Question

Consider the following statements
Statement I The vector equation of a plane in normal form is $r \cdot \hat{ n }=d$, where $\hat{ n }$ is the unit normal vector to the plane and $d$ is the distance from origin to the plane.
Statement II The cartesian equation of a plane in normal form is $l x+m y+n z=d$, where $l, m, n$ are direction cosines of normal vector and $d$ is the distance from origin to the plane.
Choose the correct option.

• A. Statement $I$ is true
• B. Statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (C)

Consider a plane whose perpendicular distance from the origin is $d(d \neq 0)$.

If $ON$ is the normal form the origin to the plane and $\hat{n}$ is the unit normal vector along $ON$. Then, $ON =d \hat{n}$. Let $P$ be any point on the plane. Therefore, $N P$ is perpendicular to $O N$.
Therefore, $ NP \cdot ON =0$...(i)
Let $r$ be the position vector of the point $P$. Then, $NP = r -d \hat{n}$ (as $ON + NP = OP$ )
Therefore, Eq. (i) becomes
$(r-d \hat{n}) \cdot d \hat{n}=0 $
$(r-d \hat{n}) \cdot \hat{n}=0 (\because d \neq 0) $
$r \cdot \hat{n}-d \hat{n} \cdot \hat{n}=0$
$r \cdot \hat{n}=d (a s \hat{n} \cdot \hat{n}=1) \ldots(i i)$
This is the vector form of the equation of the plane.
Cartesian Form
Now, Eq. (ii) gives the vector equation of plane, where n̂ is the unit vector normal to the plane. Let $P(x, y, z)$ be any point on the plane. Then,
$O P=r=x \hat{i}+y \hat{j}+z \hat{k}$
Let $l, m, n$ be the direction cosines of $\hat{n}$. Then,
$\hat{n}=l \hat{i}+m \hat{j}+n \hat{k}$
Therefore, Eq. (ii) gives
$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(l \hat{i}+m \hat{j}+n \hat{k})=d$
i.e., $ l x+m y+n z=d$ ....(iii)
This is the cartesian equation of the plane in the normal form.
Note Eq. (iii) shows that, if $r \cdot(a \hat{i}+b \hat{j}+c \hat{k})=d$ is the vector equation of a plane, then $a x+b y+c z=d$ is the cartesian equation of the plane. where $a, b$ and $c$ are the direction ratios of the normal to the plane.