27.14 - Intersection of Planes - More cases

In the previous section, we saw case I related to the intersection of two planes. In this section, we will see the solved examples related to the remaining cases.

Solved example 27.50
Find the equation of the plane through the line of intersection of the planes
$\small{x+y+z=1~\text{and}~2x+3y+4z=5}$ which is perpendicular to the plane $\small{x-y+z=0}$
Solution:
This problem belongs to Case II
1. We are given two planes:
• $\small{\pi_1~\text{whose equation is}~x+y+z=1}$
• $\small{\pi_2~\text{whose equation is}~2x+3y+4z=5}$

2. We are asked to find the equation of the plane $\small{\pi_3}$ which satisfies two conditions:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be perpendicular to a given plane.
• This given plane is:
$\small{\pi_4~\text{whose equation is}~x-y+z=0}$

• All the four planes are shown in fig.27.23 below:

Fig.27.23

• The blue plane is perpendicular to the brown plane.  

3. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two intersecting planes are:
    ♦ $\small{x+y+z=1}$
    ♦ $\small{2x+3y+4z=5}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{y+z=1}$
    ♦ $\small{3y+4z=5}$
• Solving the two equations, we get: y = −1 and z = 2
• So (0,−1,2) is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{x+y=1}$
    ♦ $\small{2x+3y=5}$
• Solving the two equations, we get: x = −2 and y = 3
• So (−2,3,0) is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

4. If we know two points on a line, we can calculate the direction ratios of that line. So the direction ratios of the line of intersection are: −2, 4, −2
• Therefore, the vector $\small{-2\hat{i}+4\hat{j}-2\hat{k}}$ is parallel to the line of intersection.
• Dividing throughout by −2, we can write:
The vector $\small{\hat{i}-2\hat{j}+\hat{k}}$ is parallel to the line of intersection.
• This vector will lie flat entirely on the required plane $\small{\pi_3}$

5. Now we consider $\small{\pi_4}$. Based on it’s given equation, we can write:
The vector $\small{\hat{i}-\hat{j}+\hat{k}}$ is perpendicular to $\small{\pi_4}$
(Imagine a vector, perpendicular to the blue plane in fig.27.23 above)

6. $\small{\pi_4}$ is perpendicular to $\small{\pi_3}$. So the vector $\small{\hat{i}-\hat{j}+\hat{k}}$ is parallel to $\small{\pi_3}$. This vector can be shifted and can be made to lie flat on $\small{\pi_3}$

7. Now we have two vectors which lie flat on $\small{\pi_3}$:
    ♦ The vector obtained in step (4)
    ♦ The vector obtained in step (6)
• The cross product of those two vectors will be a vector perpendicular to $\small{\pi_3}$
• The cross product is: $\small{-\hat{i}+\hat{k}}$
(The reader should write all steps related to the cross product)

8. Now we can write the equation of $\small{\pi_3}$. We have two parameters:
(i) From step (7), the vector perpendicular to $\small{\pi_3}$
(ii) From step (3), we can pick any one of the two points. That point will lie on $\small{\pi_3}$
• So we can use the formula: $\small{A\left(x-x_1 \right)+A\left(y-y_1 \right)+A\left(z-z_1 \right)=0}$
See section 27.10
• In our present case,
    ♦ A = −1, B = 0 and C = 1
    ♦ $\small{\left(x_1,y_1,z_1 \right) = \left(0,-1,2 \right)}$

9. Therefore, the equation of $\small{\pi_3}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{(-1)\left(x-0 \right)+(0)\left(y-(-1) \right)+(1)\left(z-2 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{-x+z-2}    & {~=~}    &{0} \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x-z+2}    & {~=~}    &{0} \\
\end{array}}$

