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
