Suppose that you stand at the point 
(5,6,0)
(5,6,0) and look in the direction of a point that is not visible because it is behind the ball. You will then be looking at a point on the sphere.
If 
(−1,−2,2)
(−1,−2,2) is not visible from 
(5,6,0)
(5,6,0), find the point on the sphere at which you are looking if you look in the direction of 
(−1,−2,2)
(−1,−2,2). Otherwise, find the point on the sphere at which you look if you are looking in the direction of 
(−1,−2,1)
(−1,−2,1). 
equation of sphere is x²+y²+z²=1 and please give the answers round off to 4 decimal places

We are given the following setup:
- You are standing at point \((5,6,0)\) and looking in a direction.
- The equation of the sphere is \(x^2 + y^2 + z^2 = 1\).
- Two directions are given: either towards point \((-1,-2,2)\) or \((-1,-2,1)\).
  
The goal is to find the intersection point(s) where the direction from your position touches the surface of the sphere.

### Step 1: Parametrize the direction vector
The direction vector from point \((5,6,0)\) to either of the points \((-1,-2,2)\) or \((-1,-2,1)\) is given by:
- For \((-1,-2,2)\), the vector is:
  \[
  \vec{v_1} = (-1 - 5, -2 - 6, 2 - 0) = (-6, -8, 2)
  \]
- For \((-1,-2,1)\), the vector is:
  \[
  \vec{v_2} = (-1 - 5, -2 - 6, 1 - 0) = (-6, -8, 1)
  \]

Thus, the parametric equation of the line is:
\[
\vec{r}(t) = \vec{r_0} + t \cdot \vec{v}
\]
where \(\vec{r_0} = (5,6,0)\) is the starting point and \(\vec{v}\) is the direction vector.

### Step 2: Find the intersection of the line and the sphere
The equation of the sphere is:
\[
x^2 + y^2 + z^2 = 1
\]
Substitute the parametric coordinates of the line into this equation:
\[
(x(t))^2 + (y(t))^2 + (z(t))^2 = 1
\]
We will handle this process for each direction vector:

#### Case 1: Direction vector \(\vec{v_1} = (-6, -8, 2)\)
The parametric coordinates of the line in this case are:
\[
x(t) = 5 - 6t, \quad y(t) = 6 - 8t, \quad z(t) = 0 + 2t
\]
Substitute these into the sphere equation:
\[
(5 - 6t)^2 + (6 - 8t)^2 + (2t)^2 = 1
\]
Simplify and solve for \(t\).

#### Case 2: Direction vector \(\vec{v_2} = (-6, -8, 1)\)
The parametric coordinates of the line in this case are:
\[
x(t) = 5 - 6t, \quad y(t) = 6 - 8t, \quad z(t) = 0 + t
\]
Substitute these into the sphere equation:
\[
(5 - 6t)^2 + (6 - 8t)^2 + (t)^2 = 1
\]
Simplify and solve for \(t\).

Let me calculate the solutions.

Suppose that you stand at the point (5,6,0) and look in the direction of a point that is not visible because it is behind the ball. You will then be looking at a point on the sphere. If (-1,-2,2) is not visible from (5,6,0), find the point on the sphere at which you are looking if you look in the direction of (-1,-2,2). Otherwise, find the point on the sphere at which you look if you are looking in the direction of (-1,-2,1). Equation of sphere is x² + y² + z² = 1.

Solve the problem of finding the point of intersection between a line and a unit sphere. Explore parametric equations, vector geometry, and the formula for the intersection of a line with a sphere. This problem is suitable for undergraduate students studying vector geometry.

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