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?

1 Answer
Apr 9, 2018

See below.

Explanation:

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)