# Distance of a Point from a Plane – Vector Form

## 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.