Solved example 27.51
Find the equation of the plane through the line of intersection of the planes
$\small{\vec{r}.\left(\hat{i}+2\hat{j}+3\hat{k} \right)-4=0~\text{and}~\vec{r}.\left(2\hat{i}+\hat{j}-\hat{k} \right)+5=0}$ and which is perpendicular to the plane $\small{\vec{r}.\left(5\hat{i}+3\hat{j}-6\hat{k} \right)+8=0}$
Solution:
This problem belongs to Case II
1. We are given two planes:
• $\small{\pi_1~\text{whose equation is}~\vec{r}.\left(\hat{i}+2\hat{j}+3\hat{k} \right)-4=0}$
• $\small{\pi_2~\text{whose equation is}~\vec{r}.\left(2\hat{i}+\hat{j}-\hat{k} \right)+5=0}$

2. We are asked to find the equation of the plane $\small{\pi_3}$ which satisfies two conditions:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be perpendicular to a given plane.
• This given plane is:
$\small{\pi_4~\text{whose equation is}~\vec{r}.\left(5\hat{i}+3\hat{j}-6\hat{k} \right)+8=0}$

3. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two intersecting planes are:
    ♦ $\small{x+2y+3z=4}$
    ♦ $\small{2x+y-z=-5}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{2y+3z=4}$
    ♦ $\small{y-z=-5}$
• Solving the two equations, we get: y = -11/5 and z = 14/5
• So $\small{\left(0,\frac{-11}{5},\frac{14}{5} \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{x+2y=4}$
    ♦ $\small{2x+y=-5}$
• Solving the two equations, we get: x = −14/3 and y = 13/3
• So $\small{\left(\frac{-14}{3},\frac{13}{3},0 \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

4. If we know two points on a line, we can calculate the direction ratios of that line. So the direction ratios of the line of intersection are:
    ♦ $\small{\frac{-14}{3}-0=\frac{-14}{3}}$
    ♦ $\small{\frac{13}{3}-\frac{-11}{5}=\frac{98}{15}}$
    ♦ $\small{0-\frac{14}{5}=\frac{-14}{5}}$
• Therefore, the vector $\small{\left(\frac{-14}{3} \right)\hat{i}+\left(\frac{98}{15} \right)\hat{j}+\left(\frac{-14}{5} \right)\hat{k}}$ is parallel to the line of intersection.
• Multiplying throughout by $\small{\frac{-15}{2}}$, we can write:
The vector $\small{35\hat{i}-49\hat{j}+21\hat{k}}$ is parallel to the line of intersection.
• This vector will lie flat entirely on the required plane $\small{\pi_3}$

5. Now we consider $\small{\pi_4}$. Based on it’s given equation, we can write:
The vector $\small{5\hat{i}+3\hat{j}-6\hat{k}}$ is perpendicular to $\small{\pi_4}$

6. $\small{\pi_4}$ is perpendicular to $\small{\pi_3}$. So the vector $\small{5\hat{i}+3\hat{j}-6\hat{k}}$ is parallel to $\small{\pi_3}$. This vector can be shifted and can be made to lie flat on $\small{\pi_3}$

7. Now we have two vectors which lie flat on $\small{\pi_3}$:
    ♦ The vector obtained in step (4)
    ♦ The vector obtained in step (6)
• The cross product of those two vectors will be a vector perpendicular to $\small{\pi_3}$
• The cross product is: $\small{231\hat{i}+315\hat{j}+350\hat{k}}$
(The reader should write all steps related to the cross product)

8. Now we can write the equation of $\small{\pi_3}$. We have two parameters:
(i) From step (7), the vector perpendicular to $\small{\pi_3}$
(ii) From step (3), we can pick any one of the two points. That point will lie on $\small{\pi_3}$
• So we can use the formula: $\small{A\left(x-x_1 \right)+A\left(y-y_1 \right)+A\left(z-z_1 \right)=0}$
See section 27.10
• In our present case,
    ♦ A = 231, B = 315 and C = 350
    ♦ $\small{\left(x_1,y_1,z_1 \right) = \left(0,\frac{-11}{5},\frac{14}{5} \right)}$

9. Therefore, the equation of $\small{\pi_3}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{(231)\left(x-0 \right)+(315)\left(y-(\frac{-11}{5}) \right)+(350)\left(z-\frac{14}{5} \right)}    & {~=~}    &{0} \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{231x+315y+350z-287}    & {~=~}    &{0} \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{7(33x+45y+50z-41)}    & {~=~}    &{0} \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{33x+45y+50z-41}    & {~=~}    &{0} \\
\end{array}}$

