Solution.
Find the equation of normal to the plane:
x − 8 4 = y − 5 3 = z − 4 0 = k , \frac{x-8}{4}=\frac{y-5}{3}=\frac{z-4}{0}=k, 4 x − 8 = 3 y − 5 = 0 z − 4 = k , where k k k is some number.
Then
C ( 4 k + 8 ; 3 k + 5 ; 4 ) − C(4k+8;3k+5;4)- C ( 4 k + 8 ; 3 k + 5 ; 4 ) − center of required spheres.
Radius of required spheres:
R = ( 4 k + 8 − 8 ) 2 + ( 3 k + 5 − 5 ) 2 + ( 4 − 4 ) 2 = = 25 k 2 = 5 k . R=\sqrt{(4k+8-8)^2+(3k+5-5)^2+(4-4)^2}=\newline
=\sqrt{25k^2}=5k. R = ( 4 k + 8 − 8 ) 2 + ( 3 k + 5 − 5 ) 2 + ( 4 − 4 ) 2 = = 25 k 2 = 5 k . Sphere x 2 + y 2 + z 2 = 1 x^2+y^2+z^2=1 x 2 + y 2 + z 2 = 1 has center ( 0 ; 0 ; 0 ) (0;0;0) ( 0 ; 0 ; 0 ) and radius 1. 1. 1.
We will have equation:
( 1 + 5 k ) 2 = ( 5 k ) 2 + 105 , where 105 (1+5k)^2=(5k)^2+105,
\newline \text{ where } 105 ( 1 + 5 k ) 2 = ( 5 k ) 2 + 105 , where 105 is distance between (0;0;0) and (8;5;4).
From here
1 + 10 k + 25 k 2 = 25 k 2 + 105 , k = 10.4 1+10k+25k^2=25k^2+105,
\newline
k=10.4 1 + 10 k + 25 k 2 = 25 k 2 + 105 , k = 10.4 So, ( 49.6 ; 36.2 ; 4 ) (49.6;36.2;4) ( 49.6 ; 36.2 ; 4 ) and ( − 33.6 ; − 26.2 ; 4 ) (-33.6;-26.2;4) ( − 33.6 ; − 26.2 ; 4 ) are the centres of spheres.
Answer. ( − 33.6 ; − 26.2 ; 4 ) , ( 49.6 ; 36.2 ; 4 ) . (-33.6;-26.2;4), (49.6;36.2;4). ( − 33.6 ; − 26.2 ; 4 ) , ( 49.6 ; 36.2 ; 4 ) .
Comments
Dear Naushad ahmad, we received answers provided in a solution of the question.
But sir answer is (4, 2 , 4) and (64/21, 27/21, 4) . Please comment