Question

A plane passes through a fixed point (a, b, c). The locus of the foot of the perpendicular to it from the origin is a sphere of radius –

• A. $\sqrt{a^{2} + b^{2} + c^{2}}$
• B. $\frac{1}{2}\sqrt{a^{2} + b^{2} + c^{2}}$
• C. $a^{2} + b^{2} + c^{2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{2}\sqrt{a^{2} + b^{2} + c^{2}}$

Answer 2

Let the foot of the perpendicular from the origin on the given plane be p(a, b, g). Since the plane passes through

A (a, b, c)

\ AP ^ OP Ž $\overset{\rightarrow}{AP}$. $\overset{\rightarrow}{OP}$ = 0

Ž [(a – a)$\widehat{i}$ + (b – b)$\widehat{j}$ + (g – c) $\widehat{k}$] . (a$\widehat{i}$ + b$\widehat{j}$ + g$\widehat{k}$) = 0

a (a – a) + b (b – b) + g (g – c) = 0

Hence, the locus of (a, b, g) is

x(x – a) + y(y – b) + z(z – c) = 0

x² + y² + z² – ax – by – cz = 0

which is a sphere of radius $\frac{1}{2}\sqrt{a^{2} + b^{2} + c^{2}}$.