Length of the perpendicular from the centre of the conicoid to the tangent plane

Given:
Conicoid equation: x^2 - y^2 - z^2 = 1
Point on the tangent plane: (1, 1, 1)

Step 1: Find the centre of the conicoid
To find the centre of the conicoid, we need to determine the values of x, y, and z that satisfy the equation x^2 - y^2 - z^2 = 1. Notice that this equation resembles the equation of a hyperboloid of one sheet. The centre of the conicoid is the point (0, 0, 0) since the coefficients of x^2, y^2, and z^2 are all 1.

Step 2: Find the equation of the tangent plane
To find the equation of the tangent plane at (1, 1, 1), we need to determine the normal vector to the plane. The normal vector can be found by taking the gradient of the conicoid equation at the given point. The gradient of the conicoid equation is given by (∂f/∂x, ∂f/∂y, ∂f/∂z), where f(x, y, z) = x^2 - y^2 - z^2.

∂f/∂x = 2x
∂f/∂y = -2y
∂f/∂z = -2z

Evaluating the gradient at (1, 1, 1):
∂f/∂x = 2(1) = 2
∂f/∂y = -2(1) = -2
∂f/∂z = -2(1) = -2

So, the normal vector to the tangent plane at (1, 1, 1) is (2, -2, -2). The equation of the tangent plane can be written as:
2(x - 1) - 2(y - 1) - 2(z - 1) = 0
2x - 2y - 2z + 2 + 2 - 2 = 0
2x - 2y - 2z + 2 = 0
x - y - z + 1 = 0

Step 3: Find the length of the perpendicular
The length of the perpendicular from the centre of the conicoid to the tangent plane can be found using the formula for the distance between a point and a plane. The formula is given by:
d = |Ax + By + Cz + D| / √(A^2 + B^2 + C^2)

In this case, the point is (0, 0, 0) and the equation of the plane is x - y - z + 1 = 0. Substituting the values into the formula, we have:
d = |0(0) + 0(0) + 0(0) + 1| / √(1^2 + (-1)^2 + (-1)^2)
d = |1| / √(1 +