4.1 Equation of plane in normal form

2.1 Vectorial form of a line passing through a given point and parallel to a given vector
2.2 Vectorial form of a line passing through two given points
2.3 Cartesian form of a line passing through a given point and parallel to a given vector
2.4 Cartesian form of a line passing through two given points

3.1 Condition for perpendicularity
3.2 Condition for parallelism
3.3 Co-planarity of two lines
3.4 Shortest distance between two lines
3.5 Distance between two skew lines
3.6 Distance between parallel lines
3.7 Perpendicular distance of a point from a line
3.8 Image of a point in a straight line

4.1 Equation of plane in normal form
4.2 Equation of a plane perpendicular to a given vector and passing through a given point
4.3 Equation of plane passing through a given point and parallel to the two given vectors
4.4 Equation of plane passing through three non-collinear points
4.5 Intercept form of plane
4.6 Equation of plane passing through the intersection of two given planes

5.1 Angle between two planes
5.2 Angle between a line and a plane
5.3 Distance of a point from a plane
5.4 Distance between two parallel planes

To find the vector equation of plane in normal form, we required the distance of plane from origin and a normal vector to the plane.

Let us consider a plane whose perpendicular distance from the origin $O$ be $d(d \ne 0)$ and $ON$ be the normal from the origin $O$ to the plane and $\widehat n$ is the unit vector normal to the plane along $\overrightarrow {ON} $.

Therefore, we can write $$\overrightarrow {ON} = d\widehat n$$ Let $P$ be any point on the plane with position vector $\overrightarrow r $ i.e., $$\overrightarrow {OP} = \overrightarrow r $$

As shown in figure, $\overrightarrow {NP} $ is perpendicular to $\overrightarrow {ON} $. Therefore, we can say that

$$\begin{equation} \begin{aligned} \overrightarrow {NP} .\overrightarrow {ON} = 0 \\ \Rightarrow \left( {\overrightarrow {OP} - \overrightarrow {ON} } \right).\overrightarrow {ON} = 0 \\ \Rightarrow \left( {\overrightarrow r - d\widehat n} \right).d\widehat n = 0 \\ \Rightarrow \overrightarrow r .d\widehat n - {d^2}\widehat n.\widehat n = 0 \\ \Rightarrow d\overrightarrow r .\widehat n - {d^2}{\left| {\widehat n} \right|^2} = 0{\text{ }}\left( {\because d \ne 0{\text{ and }}\left| {\widehat n} \right| = 1} \right) \\ \Rightarrow \overrightarrow r .\widehat n - d = 0 \\ \overrightarrow r .\widehat n = d \\\end{aligned} \end{equation} $$ which is the vector equation of plane in normal form.

To find the cartesian equation of plane in normal form, let us assume a point $P(x,y,z)$ on the plane. We can write, $$\overrightarrow {OP} = \overrightarrow r = x\widehat i + y\widehat j + z\widehat k$$ If $l$, $m$ and $n$ be the direction cosines of the normal to the given plane and $d$ be the length of perpendicular from origin to the plane, then $$\widehat n = l\widehat i + m\widehat j + n\widehat k$$ Now, put all these values in the vector equation of plane in normal form, we get

$$\begin{equation} \begin{aligned} \overrightarrow r .\widehat n = d \\ \left( {x\widehat i + y\widehat j + z\widehat k} \right).\left( {l\widehat i + m\widehat j + n\widehat k} \right) = d \\ lx + my + nz = d \\\end{aligned} \end{equation} $$ which is the cartesian equation of plane in normal form.

Note: Instead of direction cosines $l$, $m$ and $n$, if direction ratios $a$, $b$ and $c$ are given.

The vector equation of plane in normal form is $$\overrightarrow r .\left( {a\widehat i + b\widehat j + c\widehat k} \right) = d$$ The cartesian equation of plane in normal form is $$ax+by+cz=d$$

This is also called the General Equation of plane.

Question 10. Find the direction cosines of the unit vector perpendicular to the plane $$\overrightarrow r .\left( {6\widehat i - 4\widehat j + 2\widehat k} \right) + 1 = 0$$ passing through origin.

Solution: The given equation of plane can be written as $$\overrightarrow r .\left( { - 6\widehat i + 4\widehat j - 2\widehat k} \right) = 1$$ As we know that the vector equation of plane in normal form is $$\overrightarrow r .\widehat n = d$$ where $\widehat n$ is the unit vector. So divide the given equation of plane by the magnitude of $\overrightarrow n = - 6\hat i + 4\hat j - 2\hat k$ i.e., $$\left| {\overrightarrow n } \right| = \left| { - 6\hat i + 4\hat j - 2\hat k} \right| = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 2} \right)}^2}} = 2\sqrt {14} $$ We get, $$\overrightarrow r .\left( { - \frac{6}{{2\sqrt {14} }}\hat i + \frac{4}{{2\sqrt {14} }}\hat j - \frac{2}{{2\sqrt {14} }}\hat k} \right) = \frac{1}{{2\sqrt {14} }}$$ which is the equation of plane in normal form. Therefore, $$\widehat n = - \frac{6}{{2\sqrt {14} }}\hat i + \frac{4}{{2\sqrt {14} }}\hat j - \frac{2}{{2\sqrt {14} }}\hat k$$ is the unit vector perpendicular to the plane passing through origin.