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:
$\small{\vec{r}.\left(7\hat{i}-5\hat{j}+4\hat{k} \right)=8}$

---

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

Previous

Contents

Next 

Copyright©2026 Higher secondary mathematics.blogspot.com

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.

---

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.

---

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:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{abc}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}}    & {~=~}    &{p}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{a^2 b^2 c^2}{b^2 c^2 + a^2 c^2 + a^2 b^2}}    & {~=~}    &{p^2}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{b^2 c^2 + a^2 c^2 + a^2 b^2}{a^2 b^2 c^2}}    & {~=~}    &{\frac{1}{p^2}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}    & {~=~}    &{\frac{1}{p^2}}
\\ \end{array}}$

---

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

Previous

Contents

Next

Copyright©2026 Higher secondary mathematics.blogspot.com

27.11 - Plane Passing Through Three Non Collinear Points

In the previous section, we saw the plane perpendicular to a given vector and passing through a given point. In this section, we will see plane passing through three non collinear points.

A plane can be uniquely defined if we have three non collinear points on that plane. The vector equation of such a plane, can be obtained in 6 steps:

1. First we write the points on the plane
    ♦ U, V and W are three points on the plane. They are the given points.
    ♦ P is an arbitrary point on the plane.
• Those four points are shown in fig.27.18 below:

Fig.27.18

• In the above fig., some vectors are drawn in bold lines, while the remaining vectors are drawn in dashed lines. The dashed lines indicate that, those vectors are hidden from view. The plane is obstructing us from viewing those vectors.
• Note that, the x and y-axes are also drawn in dashed lines. They are below the plane and hence hidden from view.

2. Now we write the position vectors of the above points
    ♦ $\small{\vec{u}}$ is the position vector of point U
    ♦ $\small{\vec{v}}$ is the position vector of point V
    ♦ $\small{\vec{w}}$ is the position vector of point W
    ♦ $\small{\vec{r}}$ is the position vector of point P

3. Consider the vectors $\small{\vec{UV}~\text{and}~\vec{UW}}$
Both of them lie on the plane. So their cross product will be perpendicular to the plane. That is:
$\small{\left(\vec{UV}\times\vec{UW}\right)}$ is perpendicular to the plane.

4. Now consider the two vectors:
• $\small{\vec{UP}~\text{and}~\left(\vec{UV}\times\vec{UW}\right)}$
    ♦ $\small{\vec{UP}}$ lies on the plane
    ♦ $\small{\left(\vec{UV}\times\vec{UW}\right)}$ is perpendicular to the plane.
• So the dot product of the two vectors will be zero. That is:
$\small{\vec{UP}.\left(\vec{UV}\times\vec{UW}\right)=0}$

5. Our next task is to find the vectors in the above equation.
• From triangle OUP, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}+\vec{UP}}    & {~=~}    &{\vec{r}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{UP}}    & {~=~}    &{\vec{r}-\vec{u}}
\\ \end{array}}$
• From triangle OUV, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}+\vec{UV}}    & {~=~}    &{\vec{v}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{UV}}    & {~=~}    &{\vec{v}-\vec{u}}
\\ \end{array}}$
• From triangle OUW, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}+\vec{UW}}    & {~=~}    &{\vec{w}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{UW}}    & {~=~}    &{\vec{w}-\vec{u}}
\\ \end{array}}$

6. Now we can substitute in (4). We get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{UP}.\left(\vec{UV}\times\vec{UW}\right)}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ \end{array}}$
• This is the vector equation of the plane.

---

We are discussing about plane passing through three given points. For such a discussion, it is very important to specify that, the three points are non collinear. The reason is simple: There are infinite number of planes which can pass through three collinear points. This is shown in fig.27.19 below:

Fig.27.19

    ♦ Plane $\small{\pi_1}$ is drawn in blue color
    ♦ Plane $\small{\pi_2}$ is drawn in green color
    ♦ Plane $\small{\pi_3}$ is drawn in orange color
• All three planes contain the three yellow points. The three yellow points are collinear, as indicated by the magenta line.
• It is like making multiple copies of the original plane and rotating the copies by using the magenta line as axis. All the copies will contain the three points.
• So it is clear that, to define a unique plane using three points, those three points should be non collinear.

