The perpendicular distance of the origin from the plane x-3y+4z-6=0 is....units
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the equation of the given plane is
2x - 3y + 4z - 6 = 0 . The vector equation of this plane is r.(2î - 3j + 4k)= 6. It's normal form is
r. (2÷29î - 3 ÷ 29j + 4 ÷ 29k ) = 6÷29 Hence the distance of the plane form the origin is 6÷29.
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