T/F the diameter of the sphere is the longest distance on it? show work
True, the diameter of a sphere is the longest distance on it.
In order to prove this, let's define a sphere with radius r. The diameter of the sphere is the line segment that passes through the center of the sphere and has both endpoints on the surface. We can represent the sphere with its equation:
(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2
Since the center of the sphere is (a, b, c), we can substitute this into the equation to get:
(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2
Now let's consider another line segment on the surface of the sphere. If this line segment is not the diameter, then it can be represented by two points (x1, y1, z1) and (x2, y2, z2) on the surface of the sphere.
Using the distance formula, we can find the length of this line segment:
√[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]
Now, let's consider the line segment that is the diameter of the sphere. We can represent this line segment with two points on the surface of the sphere, say (x3, y3, z3) and (x4, y4, z4).
Using the distance formula again, we can find the length of this line segment:
√[(x4 - x3)^2 + (y4 - y3)^2 + (z4 - z3)^2]
Now, let's compare the lengths of these two line segments:
√[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2] vs √[(x4 - x3)^2 + (y4 - y3)^2 + (z4 - z3)^2]
Since both line segments lie on the surface of the sphere, their endpoints satisfy the equation of the sphere. Therefore, we have:
(x1 - a)^2 + (y1 - b)^2 + (z1 - c)^2 = r^2 and (x2 - a)^2 + (y2 - b)^2 + (z2 - c)^2 = r^2
(x3 - a)^2 + (y3 - b)^2 + (z3 - c)^2 = r^2 and (x4 - a)^2 + (y4 - b)^2 + (z4 - c)^2 = r^2
Substituting these equations into the lengths of the line segments, we have:
√[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2] = √[(x4 - x3)^2 + (y4 - y3)^2 + (z4 - z3)^2] = r
Therefore, the lengths of both line segments are equal, and since the diameter passes through the center of the sphere, it is the longest distance on the surface of the sphere. Thus, the statement is true.