Find the value of a for which the plane x+y+z=a touches the sphere x^2+y^2+z^2-2x-2y-2z-6=0
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equation of sphere => (x-1)²(y-1)²(z-1)² = 3²
centre of sphere is (1,1,1)
radius = 3
if plane touches the sphere then perpendicular distance from centre of sphere to the plane is equal to the radius of sphere
equation of plane is x+y+z-a = 0
|(1+1+1-a)/√(1²+1²+1²)| = 3
|(3-a)/√3| = 3
(3-a)/√3 = +3 or -3
3-a = 3√3
a = 3(1-√3)
or
3-a = -3√3
a = 3(1+√3)
centre of sphere is (1,1,1)
radius = 3
if plane touches the sphere then perpendicular distance from centre of sphere to the plane is equal to the radius of sphere
equation of plane is x+y+z-a = 0
|(1+1+1-a)/√(1²+1²+1²)| = 3
|(3-a)/√3| = 3
(3-a)/√3 = +3 or -3
3-a = 3√3
a = 3(1-√3)
or
3-a = -3√3
a = 3(1+√3)
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