Respuesta :
We are concern about the vector (x-x0, y-y0, z-z0) equaling to 5.
This happens when √(x-x0)^2 + (y-y0)^2 + (z-z0)^2 = 5
Square both side:
(x-x0)^2 + (y-y0)^2 + (z-z0)^2 = 25 , a sphere
Which you will recognize as a circle of radius one centered at (x0, y0, z0)
Answer:
Point [tex]\left ( x,y,z\right )[/tex] represents a sphere
Step-by-step explanation:
Take [tex]r=\left ( x,y,z \right )\,,\,r_0=\left ( x_0,y_0,z_0 \right )[/tex]
We need to describe the set of all points (x, y, z) such that [tex]\left | r-r_0 \right |=5[/tex]
Solution :
For [tex]r-r_0=\left ( x,y,z \right )-\left ( x_0,y_0,z_0 \right )[/tex], we will subtract the respective elements .
[tex]r-r_0=\left ( x,y,z \right )-\left ( x_0,y_0,z_0 \right )=\left ( x-x_0,y-y_0,z-z_0 \right )[/tex]
Therefore, [tex]\left | r-r_0 \right |=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}[/tex]
As [tex]\left | r-r_0 \right |=5[/tex], we get
[tex]\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}=5[/tex]
On squaring both sides, we get
[tex]\left ( \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2} \right )^2=5^2\\(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=25[/tex]
The general equation of a sphere is (x - a)² + (y - b)² + (z - c)² = r², where (a, b, c) denotes the center of the sphere and r represents the radius .
So, [tex](x-x_0)^2+(y-y_0)^2+(z-z_0)^2=25[/tex] represents a sphere with center as [tex]\left ( x_0,y_0,z_0 \right )[/tex] and radius equal to 5 units