2 + y ^ 2 + Z ^ 2 = 2, x + y + Z = 0 at point (1,0. - 1) is a normal plane equation

Finding the tangent equation of curve X ^ 2 + y ^ 2 + Z ^ 2 = 2, x + y + Z = 0 at point (1,0. - 1) is a normal plane equation

Let f (x, y, z) = x ^ 2 + y ^ 2 + Z ^ 2-2,
Then the partial derivatives of F to x, y and Z are 2x, 2Y and 2Z respectively,
The normal vector of tangent plane is (2,0, - 2) by substituting the coordinates of point (1,0, - 1),
Therefore, the tangent plane equation is 2 (x-1) - 2 (Z + 1) = 0, which is reduced to x-z-2 = 0,
Therefore, the tangent equation is {x + y + Z = 0, x-z-2 = 0,
That is, (x-1) / 1 = Y / (- 2) = (Z + 1) / 1,
The normal plane equation is 1 * (x-1) - 2 * (y-0) + 1 * (Z + 1) = 0, that is, x-2y + Z = 0