Find the distance of the plane 2x – 3 y + 4 z – 6 = 0 from the origin.

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Solution

The equation of plane is
2 x – 3 y + 4 z – 6 = 0 ...(1)
Direction ratios of the normal to the plane (1) are 2, – 3, 4.
Dividing each by

square root of left parenthesis 2 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared plus left parenthesis 4 right parenthesis squared end root space equals space square root of 4 plus 9 plus 16 end root space equals space square root of 29 comma space

we get the direction cosines of the normal as

fraction numerator 2 over denominator square root of 29 end fraction comma space minus fraction numerator 3 over denominator square root of 29 end fraction comma space fraction numerator 4 over denominator square root of 29 end fraction

.
∴ dividing (1) throughout by

square root of 29 comma

we get,

fraction numerator 2 over denominator square root of 29 end fraction straight x plus fraction numerator negative 3 over denominator square root of 29 end fraction straight y space plus space fraction numerator 4 over denominator square root of 29 end fraction straight z space equals space fraction numerator 6 over denominator square root of 29 end fraction

This is of the form lx + my + nz = p, where p is the distance of the plane from the origin.
∴ distance of plane from the origin =

fraction numerator 6 over denominator square root of 29 end fraction.

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