Reduce the equation `2x-3y-6z=14` to the normal form and hence fine the length of perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

Explanation

The given equation of the plane is `2 x-3 y-6=14`

Now,` \sqrt{2^{2}+(-3)^{2}+(-6)^{2}}=\sqrt{4+9+36}=\sqrt{49}=7`

Dividing eqn(1) by `7` , we get

`\frac{2}{7} {x}-\frac{3}{7} {y}-\frac{6}{7} {z}=2 \quad \text {........(2) }`


The cartesian equation of the normal form of a plane is


`{lx}+{my}+{nz}={p} \quad \text {........(3) }`
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