Find the radius and the center of the circular section of the sphere |r| = 26 cut off by the plane r ·(2i+6j+3k) = 70 ?

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Answer 1 · 2018-01-30T20:40:24

See below.

Calling

#vec n = (2,6,3)#
#r = (x,y,z)#
#r_0 = 26#
#c_0 = 70#

we have

#S -> norm r = r_0#
#Pi-> << r, vec n >> = c_0#
#C-> Pi nn S#

The center of #C# can be obtained as the projection of #S# center which is #(0,0,0)# onto #Pi# or as the intersection

#p_c = L nn Pi#

where

#L-> r = lambda vec n# then substituting

#<< lambda vec n, vec n >> = c_0 rArr lambda = c_0/norm(vec n)^2# and then

#p_c = c_0/norm(vec n)^2 vec n#

The radius of #C# can be calculated according

#r_c^2 = norm (p_c)(r_0-norm(p_c))# or

#r_c = sqrt(c_0/norm(vec n)(r_0-c_0/norm(vec n))) = 4 sqrt10#