Question

find the angle of intersection at (4,-3,2) of spheres x^2+y^2+z^2=29 and x^2+y^2+z^2+4x-6y-8z-47=0.

Answer 1

Let's find gradient

x^2+y^2+z^2+4x-6y-8z-47=0.

{4 + 2 x, -6 + 2 y, -8 + 2 z}
{12,-12,-4}

{2 x, 2 y, 2 z}
{8,-6,4}


let's plug in number

\[\frac{(\{12,-12,-4\}.\{8,-6,4\})}{8 \sqrt{551}}\text{==}\text{Cos}[a]\]

\[\text{ArcCos}\left[\sqrt{\frac{19}{29}}\right]\]=0.62762

Answer 2

approximately 35 degree

Answer 3

thanks