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?

Apr 9, 2018

See below.

Explanation:

Given a line

$L \to p = {p}_{1} + \lambda \vec{v}$ and a plane

$\Pi \to \left\langlep - {p}_{1} , \vec{n}\right\rangle = 0$

if $L \in \Pi \Rightarrow \lambda \left\langle\vec{v} , \vec{n}\right\rangle = 0$

now given a sphere

$\Sigma \to \left\lVert p - {p}_{0} \right\rVert = r$

the normal vector to $\Sigma$ is

$\vec{\sigma} = \frac{p - {p}_{0}}{\left\lVert p - {p}_{0} \right\rVert}$

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

${p}_{0} = \left(- 1 , 2 , - 3\right)$
$r = \sqrt{21}$
${p}_{1} = \left(\frac{13}{3} , 1 , - \frac{2}{3}\right)$
$\vec{v} = \left(3 , 6 , 0\right)$

and after solving we get

${t}_{1} = \left(1 , 1 , 1\right)$ and
${t}_{2} = \left(3 , 0 , - 4\right)$

the two tangent points, so the planes are

${\Pi}_{1} \to \left\langlep - {p}_{1} , {\vec{\sigma}}_{1}\right\rangle = 0$
${\Pi}_{2} \to \left\langlep - {p}_{1} , {\vec{\sigma}}_{2}\right\rangle = 0$

with

${\vec{\sigma}}_{1} = \left(2 , - 1 , 4\right)$
${\vec{\sigma}}_{2} = \left(4 , - 2 , - 1\right)$