Solved example 27.52
Find the equation of the plane through the line of intersection of the planes
$\small{\vec{r}.\left(\hat{i}+\hat{j}+\hat{k} \right)=1~\text{and}~\vec{r}.\left(2\hat{i}+3\hat{j}-\hat{k} \right)+4=0}$ and parallel to x-axis.
Solution:
This problem belongs to Case III
1. We are given two planes:
• $\small{\pi_1~\text{whose equation is}~\vec{r}.\left(\hat{i}+\hat{j}+\hat{k} \right)=1}$
• $\small{\pi_2~\text{whose equation is}~\vec{r}.\left(2\hat{i}+3\hat{j}-\hat{k} \right)+4=0}$

2. We are asked to find the equation of the plane $\small{\pi_3}$ which satisfies two conditions:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be parallel to the x-axis.

3. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two intersecting planes are:
    ♦ $\small{x+y+z=1}$
    ♦ $\small{2x+3y-z=-4}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{y+z=1}$
    ♦ $\small{3y-z=-4}$
• Solving the two equations, we get: y = -3/4 and z = 7/4
• So $\small{\left(0,\frac{-3}{4},\frac{7}{4} \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{x+y=1}$
    ♦ $\small{2x+3y=-4}$
• Solving the two equations, we get: x = 7 and y = -6
• So $\small{\left(7,-6,0 \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

4. If we know two points on a line, we can calculate the direction ratios of that line. So the direction ratios of the line of intersection are:
    ♦ $\small{7-0=7}$
    ♦ $\small{-6-\frac{-3}{4}=\frac{-21}{4}}$
    ♦ $\small{0-\frac{7}{4}=\frac{-7}{4}}$
• Therefore, the vector $\small{\left(7 \right)\hat{i}-\left(\frac{21}{4} \right)\hat{j}-\left(\frac{7}{4} \right)\hat{k}}$ is parallel to the line of intersection.
• Multiplying throughout by $\small{4}$, we can write:
The vector $\small{28\hat{i}-21\hat{j}-7\hat{k}}$ is parallel to the line of intersection.
• This vector will lie flat entirely on the required plane $\small{\pi_3}$

5. Now we consider the x-axis.
The vector equation of x-axis is: $\small{\hat{i}}$

6. $\small{\pi_3}$ is to be parallel to the x-axis.
That is., $\small{\pi_3}$ will never intersect the x-axis. So the vector $\small{\hat{i}}$ is parallel to $\small{\pi_3}$. This vector can be shifted and can be made to lie flat on $\small{\pi_3}$

7. Now we have two vectors which lie flat on $\small{\pi_3}$:
    ♦ The vector obtained in step (4)
    ♦ The vector obtained in step (6)
• The cross product of those two vectors will be a vector perpendicular to $\small{\pi_3}$
• The cross product is: $\small{-7\hat{j}+21\hat{k}}$
(The reader should write all steps related to the cross product)

8. Now we can write the equation of $\small{\pi_3}$. We have two parameters:
(i) From step (7), the vector perpendicular to $\small{\pi_3}$
(ii) From step (3), we can pick any one of the two points. That point will lie on $\small{\pi_3}$
• So we can use the formula: $\small{A\left(x-x_1 \right)+A\left(y-y_1 \right)+A\left(z-z_1 \right)=0}$
See section 27.10
• In our present case,
    ♦ A = 0, B = −7 and C = 21
    ♦ $\small{\left(x_1,y_1,z_1 \right) = \left(7,-6,0 \right)}$

9. Therefore, the equation of $\small{\pi_3}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{(0)\left(x-7 \right)+(-7)\left(y-(-6) \right)+(21)\left(z-0 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{-7y+21z-42}    & {~=~}    &{0} \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{-7(y-3z+6)}    & {~=~}    &{0} \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y-3z+6}    & {~=~}    &{0} \\
\end{array}}$

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In the next section, we will see coplanarity of two lines.

