find angel between spheres x^2 +y^2 +z^2 =29, x^2 +y^2 +z^2 +4x−6y−8z−47=0 (4,−3,2)


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Question Number 47034 by 23kpratik last updated on 04/Nov/18

$f i n d a n g e l b e t w e e n s p h e r e s x^{2} + y^{2} + z^{2} = 29, x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z - 47 = 0 (4, - 3, 2)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18
	$point (4,−3,2) satisfy both eqn of sphere now let p_1 be the tangent plane though(4,−3,2) with respect to first sphere and similarly p_2 be the tangent plane with respect to 2nd sphere determinanation of eqn plane p_1 and plane p_2 and angle between p_1 and p_2 is the required angle first sphere eqn... x^2 +y^2 +z^2 =29 centre c_1 (0,0,0) radius r_1 ((√(29)) ) direction ratio bdtween (0,0,0) and (4,−3,2) is formula {(x_2 −x_1 ),(y_2 −y_1 ),(z_2 −z_1 )}=4,−3,2 so eqn of plane p_(1 ) is formula A(x−x_1 )+B(y−y_1 )+C(z−z_1 )=0 A,B,C are direction ratio plane p_1 is 4(x−4)−3(y+3)+2(z−2)=0 4x−3y+2z−16−9−4=0 4x−3y+2z=29 ←plane p_1 the 2nd sphre x^2 +y^2 +z^2 +4x−6y−8z−47=0 centre (−2,3,4) direction ratio between (4,−3,2) and(−2,3,4) (6,−6,−2) eqn of second plane p_2 tangent to second sphere is A(x−x_1 )+B(y−y_1 )+C(z−z_1 )=0 6(x−4)−6(y+3)−2(z−2)=0 6x−6y−2z−24−18+4=0 6x−6y−2z=38←plane p_2 let required angle is θ formula cosθ=((a_1 a_2 +b_1 b_2 +c_1 c_2 )/((√(a_1 ^2 +b_1 ^2 +c_1 ^2 )) ×(√(a_2 ^2 +b_2 ^2 +c_2 ^2 )))) cosθ=(((6×4)+(−6×−3)+(−2×2))/((√((4)^2 +(−3)^2 +(2)^2 )) ×(√((6)^2 +(−6)^2 +(−2)^2 )) )) cosθ=((38)/((√(29)) ×(√(76)))) θ=cos^(−1) (((38)/((√(29)) ×(√(76)))))$
	$p o i n t (4, - 3, 2) s a t i s f y b o t h e q n o f s p h e r e$ $n o w l e t p_{1} b e t h e t a n g e n t p l a n e t h o u g h (4, - 3, 2)$ $w i t h r e s p e c t t o f i r s t s p h e r e a n d s i m i l a r l y$ $p_{2} b e t h e t a n g e n t p l a n e w i t h r e s p e c t t o 2 n d s p h e r e$ $d e t e r m i n a n a t i o n o f e q n p l a n e p_{1} a n d p l a n e p_{2}$ $a n d a n g l e b e t w e e n p_{1} a n d p_{2} i s t h e r e q u i r e d a n g l e$ $f i r s t s p h e r e e q n . . .$ $x^{2} + y^{2} + z^{2} = 29 c e n t r e c_{1} (0, 0, 0) r a d i u s r_{1} (\sqrt{29})$ $d i r e c t i o n r a t i o b d t w e e n (0, 0, 0) a n d (4, - 3, 2) i s$ $f o r m u l a {(x_{2} - x_{1}), (y_{2} - y_{1}), (z_{2} - z_{1})} = 4, - 3, 2$ $s o e q n o f p l a n e p_{1} i s$ $f o r m u l a A (x - x_{1}) + B (y - y_{1}) + C (z - z_{1}) = 0$ $A, B, C a r e d i r e c t i o n r a t i o$ $p l a n e p_{1} i s 4 (x - 4) - 3 (y + 3) + 2 (z - 2) = 0$ $4 x - 3 y + 2 z - 16 - 9 - 4 = 0$ $4 x - 3 y + 2 z = 29 \leftarrow p l a n e p_{1}$ $t h e 2 n d s p h r e x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z - 47 = 0$ $c e n t r e (- 2, 3, 4)$ $d i r e c t i o n r a t i o b e t w e e n (4, - 3, 2) a n d (- 2, 3, 4)$ $(6, - 6, - 2)$ $e q n o f s e c o n d p l a n e p_{2} t a n g e n t t o s e c o n d s p h e r e$ $i s A (x - x_{1}) + B (y - y_{1}) + C (z - z_{1}) = 0$ $6 (x - 4) - 6 (y + 3) - 2 (z - 2) = 0$ $6 x - 6 y - 2 z - 24 - 18 + 4 = 0$ $6 x - 6 y - 2 z = 38 \leftarrow p l a n e p_{2}$ $l e t r e q u i r e d a n g l e i s θ$ $f o r m u l a$ $c o s θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \times \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$ $c o s θ = \frac{(6 \times 4) + (- 6 \times - 3) + (- 2 \times 2)}{\sqrt{{(4)}^{2} + {(- 3)}^{2} + {(2)}^{2}} \times \sqrt{{(6)}^{2} + {(- 6)}^{2} + {(- 2)}^{2}}}$ $c o s θ = \frac{38}{\sqrt{29} \times \sqrt{76}}$ $θ = {c o s}^{- 1} (\frac{38}{\sqrt{29} \times \sqrt{76}})$
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