Question

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3 y – z = 5

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Solution

The equation of plane is
2x + 3y – z = 5
Dividing both sides by

square root of left parenthesis 2 right parenthesis squared plus left parenthesis 3 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared end root space equals space square root of 4 plus 9 plus 1 end root space equals space square root of 14 comma space space we space get comma

fraction numerator 2 over denominator square root of 14 end fraction straight x plus fraction numerator 3 over denominator square root of 14 end fraction straight y space minus space fraction numerator 1 over denominator square root of 14 end fraction straight z space equals space fraction numerator 5 over denominator square root of 14 end fraction

It is of form lx + my + nz = p where

straight l equals space fraction numerator 2 over denominator square root of 14 end fraction comma space space straight m space equals space fraction numerator 3 over denominator square root of 14 end fraction comma space space straight n space equals space minus fraction numerator 1 over denominator square root of 14 end fraction comma space space straight p space equals space fraction numerator 5 over denominator square root of 14 end fraction

∴ direction cosines of the normal to the plane are

fraction numerator 2 over denominator square root of 14 end fraction comma space fraction numerator 3 over denominator square root of 14 end fraction comma space minus fraction numerator 1 over denominator square root of 14 end fraction

and distance from origin =

fraction numerator 5 over denominator square root of 14 end fraction.

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