The sphere $x^2+y^2+z^2-2x+6y+14z+3=0$ meets the line joining $A(2,-1,4),B(5,5,5)$ in the points $C$ and $D$. Prove that $AC:CB=-AD:DB=1:2$.
Any hints will be appreciated.
Edit1: I know if we find the equation of plane tangent to sphere and $\perp$ to line $AB$, then it has formula. But how to get equation of plane?
Edit2: There is misprint in the book, the coordinates of $A$ are $(2,-1,-4)$.
On
The line through $A(2,-1,4),B(5,5,5)$ has a direction given by $\vec u=B-A=\{3,6,1\}$
so the line $AB$ has equation $A+s\vec v =\{2+3s,-1+6s,4+s\}$
plug in the equation of the sphere
$(s+4)^2+(3 s+2)^2+(6 s-1)^2-2 (3 s+2)+6 (6 s-1)+14 (s+4)+3=0$
$2 \left(23 s^2+26 s+35\right)=0$
Which has no real solutions
Hope this helps
On
In answer to your specific second question about tangent planes, every tangent plane to the sphere is at a distance equal to the sphere’s radius from its center. The family of planes perpendicular to $AB$ has equations of the form $(B-A)\cdot(x,y,z)=d$, i.e., $3x+6y+z=d$. Find the center and radius of the sphere and use the formula for the distance of a point to a plane to get a quadratic equation for $d$.
Hint : The equation of the line through $A$ and $B$ is $(x,y,z)=(2+3t,-1+6t,4+t)$. Substitute this into the equation for the circle, solve the quadratic to find the points $C$ and $D$ ... etc ...