Perpendicular Distance Of A Point From A Plane - Practice Questions & MCQ

The perpendicular distance (D) from a point having position vector $\overrightarrow{\mathbf{a}}$ to the plane $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d$ is given by
$
\mathbf{D}=\frac{|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}-d|}{|\overrightarrow{\mathbf{n}}|}
$

Consider a point P with position vector $\overrightarrow{\mathbf{a}}$ and a plane $\pi_1$ whose equation is $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d$

Let $R$ be a point in the plane such that $\overrightarrow{P R}$ is orthogonal to the plane $\pi_1$. sincel line $P R$ passes through $P(a)$ and is parallel to the vector $\overrightarrow{\mathbf{n}}$ which is normal to the plane $\pi_1$. So, the vector equation of line $P R$ is
$
\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{n}}
$

Point $R$ is the intersection of Eq. (i) and the given plane $\pi_1$.
$
\begin{array}{lc}
\therefore & (\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{n}}) \cdot \overrightarrow{\mathbf{n}}=d \\
\Rightarrow & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}+\lambda \overrightarrow{\mathbf{n}} \cdot \overrightarrow{\mathbf{n}}=d \\
\Rightarrow & \lambda=\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}
\end{array}
$

On putting the value of λ in Eq. (i), we obtain the position vector of R given by

$\begin{aligned} \overrightarrow{\mathbf{r}} & =\overrightarrow{\mathbf{a}}+\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}} \\ \overrightarrow{\mathbf{P R}} & =\text { Position vector of } R-\text { Position vector of } P \\ & =\overrightarrow{\mathbf{a}}+\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}}-\overrightarrow{\mathbf{a}} \\ \overrightarrow{\mathbf{P R}} & =\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}} \\ \Rightarrow \quad|\overrightarrow{\mathbf{P R}}| & =\left|\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}}\right| \\ \Rightarrow \quad|\overrightarrow{\mathbf{P R}}| & =\frac{|d-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}})|}{|\overrightarrow{\mathbf{n}}|} \\ \text { or } \quad D & =\frac{|(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}})-d|}{|\overrightarrow{\mathbf{n}}|}\end{aligned}$
Cartesian Form

Let P(x₁, y₁, z₁) be the given point with position vector and ax + by + cz + d = 0 be the Cartesian equation of the given plane. Then

$
\begin{aligned}
\vec{a} & =x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \\
\vec{n} & =\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k}
\end{aligned}
$

Hence, the Vector form of the perpendicular from P to the plane is
$
\left|\frac{\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right) \cdot(a \hat{i}+b \hat{j}+c \hat{k})-(-d)}{\sqrt{a^2+b^2+c^2}}\right|=\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|
$

Distance Between The Parallel Planes

The distance between the two parallel planes $a x+b y+c z+d_1=0$ and $a x+b y+c z+d_2=0$ is given by
$
D=\left|\frac{\left(d_2-d_1\right)}{\sqrt{a^2+b^2+c^2}}\right|
$

Let $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ be any point in the plane $a x+b y+c z+d_1=0$.
Then the distance of the point $P$ from plane $a x+b y+c z+d_2=0$ is
$
D=\left|\frac{a x_1+b y_1+c z_1+d_2}{\sqrt{a^2+b^2+c^2}}\right|
$

Also,
$
\begin{array}{ll}
\text { Also, } & a x_1+b y 1+c z_1+d_1=0 \\
\Rightarrow & \mathbf{D}=\left|\frac{\left(\mathbf{d}_2-\mathbf{d}_1\right)}{\sqrt{\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2}}\right|
\end{array}
$