Distance of a Point from a Plane – Vector Form
Vector Form: Let \(\pi \left( \vec{r}.\vec{n}=d \right)\) be the given plane and \(P\left( {\vec{a}} \right)\) be the given point.
Let PM be the length of the perpendicular from P to the Plane π.
Since line P passes through \(P\left( {\vec{a}} \right)\) and is parallel to vector \(\vec{n}\), which is normal to the Plane π, the vector equation of line PM is \(\vec{r}=\vec{a}+\lambda \vec{n}\) … (i)
Point M is the intersection of (i) and the given plane π.
Therefore, \(\left( \vec{a}+\lambda \vec{n} \right).\vec{n}=d\),
\(\vec{a}.\vec{n}+\lambda \vec{n}.\vec{n}=d\),
\(\lambda =\frac{d-\left( \vec{a}.\vec{n} \right)}{|\overrightarrow{n}{{|}^{2}}}\).
Putting the value of λ in (i), we obtain the position vector of M given by
\(\vec{r}=\vec{a}+\left( \frac{d-\vec{a}.\vec{n}}{{{\left| {\vec{n}} \right|}^{2}}} \right)\vec{n}\).
\(\overrightarrow{PM}=\vec{a}+\left( \frac{d-\left( \vec{a}.\vec{n} \right)}{{{\left| {\vec{n}} \right|}^{2}}} \right)\vec{n}-\vec{a}\).
\(=\left( \frac{d-\left( \vec{a}.\vec{n} \right)}{{{\left| {\vec{n}} \right|}^{2}}} \right)\vec{n}\).
\(\left| \overrightarrow{PM} \right|=\left| \frac{d-\left( \vec{a}.\vec{n} \right)}{{{\left| {\vec{n}} \right|}^{2}}} \right|\vec{n}\).
\(\left| \overrightarrow{PM} \right|=\frac{\left| d-\left( \vec{a}.\vec{n} \right) \right|\left| {\vec{n}} \right|}{{{\left| {\vec{n}} \right|}^{2}}}\).
\(=\frac{\left| d-\left( \vec{a}.\vec{n} \right) \right|}{\left| {\vec{n}} \right|}\) Which is the required Length.