Find the shortest distance from origin to the plane x -2y -2z = 3.

1.	Find the shortest distance from origin to the plane x -2y -2z = 3.
Answer» Shortest distance from the origin (0, 0) to the plane x- 2y - 2z = 3 is \(d = \left\| \frac{0-2\times 0 - 2\times 0 -3}{\sqrt{1^2 + (-2)^2 + (-2)^2}}\right\| = \left\|\frac{-3}{\sqrt{1 + 4 + 4}}\right\|\) \(= \frac 3{\sqrt 9} = \frac 33 = 1\) unit

Answer»

Shortest distance from the origin (0, 0) to the plane x- 2y - 2z = 3 is

\(d = \left| \frac{0-2\times 0 - 2\times 0 -3}{\sqrt{1^2 + (-2)^2 + (-2)^2}}\right| = \left|\frac{-3}{\sqrt{1 + 4 + 4}}\right|\)

\(= \frac 3{\sqrt 9} = \frac 33 = 1\) unit

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Find the shortest distance from origin to the plane x -2y -2z = 3.

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