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27.13 - Plane Passing Through the Intersection of two Given Planes

In the previous section, we saw the intercept form. In this section, we will see the plane passing through the intersection of two given planes.

Some basic details can be written in 6 steps:
1. Fig.27.22 below shows three planes $\small{\pi_1,~\pi_2~\text{and}~\pi_3}$
    ♦ $\small{\pi_1}$ is shown in yellow color
    ♦ $\small{\pi_2}$ is shown in violet color
    ♦ $\small{\pi_3}$ is shown in brown color

Fig.27.22

2. $\small{\pi_1~\text{and}~\pi_2}$ are the given planes
    ♦ $\small{\pi_1}$ has the equation: $\small{\vec{r}.\hat{n_1} = d_1}$
    ♦ $\small{\pi_2}$ has the equation: $\small{\vec{r}.\hat{n_2} = d_2}$

3. If two planes intersect, that intersection will be along a line. In our present case, $\small{\pi_1~\text{and}~\pi_2}$ intersect along the red line.

4. We want the equation of a third plane $\small{\pi_3}$, which passes through the line of intersection.

5. Here a problem arises:
In addition to $\small{\pi_1~\text{and}~\pi_2}$, infinite number of planes can pass through the red line. All of them qualify to be called as $\small{\pi_3}$. It is like making multiple copies of one $\small{\pi_3}$ and rotating those copies by considering the red line as axis.

6. To overcome this problem, we provide an additional condition. Based on this additional condition, three cases can arise. In all those three cases, the first condition is the same.
Case I:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must pass through a particular point $\small{U \left(x_1,y_1,z_1 \right)}$.
Case II:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be perpendicular to a fourth plane.
Case III:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be parallel to an axis.

---

Now we will see some solved examples

Solved example 27.47
Find the vector equation of the plane passing through the intersection of the planes:
$\small{\vec{r}.\left(\hat{i}+\hat{j}+\hat{k} \right)=6~\text{and}~\vec{r}.\left(2\hat{i}+3\hat{j}+4\hat{k} \right)=-5}$ and the point (1,1,1).
Solution:
This problem belongs to Case I
1. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two given planes are:
    ♦ $\small{x+y+z=6}$
    ♦ $\small{2x+3y+4z=-5}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{y+z=6}$
    ♦ $\small{3y+4z=-5}$
• Solving the two equations, we get: y = 29 and z = −23
• So (0,29,−23) is a point on the line of intersection.
• Check:
    ♦ $\small{0+29+(-23)=6}$
    ♦ $\small{0+3(29)+4(-23)=87-92=-5}$
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{x+y=6}$
    ♦ $\small{2x+3y=-5}$
• Solving the two equations, we get: x = 23 and y = −17
• So (23,−17,0) is a point on the line of intersection.
• Check:
    ♦ $\small{23+(-17)+0=6}$
    ♦ $\small{2(23)+3(-17)+0=46-51=-5}$
(The reader must write all steps related to the solution of the two equations and also the checking steps)

2. The required plane must pass through (1,1,1). So we have a total of three points on the required plane. We can use the formula:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    
\\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     
\\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     
\\ \end{array}\right|~=~0$

3. In our present case,
    ♦ $\small{\left(x_1,y_1,z_1 \right)}$ can be taken as $\small{\left(1,1,1 \right)}$
    ♦ $\small{\left(x_2,y_2,z_2 \right)}$ can be taken as $\small{\left(0,29,-23 \right)}$
    ♦ $\small{\left(x_3,y_3,z_3 \right)}$ can be taken as $\small{\left(23,−17,0 \right)}$
• So the determinant form is:
$\left|\begin{array}{r}                             x-1     &{    y-1     }    &{    z-1      }    
\\ 0-1      &{    29-1     }    &{   -23-1      }     
\\ 23-1      &{    -17-1     }    &{ 0-1        }     
\\ \end{array}\right|~=~0$

• This is same as:
$\left|\begin{array}{r}                             x-1     &{    y-1     }    &{    z-1      }    
\\ -1      &{    28     }    &{   -24      }     
\\ 22      &{    -18     }    &{ -1        }     
\\ \end{array}\right|~=~0$