---

The Cartesian equation of the plane can be derived in 6 steps:
1. First we write the coordinates of the four points U, V, W and P:
• $\small{U\left(x_1,y_1,z_1 \right)}$
• $\small{V\left(x_2,y_2,z_2 \right)}$
• $\small{W\left(x_3,y_3,z_3 \right)}$
• $\small{P\left(x,y,z\right)}$

2. Based on the coordinates, we can write the position vectors:
• $\small{\vec{u}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}}$
• $\small{\vec{v}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}}$
• $\small{\vec{w}=x_3\hat{i}+y_3\hat{j}+z_3\hat{k}}$
• $\small{\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}}$

3. Based on the position vectors, we get:
$\small{\vec{r}-\vec{u} = \left(x-x_1 \right)\hat{i}+\left(y-y_1 \right)\hat{j}+\left(z-z_1 \right)\hat{k}}$
$\small{\vec{v}-\vec{u} = \left(x_2-x_1 \right)\hat{i}+\left(y_2-y_1 \right)\hat{j}+\left(z_2-z_1 \right)\hat{k}}$
$\small{\vec{w}-\vec{u} = \left(x_3-x_1 \right)\hat{i}+\left(y_3-y_1 \right)\hat{j}+\left(z_3-z_1 \right)\hat{k}}$

4. Now we can calculate the cross product:
$\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|$
• Expanding the determinant, we get:
$\small{\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)}$

$\small{~=\left[\left(y_2-y_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\hat{i}}$

$\small{~-\left[\left(x_2-x_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\hat{j}}$

$\small{~+\left[\left(x_2-x_1 \right)\left(y_3-y_1 \right)-\left(y_2-y_1 \right)\left(x_3-x_1 \right) \right]\hat{k}}$

5. Now we can substitute the various vectors in the vector equation:
$\small{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}$

• The L.H.S involves the dot product of two vectors:
(i) $\small{\left(\vec{r}-\vec{u} \right)}$. We calculated this in (3) above
(ii) $\small{\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}$. We calculated this in (4) above.

• So the dot product is:
$\small{\left[\left(y_2-y_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\left(x-x_1 \right)}$

$\small{~-\left[\left(x_2-x_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\left(y-y_1 \right)}$

$\small{~+\left[\left(x_2-x_1 \right)\left(y_3-y_1 \right)-\left(y_2-y_1 \right)\left(x_3-x_1 \right) \right]\left(z-z_1 \right)}$

6. Now we can compile the results:
• The dot product obtained in (5) above, is the L.H.S of the vector equation. It can be written in determinant form.
• The R.H.S of the vector equation is zero.
• So substituting in the vector equation, we get:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|~=~0$
• This is the Cartesian form

---

Now we will see some solved examples

Solved example 27.41
Find the vector and Cartesian equations of the planes that passes through three points (2,5,−3), (−2,−3,5), (5,3,−3)
Solution:
1. Let the points be:
U(2,5,−3), V(−2,−3,5), W(5,3,−3)
• First we check whether the three points are collinear:
(i) For the line UV, the direction ratios are: −4,−8,8
(ii) For the line VW, the direction ratios are: 7,6,−8
(iii) Taking ratios, we get:
$\small{\frac{-4}{7}\ne\frac{-8}{6}\ne\frac{8}{-8}}$
(iv) The ratio is not a constant. That means, the two lines are not parallel.
(v) But point V is common to both the lines. That means, we cannot travel from U to W through V, along a straight line. That means, the three points U, V and W are non collinear.

2. Since the given three points are non collinear, there will be a unique plane.

3. Based on the coordinates, we get the position vectors:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}}    & {~=~}    &{2\hat{i}+5\hat{j}-3\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{v}}    & {~=~}    &{-2\hat{i}-3\hat{j}+5\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{w}}    & {~=~}    &{5\hat{i}+3\hat{j}-3\hat{k}}
\\ \end{array}}$

4. Based on the position vectors, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{v}-\vec{u}}    & {~=~}    &{-4\hat{i}-8\hat{j}+8\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{w}-\vec{u}}    & {~=~}    &{3\hat{i}-2\hat{j}}
\\ \end{array}}$

5. So the vector form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(2\hat{i}+5\hat{j}-3\hat{k} \right) \right].\left[\left(-4\hat{i}-8\hat{j}+8\hat{k} \right)\times\left(3\hat{i}-2\hat{j} \right)\right]}    & {~=~}    &{0}
\\ \end{array}}$

