if r − kx, y, zl, a − ka1, a2, a3 l, and b − kb1, b2, b3l, show that the vector equation sr 2 ad sr 2 bd − 0 represents a sphere, and find its center and radius.

Answer:the center of the sphere is (1/k, 1/k, 1/k) (or (r/k, r/k, r/k) if r is taken as 1), and the radius is sqrt((1 - ka1)^2 + (1 - ka2)^2 + (1 - ka3)^2) / k.

Step-by-step explanation:

The given vector equation, s(r - kx, y, z) = s(r - ka1, a2, a3) = s(r - kb1, b2, b3) - 0, represents a sphere in three-dimensional space. To find its center and radius, we need to analyze the equation.

Let's compare the given vector equation to the standard equation of a sphere:

(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2.

From the given equation, we can identify the following:

1. Center: The center of the sphere can be found by equating the expressions inside the parentheses to zero.

Setting r - kx = 0, we find that x = r/k.

Setting r - ky = 0, we find that y = r/k.

Setting r - kz = 0, we find that z = r/k.

Therefore, the center of the sphere is (r/k, r/k, r/k), or simply (1/k, 1/k, 1/k) if r is taken as 1.

2. Radius: The radius of the sphere can be found by calculating the distance between the center and any point on the sphere.

Considering the points (r - ka1, a2, a3) and (r - kb1, b2, b3), we can calculate the distance between the center (1/k, 1/k, 1/k) and any of these points.

Using the distance formula, the distance between two points (x1, y1, z1) and (x2, y2, z2) is given by:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2).

Therefore, the radius of the sphere is d = sqrt((1/k - ka1)^2 + (1/k - a2)^2 + (1/k - a3)^2), which simplifies to sqrt((1 - ka1)^2 + (1 - ka2)^2 + (1 - ka3)^2) / k.