4. Expanding the determinant, we get:
$\small{-23 (20 x + 23 y + 26 z - 69)=0}$
$\small{\Rightarrow 20 x + 23 y + 26 z  = 69}$
(The reader must write all steps related to the expansion of the determinant)

• This is the Cartesian equation of the required plane.
• So the vector equation will be:
$\small{\vec{r}.\left(20\hat{i}+23\hat{j}+26\hat{k} \right)=69}$

Solved example 27.48
Find the vector equation of the plane passing through the intersection of the planes:
$\small{\vec{r}.\left(2\hat{i}+2\hat{j}-3\hat{k} \right)=7~\text{and}~\vec{r}.\left(2\hat{i}+5\hat{j}+3\hat{k} \right)=9}$ and the point (2,1,3).
Solution:
This problem belongs to Case I
1. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two given planes are:
    ♦ $\small{2x+2y-3z=7}$
    ♦ $\small{2x+5y+3z=9}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{2y-3z=7}$
    ♦ $\small{5y+3z=9}$
• Solving the two equations, we get: y = 16/7 and z = −(17/21)
• So $\small{\left(0,\frac{16}{7},\frac{-17}{21} \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{2x+2y=7}$
    ♦ $\small{2x+5y=9}$
• Solving the two equations, we get: x = 17/6 and y = 2/3
• So $\small{\left(\frac{17}{6},\frac{2}{3},0 \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

2. The required plane must pass through (2,1,3). So we have a total of three points on the required plane. We can use the formula:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    
\\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     
\\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     
\\ \end{array}\right|~=~0$

3. In our present case,
    ♦ $\small{\left(x_1,y_1,z_1 \right)}$ can be taken as $\small{\left(2,1,3 \right)}$
    ♦ $\small{\left(x_2,y_2,z_2 \right)}$ can be taken as $\small{\left(0,\frac{16}{7},\frac{-17}{21} \right)}$
    ♦ $\small{\left(x_3,y_3,z_3 \right)}$ can be taken as $\small{\left(\frac{17}{6},\frac{2}{3},0 \right)}$
• So the determinant form is:
$\left|\begin{array}{r}                             x-2     &{    y-1     }    &{    z-3      }    
\\ 0-2      &{    \frac{16}{7}-1     }    &{   \frac{-17}{21}-3      }     
\\ \frac{17}{6}-2      &{    \frac{2}{3}-1     }    &{ 0-3        }     
\\ \end{array}\right|~=~0$

• This is same as:
$\left|\begin{array}{r}                             x-2     &{    y-1     }    &{    z-3      }    
\\ -2      &{    \frac{9}{7}     }    &{   \frac{80}{21}      }     
\\ \frac{5}{6}      &{    \frac{-1}{3}     }    &{ -3        }     
\\ \end{array}\right|~=~0$

4. Expanding the determinant, we get:
$\small{\frac{-17}{126} (38 x + 68 y + 3 z - 153)=0}$
$\small{\Rightarrow 38 x + 68 y + 3 z  = 153}$
(The reader must write all steps related to the expansion of the determinant)

• This is the Cartesian equation of the required plane.
• So the vector equation will be:
$\small{\vec{r}.\left(38\hat{i}+68\hat{j}+3\hat{k} \right)=153}$

Solved example 27.49
Find the vector equation of the plane passing through the intersection of the planes:
$\small{3x-y+2z-4=0~\text{and}~x+y+z-2=0}$ and the point (2,2,1).
Solution:
This problem belongs to Case I
1. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two given planes are:
    ♦ $\small{3x-y+2z=4}$
    ♦ $\small{x+y+z=2}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{-y+2z=4}$
    ♦ $\small{y+z=2}$
• Solving the two equations, we get: y = 0 and z = 2
• So $\small{\left(0,0,2 \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{3x-y=4}$
    ♦ $\small{x+y=2}$
• Solving the two equations, we get: x = 3/2 and y = 1/2
• So $\small{\left(\frac{3}{2},\frac{1}{2},0 \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

2. The required plane must pass through (2,2,1). So we have a total of three points on the required plane. We can use the formula:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    
\\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     
\\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     
\\ \end{array}\right|~=~0$

