Show that the plane #2x-4y-z+9=0# touches the sphere which passes through #(1,1,6)# and whose centre is #(2,-3,4)#. Also, find the point of contact.?

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Answer 1 · 2017-06-21T14:52:20

See below.

With #p = (x,y,z)#

#Pi -> << vec n, p - p_0 >> = 2x-4y-z+9=0#

here

#vec n = (2,-4,-1)#
#p_0 = (0,0,9)#

#S-> norm(p-p_1) = norm(p_1-p_2) = r#

here

#p_1=(2,-3,4)#
#p_2=(1,1,6)#

Now if #Pi# and #S# are tangent then

#p_t = p_1-r vec n/norm(vec n) in Pi#

Computing #p_t = (0,1,5)#

and as can easily be verified

#<< vec n , p_t-p_0 >> = 0# so the tangency point is

#p_t = (0,1,5)#

Answer 2 · 2017-06-28T04:06:40

Distance from the Plane to Centre = Radius of the Given Sphere
Point of contact #(0, 1, 5)#

The Sphere passes through the point #(1, 1, 6)#.
Hence, The distance from it's centre #(2, -3, 4)# to this point is the radius of the sphere.

#:. r = sqrt((2-1)^2 + (-3-1)^2 + (4-6)^2)#
#=sqrt(21)#

Now, the distance between the given plane #2x-4y-z+9=0# to the centre of the sphere is :

#= |(2.2 + (-4).(-3) + (-1).4 + 9)|/(sqrt(2^2 + (-4)^2 + (-1)^2))#
#= |4+12-4+9|/(sqrt(4+16+1))#
#=21/sqrt21#
#=sqrt21#

As the distance from the plane to centre is equal to the distance of the radius of the sphere, it touches the sphere.

The point of contact is:
#vec u = (2, -3, 4)-(2, -4, -1)#
#= (0, 1, 5)#