Answer to Question #132815 in Analytic Geometry for Harshita

Question

Show that the two spheres x²+y²+z²+6y+2z+8=0 and x²+y²+z²+6x+8y+4z+20=0

 are orthogonal.

Answer 1

$2*u_1*u_2+2*v_1*v_2+2*w_1*w_2=d_1+d_2$ --- it is a orthogonality condition

$x^2+y^2+z^2+6*y+2*z+8=0$ --- the first sphere

$u_1 = 0/2 = 0$ ; $v_1 = 6/2 = 3$ ; $w_1 = 2/2 = 1$ ; $d_1 = 8$

$x^2 + y^2+z^2+6*x+8*y+4*z+20 = 0$ --- the second sphere

$u_2 = 6/2 = 3$ ; $v_2 = 8/2 = 4$ ; $w_2 = 4/2 = 2$ ; $d_2 = 20$

As the first sphere is orthogonal to the second sphere, we must have

$2*0*3+2*3*4+2*2*1=8+20$

$24+4=28$

$28=28$ - equality is true

so these two spheres are orthogonal