Question

shortest distance from the origin to the plane x - 2y - 2z = 3 is​

Answer 1

Given:

the plane x - 2y - 2z = 3

To find:

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

Solution:

We have given the point as the origin from where we have to find the distance of the point from this plane which equation is x - 2y - 2z = 3.

And in the coordinate geometry the distance of the point (x₁, y₁, z₁) from the plane of the equation ax+by+cz+d = 0 is given by:

$\frac{ax_{1}+by_{1}+cz_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}} }$

Here the coordinates of the point are (0, 0, 0) and the equation of the plane is x-2y-2z = 3 ⇒ x-2y-2z-3 = 0

$|\frac{ 0-0-0-3}{\sqrt{1^{2}+(-2)^{2}+(-2)^{2}} }|$

$|\frac{-3}{\sqrt{1 + 4 + 4} }|$

$|\frac{-3}{\sqrt{9}}|$

$|\frac{-3}{3}|$

1

The shortest distance from the origin to the plane x - 2y - 2z = 3  is 1

Answer 2

SOLUTION

TO DETERMINE

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

FORMULA TO BE IMPLEMENTED

The shortest distance from the origin to the plane ax + by + cz = d is

$\displaystyle \sf{ = \bigg| \frac{d}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } } \bigg| }$

EVALUATION

Here the given equation of the plane is x - 2y - 2z = 3

Hence the shortest distance from the origin to the plane x - 2y - 2z = 3 is

$\displaystyle \sf{ = \bigg| \frac{3}{ \sqrt{ {1}^{2} + {( - 2)}^{2} + {( - 2)}^{2} } } \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \frac{3}{ \sqrt{1 + 4 + 4 } } \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \frac{3}{ \sqrt{9 } } \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \frac{3}{ 3 } \bigg| } \: \: unit$

$\displaystyle \sf{ = 1} \: \: unit$

FINAL ANSWER

The shortest distance from the origin to the plane x - 2y - 2z = 3 is 1 unit

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