Find the radius and centre of the circular section of sphere |r|= 4, cut off by the plane
r.(2i-j+4k)= 3.
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Expert's answer
2021-12-02T16:10:43-0500
Let S be the sphere in R3 with center O(x0,y0,z0) and radius r, and let P be the plane with equation Ax+By+Cz+D=0, so that n=(A,B,C)is a normal vector of P.
If P0 is an arbitrary point on P, the signed distance from the center of the sphere O to the plane P is
ρ=∥n∥(O−P0)n=A2+B2+C2Ax0+By0+Cz0−D.
The intersection S∩P is a circle if and only if r<ρ<r, and in that case, the circle has radius rc=r2−ρ2 and center
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