Find the radius and the center of the circular section of the sphere |r| = 17 cut off by the plane r·(i+2j+2k) = 24. Solution
what is the normal to the plane that passes through (0,0,0)? If the distance from the plane to the origin is d, then
d^2+r^2 = 17^2
c) Find the radius and the center of the circular section of the sphere |r| = 17 cut off by the plane r·(i+2j+2k) = 24.
To find the radius and center of the circular section of the sphere, we will first find the equation of the circle formed by the intersection of the sphere and the plane.
The equation of the sphere is given as |r| = 17, which implies that the distance of any point r in space from the origin is 17 units.
The equation of the plane is r · (i+2j+2k) = 24, where · represents the dot product. Let's call this plane P.
To find the equation of the circle formed by the intersection of the sphere and the plane P, we need to find the coordinates of the center and the radius of the circle.
Step 1: Find the normal vector of the plane P.
The vector i+2j+2k is the direction vector of the plane P. To find the normal vector (perpendicular to the plane), we can take the cross product of the direction vector with any other vector on the plane, such as the vector r=(x, y, z). Taking the cross product gives us the normal vector N:
N = (i+2j+2k) x (x, y, z)
= (2z-y)i + (x-2z)j + (y-x)k
Step 2: Substitute the value of N into the equation of the plane to find the center of the circle.
We substitute (2z-y)i + (x-2z)j + (y-x)k for the vector i+2j+2k in the equation r · (i+2j+2k) = 24:
(x, y, z) · [(2z-y)i + (x-2z)j + (y-x)k] = 24
Expanding this equation, we get:
2xz - xy + x(x-2z) + y(y-x) + z(2z-y) = 24
2xz - xy + x^2 - 2xz + y^2 - xy + 2z^2 - yz = 24
Simplifying gives:
x^2 + y^2 + z^2 - 3xy - 3yz = 24
Step 3: Find the radius of the circle.
The radius of the circle is the distance from the center of the circle to any point on the circle. To find this, we substitute the coordinates of the center (x, y, z) into the equation of the sphere |r| = 17:
|x, y, z| = 17
sqrt(x^2 + y^2 + z^2) = 17
Since we have already found the equation of the circle (x^2 + y^2 + z^2 - 3xy - 3yz = 24), we can substitute the x, y, and z values from there into this equation to solve for the radius.
By solving the equations obtained from Step 2 and Step 3 simultaneously, we can find the center and the radius of the circular section of the given sphere.