What is the equation of those tangent planes to the sphere #x^2+y^2+z^2+2x-4y+6z-7=0# which intersect the line #6x-3y-23=0#, #3z+2=0#?

Question

1317 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-09T15:30:04

See below.

Given a line

#L -> p = p_1 + lambda vec v# and a plane

#Pi-> << p-p_1, vec n >> = 0#

if #L in Pi rArr lambda << vec v, vec n >> = 0#

now given a sphere

#Sigma-> norm(p-p_0) = r#

the normal vector to #Sigma# is

#vec sigma = (p-p_0)/norm(p-p_0)#

then a tangent plane to #Sigma# should obey

#{(<< p-p_1, (p-p_0)/norm(p-p_0) >> = 0),(<< vec v , (p-p_0)/norm(p-p_0) >> = 0), (norm(p-p_0) = r):}#

or

#{(normp^2 - << p, p_0+p_1 >> + << p_0, p_1 >> = 0), (<< vec v, p >> = << vec v, p_0 >> = 0), ( norm(p-p_0) = r):}#

here

#p_0 = (-1,2,-3)#
#r = sqrt 21#
#p_1 = (13/3, 1,-2/3)#
#vec v = (3,6,0)#

and after solving we get

#t_1 = (1,1,1)# and
#t_2 = (3,0,-4)#

the two tangent points, so the planes are

#Pi_1-> << p-p_1, vec sigma_1 >> = 0#
#Pi_2-> << p-p_1, vec sigma_2 >> = 0#

with

#vec sigma_1 = (2,-1,4)#
#vec sigma_2=(4,-2,-1)#