Question

5. If F = (x, y, z), a = (a1, 22, a3) and 6 = {b1,b2,63), show that the vector equation (F- ā) . (-6)=0 is the equation of a sphere. What are the coordinates of the center of this sphere? What is the radius of the sphere?

Answer 1

5A) Given - T = («,4,7), à = (2, , ) 1 = (6, 62, 63) vector equation sch ( -a ), (7-6) = 0. (7-2) = (2,4, 2) – (a, a, ay) (8-3) = (x, y, z) - ( 6, 62, 63) = (x-b, t-ba, z-63) (8-2). (8-5) = (x-a,,g-a2, 2-a3). (x-bi, y-ba, 2-33) = (-a) (2-6,) + (9-9₂) (G-be) + (z-a3) (2-63) = 0 vector equation (x-a,)(x-b, ) + (4-02) (4-62) & (za-ag) (z-bz)=0 o

The above equation and equation of Sphere To find - 8) Co ordinates of the Sphere a) radius of the Sphere Generd equation of Sphere els [(-a) + (9-b) + (z -c)² = 2 2 2 where (a, b, c) are center coordinate ef and ro radius eq 0 becomes xZ-a, x-b, x ta, b, tegnagyobay tazby + z ² – Azz-bzZ+Q2 63 = 0 *747z?- (q, tb.) x + (Qz+ba]y - (aztba) z ta, b, tazb2 taz bz = 0 = 2 (a, b) x +49-2 (an then by tas, taabat as bz=0.

The above equation underlined ferm are in the form of x-ray if we add it then it becomes (x+y)2 and again substract with ye - u rry= (x - y - y +(2_ Q3 +63 ) = (, ) 70.44) taak taz 63 +(Ca, to, 't Cantb.) 7(autta)?) ta, b, tazbz tag bz The above aquation is in the form of Sphere equation (eg (2) Compare with eq ②

artbz az a to be auto come a = 2 are ? Center Coordinates (a, tb, aztba ) az +63 - / and s= }((a, tb,) + Castbe) + (ag tha)) ta, b, taabeta, bz is) sediud of the sphere (8) (ca:+6) +(02 + by] + Cag toy ta, b, taabat ag ba (cohole root) 8 = 14 (la, to, 3 Cartba) * Castoy}] +a, b, that they