3. In our present case,
    ♦ $\small{\left(x_1,y_1,z_1 \right)}$ can be taken as $\small{\left(2,2,1 \right)}$
    ♦ $\small{\left(x_2,y_2,z_2 \right)}$ can be taken as $\small{\left(0,0,2 \right)}$
    ♦ $\small{\left(x_3,y_3,z_3 \right)}$ can be taken as $\small{\left(\frac{3}{2},\frac{1}{2},0 \right)}$
• So the determinant form is:
$\left|\begin{array}{r}                             x-2     &{    y-2     }    &{    z-1      }    
\\ 0-2      &{    0-2     }    &{   2-1      }     
\\ \frac{3}{2}-2      &{    \frac{1}{2}-2     }    &{ 0-1        }     
\\ \end{array}\right|~=~0$

• This is same as:
$\left|\begin{array}{r}                             x-2     &{    y-2     }    &{    z-1      }    
\\ -2      &{    -2     }    &{   1      }     
\\ \frac{-1}{2}      &{    \frac{-3}{2}     }    &{ -1        }     
\\ \end{array}\right|~=~0$

4. Expanding the determinant, we get:
$\small{\frac{1}{2} (7 x  -5 y + 4 z - 8)=0}$
$\small{\Rightarrow 7 x  -5 y + 4 z  = 8}$
(The reader must write all steps related to the expansion of the determinant)

• This is the Cartesian equation of the required plane.
• So the vector equation will be:
$\small{\vec{r}.\left(7\hat{i}-5\hat{j}+4\hat{k} \right)=8}$

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In the next section, we will see the other cases.

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27.12 - Intercept Form of The Equation of a Plane

In the previous section, we saw the plane passing through three non collinear points. In this section, we will see intercept form.

Some basic details can be written in 2 steps:
1. Consider our familiar 2D plane with the x and y axes.
We can write three facts:
(i) A line parallel to the x-axis will never intersect the x-axis.
• But that line will definitely intersect the y-axis.
(ii) A line parallel to the y-axis will never intersect the y-axis.
• But that line will definitely intersect the x-axis.
(iii) A line parallel to neither x-axis nor y-axis will definitely intersect both the axes.

2. In a similar way, for the 3D space, we can write four facts:
(i) A plane parallel to the XOY plane will never intersect the x or y-axes
• But that plane will definitely intersect the z-axis.
(ii)  A plane parallel to the XOZ plane will never intersect the x or z-axes
• But that plane will definitely intersect the y-axis.
(iii) A plane parallel to the YOZ plane will never intersect the y and z-axes
• But that plane will definitely intersect the x-axis.
(iv) A plane which is not parallel to any one of XOY, XOZ or YOZ planes, will definitely intersect all the three axes.

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In this section, we consider the plane which is written in 2(iv) above. It’s Cartesian  equation can be derived in 4 steps:
1. Let the equation of the plane be:
$\small{Ax + By + Cz + D = 0,~(D \ne 0)}$

2. Let the plane make intercepts a,b,c on the x, y and z-axes respectively. This is shown in the fig.27.20 below:

Fig.27.20

• Based on the intercepts, we can write:
    ♦ The plane meets the x-axis at P(a,0,0)
    ♦ The plane meets the y-axis at Q(0,b,0)
    ♦ The plane meets the z-axis at R(0,0,c)

3. Any point on the plane will satisfy the equation of the plane.
• Substituting the coordinates of P in the equation of the plane, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A(a) + B(0) + C(0) + D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{A(a)}    & {~=~}    &{-D}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{A}    & {~=~}    &{\frac{-D}{a}}
\\ \end{array}}$

• Substituting the coordinates of Q in the equation of the plane, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A(0) + B(b) + C(0) + D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{B(b)}    & {~=~}    &{-D}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{B}    & {~=~}    &{\frac{-D}{b}}
\\ \end{array}}$

• Substituting the coordinates of R in the equation of the plane, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A(0) + B(0) + C(c) + D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{C(c)}    & {~=~}    &{-D}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{C}    & {~=~}    &{\frac{-D}{c}}
\\ \end{array}}$

4. Substituting the above values of A, B and C in the original equation, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\frac{-D}{a} \right)x+ \left(\frac{-D}{b} \right)y+ \left(\frac{-D}{c} \right)z+ D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}-1}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}    & {~=~}    &{1}
\\ \end{array}}$
• This is the equation of the plane in intercept form.

