Question

If the equation of plane is at 3p distance from origin which intersect the axis at A, B, C then the centroid of ΔABC from an equation ____

(A) (1/x²) + (1/y²) + (1/z²) = – (1/p²)

(B) (1/x²) + (1/y²) + (1/z²) = (1/p²)

(C) (1/x²) + (1/y²) + (1/z²) + (1/p²) = 1

(D) (1 / x²) + (1 / y²) + (1 / z²) = 0

Answer 1

The correct option (B) (1/x²) + (1/y²) + (1 / z²) = (1/p²)

Explanation:

plane intersect A(a, 0, 0), B(0, b, 0) and C(0, 0, c)

eqⁿ of plane is (x/a) + (y/b) + (z/c) = 1

perpendicular distance from (0, 0, 0) is 3p

∴ [(|– 1|)/√{(1/a)² + (1/b)² + (1/c)²}] = 3p

∴ (1/a2) + (1/b²) + (1/c²) = (1/9p²) ...(1)

entroid of ΔABC, [(a/3), (b/3), (c/3)].

consider option, (1/x²) + (1/y²) + (1/z²) = (1p²) 

∴ [1/(a²/9)] + [1/(b2/9)] + [1/(c²/9)] = (1p²)  

 ∴ 9[(1/a²) + (1/b²) + (1/c2)] = (1/p²) 

∴ (1/a^₂) + (1/b²) + (1/c²) = (1/9p²)

∴ from (1), option b is the correct answer.