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प्रश्न
Show that the points (3, 2, 2), (–1, 4, 2), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4). Find also its radius.
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उत्तर
Let the points be A (3, 2, 2), B (\[-\] 1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere
whose centre be P (1, 3, 4).
Since, AP, BP, CP and DP are radii.
Hence, AP = BP = CP = DP
\[AP = \sqrt{\left( 1 - 3 \right)^2 + \left( 3 - 2 \right)^2 + \left( 4 - 2 \right)^2}\]
\[ = \sqrt{\left( - 2 \right)^2 + \left( 1 \right)^2 + \left( 2 \right)^2}\]
\[ = \sqrt{4 + 1 + 4}\]
\[ = \sqrt{9}\]
\[ = 3\]
\[BP = \sqrt{\left( 1 + 1 \right)^2 + \left( 3 - 4 \right)^2 + \left( 4 - 2 \right)^2}\]
\[ = \sqrt{\left( 2 \right)^2 + \left( - 1 \right)^2 + \left( 2 \right)^2}\]
\[ = \sqrt{4 + 1 + 4}\]
\[ = \sqrt{9}\]
\[ = 3\]
\[CP = \sqrt{\left( 1 - 0 \right)^2 + \left( 3 - 5 \right)^2 + \left( 4 - 6 \right)^2}\]
\[ = \sqrt{\left( 1 \right)^2 + \left( - 2 \right)^2 + \left( - 2 \right)^2}\]
\[ = \sqrt{1 + 4 + 4}\]
\[ = \sqrt{9}\]
\[ = 3\]
\[DP = \sqrt{\left( 1 - 2 \right)^2 + \left( 3 - 1 \right)^2 + \left( 4 - 2 \right)^2}\]
\[ = \sqrt{\left( - 1 \right)^2 + \left( 2 \right)^2 + \left( 2 \right)^2}\]
\[ = \sqrt{1 + 4 + 4}\]
\[ = \sqrt{9}\]
\[ = 3\]
Here, we see that AP = BP = CP = DP
Hence, A (3, 2, 2), B (\[-\]1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere whose radius is 3 .
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