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Q. The direction cosines of the normal to the plane $ 6x-3y-2z=1 $ are:

KEAMKEAM 2001

Solution:

Direction cosines of the normal to the plane are $ \frac{6}{\sqrt{36+9+4}},\frac{-3}{\sqrt{36+9+4}},\frac{-2}{\sqrt{36+9+4}} $ $ \equiv \left( \frac{6}{7},\frac{-3}{7},\frac{-2}{7} \right) $