Reduce the equation of the plane `x-2y-2z=12` to normal form and hence find the length of the perpendicular for the origin to the plane. Also, find the direction cosines of the normal to the plane.

Explanation

General equation of a plane in normal form


`vec{r} cdot hat{n}=d `


`x-2 y-2 z=12 `


`(x hat{i}+y hat{j}+2 hat{k}) cdot(hat{i}-2 hat{j}-2 hat{k})=12 `


`vec{r}(hat{i}-2 hat{j}-2 hat{k})=12 `


`Rightarrow frac{vec{r}(hat{i}-2 hat{j}-2 hat{k})}{3}=4 `


`Rightarrow vec{r} cdot(frac{1}{3} hat{i}-frac{2}{3} hat{j}-frac{2}{3} hat{k))=4 .`


length of perpendicular `=4`.


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