The coordinate of a point equidistant from th | vedclass

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(C) The required point is the center of the sphere passing through the given four points.Let the equation of the sphere be $x^2 + y^2 + z^2 + 2ux + 2v

Vedclass · Accepted Answer

$\left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2} \right)$

Vedclass · Answer

(C) The required point is the center of the sphere passing through the given four points.Let the equation of the sphere be $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 \dots (i)$.Since the sphere passes through $(0, 0, 0)$,we get $d = 0$.Since it passes through $(a, 0, 0)$,we have $a^2 + 2ua = 0 \implies u = -a/2$.Since it passes through $(0, b, 0)$,we have $b^2 + 2vb = 0 \implies v = -b/2$.Since it passes through $(0, 0, c)$,we have $c^2 + 2wc = 0 \implies w = -c/2$.The center of the sphere is $(-u, -v, -w) = (a/2, b/2, c/2)$.This point is equidistant from all four given points.

The coordinate of a point equidistant from the points $(0, 0, 0), (a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ is

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