If $A (- 1, 4, - 3)$ is one end of the diameter $A B$ of the spere $x^{2} + y^{2} + 2^{2} - 3 x - 2 y + 2 z - 15 = 0$ , then find the co-ordinates of $B$ .

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Solution

The equation of the given sphere is
$x^{2} + y^{2} + z^{2} - 3 x - 2 y + 2 z - 15 = 0$
$2 u =$ Co.efficient of $x = - 3 \Rightarrow u = \frac{- 3}{2}$
$2 v =$ Co.efficient of $y = - 2 \Rightarrow v = \frac{- 2}{2} = - 1$
$2 w =$ Co.efficient of $z = 2 \Rightarrow w = \frac{2}{2} = 1$
$d = - 15$
$∴$ centre is $= (- u, - v, - w) = (\frac{- 3}{2}, 1, - 1)$
Given one end of the diameter is $A (- 1, 4, 3)$ and centre (mid point of the diameter $A B$ ) is $(\frac{- 3}{2}, 1, - 1)$
Let the other end be $B (x, y, z)$
$\frac{x - 1}{2} = \frac{3}{2} \Rightarrow x - 1 = 2 \Rightarrow x = 4$
$\frac{x + 4}{2} = 1 \Rightarrow y + 4 = 2 \Rightarrow y = - 2$
$\frac{z - 3}{2} = 1 \Rightarrow z - 3 = - 2 \Rightarrow z = 1$

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