# Distance of a point from a plane

I know that distance of a plane $π_1$ (whose equation is $\vec{r}\cdot \hat{n}= d$) from a point $P$ (with position vector $\vec{a}$) in plane $π_2$ is given by $PQ = |d - \vec{a}\cdot\hat{n}|$ where $\hat{n}$ is unit vector perpendicular to plane $π_1$. But I am not able to solve the following question:

If the equation of the plane $π_2$ is in the form $\vec{r}\cdot\vec{N} = d$, where $\vec{N}$ is normal to the plane, then the perpendicular distance is $$\frac{|\vec{a}\cdot \vec{N}-d|}{|\vec{N}|}.$$

• If the equation of plane is of form $\vec{r}.\vec{N}=d$ then the distance from origin to plane is $\frac{d}{N}$. – gambler101 Aug 16 '16 at 8:15

Hint:

• If you have equation $\vec{r}\cdot\vec{N} = d$ where $\vec{N}$ is normal, but of unit length, then you can transform this equation into $\vec{r}\cdot\hat{n} = d'$ for some $\hat{n}$ and $d'$ dependent on $\vec{N}$ and $d$.
• Using this new equation you can find the actual distance between your point and the plane dependent on $\hat{n}$ and $d'$.
• Going back you can express this distance in terms of $\vec{N}$ and $d$, simplify it and check if what you get is the desired equation.

I hope this helps $\ddot\smile$

Thanks for your guidance sir. Following is my attempt for solving;

Equation of $π_2≡ \vec{r}\cdot\vec{N}= d$

⇒ $\vec{r}\cdot\frac{\vec{N}}{|\vec{N}|}= \frac{d}{|\vec{N}|}$

⇒ $\vec{r}\cdot\hat{n}= \frac{d}{|\vec{N}|} = d_1$ ... (i) Now plane $π_1$ is parallel to $π_2$ and considering a point A on $π_1$ with position vector $\vec{a}$ we have $π_1$≡ $(\vec{r}-\vec{a})\cdot\hat{n}$= 0

⇒ $\vec{r}\cdot\hat{n}-\vec{a}\cdot\hat{n}$= 0 ⇒ $\vec{r}\cdot\hat{n}=\vec{a}\cdot\hat{n}$ = $d_2$ ... (ii)

Therefore distance between parallel planes p = $d_2 - d_1$

⇒ p = $\vec{a}\cdot\hat{n} - \frac{d}{|\vec{N}|}$

⇒ p = $\vec{a}\cdot\frac{\vec{N}}{|\vec{N}|} - \frac{d}{|\vec{N}|}$

⇒ p = $\frac{\vec{a}\cdot\vec{N} - d}{|\vec{N}|}$