What is the distance between the plane #x+y+z=0# and the sphere of unit radius with centre #(1, 3, 5)# ?

1 Answer
George C.
Dec 16, 2017

#3sqrt(3)-1#

Explanation:

The normal to the plane #x+y+z=0# running through the point #(1, 3, 5)# can be described parametrically by:

#p(t) = (1,3,5)+t(1,1,1) = (t+1, t+3, t+5)#

So when #t=-3# we have:

#p(-3) = (-2, 0, 2)#

which is on the plane #x+y+z = 0#

The distance between #(-2, 0, 2)# and #(1,3,5)# is:

#sqrt(3^2+3^2+3^2) = sqrt(27) = 3sqrt(3)#

The line segment joining these two points intersects the sphere at a distance of #1# from the centre. So the distance between this point of intersection and the plane is #3sqrt(3)-1#

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