CBSE (Arts) Class 12CBSE
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Plane - Equation of a Plane in Normal Form

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notes

Consider a plane whose perpendicular distance from  the origin is d (d ≠ 0). in following fig. 

If `vec (ON)` is the normal from the origin to the plane, and `hat n ` is the unit normal vector along `vec (ON)`. Then `vec (ON)` = d . `hat n` . Let P be any point on the plane. Therefore , `vec (NP)` is perpendicular to `vec (ON)`.
Therefore, `vec (NP) . vec (ON) = 0`  ...(1)
Let `vec r` be the position vector of the point P, then `vec (NP) = vec r - d . hat n`     (as `vec (ON) + vec (NP) = vec (OP)`)
Therefore, (1) becomes 
`(vec r - d . hat n) . d hat n = 0`
or `(vec r - d.hat n). hat n =0`  (d ≠ 0)
or `vec r . hat n - d  hat n . hat n = 0`
i.e., `vec r .  hat n = d`        (as `hat n . hat n = 1`)          ...(2)
This is the vector form of the equation of the plane. 

Cartesian form
Equation (2) gives the vector equation of a plane, where `hat n` is the unit vector normal to the plane. Let P(x, y, z) be any point on the plane.  Then 
`vec (OP) = vec r = x hat i + y hat j + z hat k`
Let l, m, n be the direction cosines of `hat n` . Then
`hat n = l hat i + m hat j + n hat k ` 
Therefore, (2) gives
`(x hat i + y hat j + z hat k) . (l hat i + m hat j + n hat k) = d`
i.e. lx + my + nz = d                             ... (3) 

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