4.6 Equation of plane passing through the intersection of two given planes

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

Let us assume the equation of two given planes in vector form i.e. $$\overrightarrow r .\widehat {{n_1}} = {d_1}$$ and $$\overrightarrow r .\widehat {{n_2}} = {d_2}$$

Let a point $P$ with position vector $\overrightarrow p $ lies on the line of intersection of two planes. Therefore, the point must satisfy both the equations i.e., $$\overrightarrow p .\widehat {{n_1}} = {d_1}$$ and $$\overrightarrow p .\widehat {{n_2}} = {d_2}$$ Now, we can say that for all real values of $\lambda $, we have $$\overrightarrow p .\left( {\widehat {{n_1}} + \lambda \widehat {{n_2}}} \right) = {d_1} + \lambda {d_2}$$

Since $\overrightarrow p $ is arbitrary, so it satisfies for all points on the line.

Hence the above derived equation

$$\overrightarrow p .\left( {\widehat {{n_1}} + \lambda \widehat {{n_2}}} \right) = {d_1} + \lambda {d_2}$$ represents the third plane in such a manner that if any vector $\overrightarrow r $ satisfies both the given equations of plane, then it also satisfies the equation of third plane passing through the intersection of other two.

Let us assume the equation of two given planes in cartesian form i.e.,

$$\begin{equation} \begin{aligned} {a_1}x + {b_1}y + {c_1}z + {d_1} = 0\;...(1) \\ {a_2}x + {b_2}y + {c_2}z + {d_2} = 0\;...(2) \\\end{aligned} \end{equation} $$

The equation of plane passing through the intersection of two given planes i.e., through the line of intersection of planes is

$$\left( {{a_1}x + {b_1}y + {c_1}z + {d_1}} \right) + k\left( {{a_2}x + {b_2}y + {c_2}z + {d_2}} \right) = 0$$

Question 13. Find the equation of plane passing through the point $(1,1,1)$ which passes through the line of intersection of the planes $x+y+z=6$ and $2x+3y+4z+5=0$.

Solution: The equation of given planes are $$x+y+z-6=0$$ and $$2x+3y+4z+5=0$$

The equation of plane passing through the intersection of two planes is given by $$\left( {x + y + z - 6} \right) + \lambda \left( {2x + 3y + 4z + 5} \right) = 0$$ As the plane also pass through $(1,1,1)$, therefore, $$\begin{equation} \begin{aligned} \left( {1 + 1 + 1 - 6} \right) + \lambda \left( {2 \times 1 + 3 \times 1 + 4 \times 1 + 5} \right) = 0 \\ \lambda = \frac{3}{{14}} \\\end{aligned} \end{equation} $$ The equation of plane is $$\begin{equation} \begin{aligned} \left( {x + y + z - 6} \right) + \frac{3}{{14}}\left( {2x + 3y + 4z + 5} \right) = 0 \\ 20x + 23y + 26z - 69 = 0 \\\end{aligned} \end{equation} $$