Equation of Plane in Normal Form

Here you will learn equation of plane in normal form with example.

Let’s begin –

Equation of Plane in Normal Form

(a) Vector Form 

The vector equation of a plane normal to unit vector \(\hat{n}\) and at a distance d from the origin is

\(\vec{r}.\hat{n}\) = d

Remark 1 : The vector equation of ON is \(\vec{r}\) = \(\vec{0}\) + \(\lambda\) \(\hat{n}\) and the position vector of N is d\(\hat{n}\) as it is at a distance d from the origin from the origin O.

(b) Cartesian Form 

If l, m, n are direction cosines of the normal to a given plane which is at a distance p from the origin, then the equation of the plane is

lx + my + nz = p

Note : The equation \(\vec{r}.\vec{n}\) = d is in normal form if \(\vec{n}\) is a unit vector and in such a case d on the right hand side denotes the distance of the plane from the origin. If \(\vec{n}\) is not a unit vector, then to reduce the equation \(\vec{r}.\vec{n}\) = d to normal form divide both sides by | \(\vec{n}\) | to obtain

\(\vec{r}\).\(\vec{n}\over |\vec{n}|\) = \(d\over |\vec{n}|\) \(\implies\) \(\vec{r}.\hat{n}\) = \(d\over |\vec{n}|\)

Example : Find the vector equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector \(2\hat{i} + \hat{j} + 2\hat{k}\).

Solution : Here, d = 8 and \(\vec{n}\) = \(2\hat{i} + \hat{j} + 2\hat{k}\)

\(\therefore\) \(\hat{n}\) = \(\vec{n}\over |\vec{n}|\) = \(2\hat{i} + \hat{j} + 2\hat{k}\over \sqrt{4 + 1 + 4}\)

= \({2\over 3}\hat{i} + {1\over 3}\hat{j} + {2\over 3}\hat{k}\)

Hence, the required equation of the plane is

\(\vec{r}\).(\({2\over 3}\hat{i} + {1\over 3}\hat{j} + {2\over 3}\hat{k}\)) = 8                    

[ By using \(\vec{r}.\hat{n}\) = d ]

or, \(\vec{r}\).(\(2\hat{i} + \hat{j} + 2\hat{k}\)) = 24

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