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Now we will see some solved examples

Solved example 27.43
Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axes respectively.
Solution:
1. The general intercept form is:
$\small{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$
• Where a, b and c are the intercepts on the x, y and z-axes respectively.

2. Substituting the given intercepts, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x}{2}+\frac{y}{3}+\frac{z}{4}}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{2}(12)+\frac{y}{3}(12)+\frac{z}{4}(12)}    & {~=~}    &{12}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{6x+4y+3z}    & {~=~}    &{12}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we multiply throughout by the L.C.M of 2, 3 and 4, which is 12.

Solved example 27.44
Find the intercepts cut off by the plane 2x + y −z = 5.
Solution:
1. The given equation can be divided throughout by 5 and rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{2x}{5}+\frac{y}{5}+\frac{(-1)z}{5}}    & {~=~}    &{\frac{5}{5}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x}{5/2}+\frac{y}{5}+\frac{z}{(-5)}}    & {~=~}    &{1}
\\ \end{array}}$

2. Now the equation is in the general intercept form.
So the intercepts are:
5/2, 5, −5

3. Fig.27.21 below shows the plane and the intercepts

Fig.27.21

Solved example 27.45
Find the equation of the plane with intercept 3 on the y axis and parallel to ZOX plane.
Solution:
1. The general intercept form is:
$\small{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$
• Where a, b and c are the intercepts on the x, y and z-axes respectively.

2. In our present case, the plane is parallel to the ZOX plane. For such a plane, there are no intercepts on the z and x axes. So we can ignore the first and last terms on the L.H.S.

3. Therefore, the required equation is:
$\small{\frac{y}{3}=1}$, which is same as y=3

Solved example 27.46
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then
$\small{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}}$.
Solution:
1. Based on the general intercept form, the equation of the given plane is:
$\small{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$

2. This can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{(abc)x}{a}+\frac{(abc)y}{b}+\frac{(abc)z}{c}}    & {~=~}    &{abc}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{(bc)x+(ac)y+(ab)z}    & {~=~}    &{abc}
\\ \end{array}}$

3. The above result can be written in vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(bc)x+(ac)y+(ab)z}    & {~=~}    &{abc}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left[(bc)\hat{i}+(ac)\hat{j}+(ab)\hat{k} \right]}    & {~=~}    &{abc}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\hat{r} \right).\left[(bc)\hat{i}+(ac)\hat{j}+(ab)\hat{k} \right]}    & {~=~}    &{abc}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\hat{r} \right).\frac{\left[(bc)\hat{i}+(ac)\hat{j}+(ab)\hat{k} \right]}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}}    & {~=~}    &{\frac{abc}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}}
\\ \end{array}}$

◼ Remarks:
4 (magenta color): Here, in the L.H.S, we do a division by the magnitude. This is to obtain the unit vector $\small{\hat{n}}$. But then, the R.H.S must also be divided by the same magnintude. 

4. The above result in (3), is in the form: $\small{\vec{r}\hat{n}=d}$
• Where:
    ♦ $\small{\hat{n}}$ is the unit vector normal to the plane
    ♦ $\small{d}$ is distance of the plane from the origin

5. So the R.H.S of the result in (3), is the distance of the plane from the origin. But this distance is given to us. It is p. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{abc}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}}    & {~=~}    &{p}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{a^2 b^2 c^2}{b^2 c^2 + a^2 c^2 + a^2 b^2}}    & {~=~}    &{p^2}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{b^2 c^2 + a^2 c^2 + a^2 b^2}{a^2 b^2 c^2}}    & {~=~}    &{\frac{1}{p^2}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}    & {~=~}    &{\frac{1}{p^2}}
\\ \end{array}}$

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In the next section, we will see plane passing through the intersection of two planes.

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