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:
$\small{\vec{r}.\left(7\hat{i}-5\hat{j}+4\hat{k} \right)=8}$
In the next section, we will see the other cases.
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