Equation of a Plane in Normal Form of three dimensional geometry.

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answered Oct 30, 2024 by PawanThaware (78.6k points)
selected Nov 5, 2024 by NidhiUpadhyay

Best answer

Let the distance of plane ABC from origin is d. ON is normal to plane ABC from origin O and n̂ is unit normal vector along \(\overrightarrow {ON}.\)

Then \(\overrightarrow {ON}\) = dn̂

Let P be any point on the plane. So, \(\overrightarrow {NP}\) is perpendicular to \(\overrightarrow {ON}.\)

Equation of a Plane in Normal Form

This is the vector form equation of the plane.

Equation of Cartesian Form

Let P(x, y, z) be any point on a plane, then

\(\overrightarrow {OP}\) = \(\vec r\)= x î + y ĵ + z k̂

Let direction cosines of n are l, m, n, then

n̂ = l î + m ĵ + nk̂

Since, \(\vec r\) n̂ = d,

(x î + y ĵ + zk̂).(l î + mĵ + nk̂) = d

⇒ lx + my + nz = d

This is the cartesian form of the plane in the normal form.

Remark:

If distance of the plane from the origin is d and l, m, n are the direction cosines of the normal to the plane, then the foot of the perpendicular is (ld, md, nd).
Equation lx + my + nz - d shows that if \(\vec r\) .(aî + bĵ + ck̂) = d is the vector equation of a plane, then ax + by+ cz = d is the cartesian equation of the plane, where a, b and c are the direction ratios of the normal to the plane.