6. The above vector form can be expanded to get the Cartesian form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(2\hat{i}+5\hat{j}-3\hat{k} \right) \right].\left[\left(-4\hat{i}-8\hat{j}+8\hat{k} \right)\times\left(3\hat{i}-2\hat{j} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left[(x-2)\hat{i}+(y-5)\hat{j}+(z+3)\hat{k} \right].\left[16\hat{i}+24\hat{j}+32\hat{k}\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(16)(x-2)+(24)(y-5)+(32)(z+3)}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{(2)(x-2)+(3)(y-5)+(4)(z+3)}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{2x - 4+3y-15+4z+12}    & {~=~}    &{0}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{2x+3y+4z-7}    & {~=~}    &{0}
\\ {~\color{magenta}    8    }    &{{\Rightarrow}}    &{2x+3y+4z}    & {~=~}    &{7}
\\ \end{array}}$

7. If we are asked to find only the Cartesian form, we can avoid all the above six steps and apply the determinant form directly:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-2     &{    y-5     }    &{    z+3      }    
\\ -2-2      &{    -3-5     }    &{   5+3      }     
\\ 5-2      &{    3-5     }    &{ -3+3        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-2     &{    y-5     }    &{    z+3      }    
\\ -4      &{    -8     }    &{   8      }     
\\ 3      &{    -2     }    &{ 0        }     \\ \end{array}\right|~=~0$

$\small{\Rightarrow(x-2)(0+16)-(y-5)(0-24)+(z+3)(8+24)=0}$

$\small{\Rightarrow(x-2)(16)-(y-5)(-24)+(z+3)(32)=0}$

$\small{\Rightarrow(x-2)(2)-(y-5)(-3)+(z+3)(4)=0}$

$\small{\Rightarrow 2x-4+3y-15+4z+12=0}$

$\small{\Rightarrow 2x+3y+4z=7}$

Solved example 27.42
Find the vector and Cartesian equations of the planes that passes through three points
(a) (1,1,−1), (6,4,−5), (−4,−2,3)
(b) (1,1,0), (1,2,1), (−2,2,−1)
Solution:
Part (a):
1. Let the points be:
U(1,1,−1), V(6,4,−5), W(−4,−2,3)
• First we check whether the three points are collinear:
(i) For the line UV, the direction ratios are: 5,3,−4
(ii) For the line VW, the direction ratios are: −10,−6,8
(iii) Taking ratios, we get:
$\small{\frac{5}{-10}=\frac{3}{-6}=\frac{-4}{8}=\frac{-1}{2}}$
(iv) The ratio is a constant. That means, the two lines are parallel.
(v) But point V is common to both the lines. That means, the three points U, V and W are collinear.

2. Since the given three points are collinear, there will be infinite planes passing through them. We cannot write a unique plane.

Part (b):
1. Let the points be:
U(1,1,0), V(1,2,1), W(−2,2,−1)
• First we check whether the three points are collinear:
(i) For the line UV, the direction ratios are: 0,1,1
(ii) For the line VW, the direction ratios are: −3,0,−2
(iii) Taking ratios, we get:
$\small{\frac{0}{-3}\ne\frac{1}{0}\ne\frac{1}{-2}}$
(iv) The ratio is not a constant. That means, the two lines are not parallel.
(v) But point V is common to both the lines. That means, we cannot travel from U to W through V, along a straight line. That means, the three points U, V and W are non collinear.

2. Since the given three points are non collinear, there will be a unique plane.

3. Based on the coordinates, we get the position vectors:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}}    & {~=~}    &{\hat{i}+\hat{j}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{v}}    & {~=~}    &{\hat{i}+2\hat{j}+\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{w}}    & {~=~}    &{-2\hat{i}+2\hat{j}-\hat{k}}
\\ \end{array}}$

4. Based on the position vectors, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{v}-\vec{u}}    & {~=~}    &{\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{w}-\vec{u}}    & {~=~}    &{-3\hat{i}+\hat{j}-\hat{k}}
\\ \end{array}}$

5. So the vector form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(\hat{i}+\hat{j} \right) \right].\left[\left(\hat{j}+\hat{k} \right)\times\left(-3\hat{i}+\hat{j}-\hat{k} \right)\right]}    & {~=~}    &{0}
\\ \end{array}}$

6. The above vector form can be expanded to get the Cartesian form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(\hat{i}+\hat{j} \right) \right].\left[\left(\hat{j}+\hat{k} \right)\times\left(-3\hat{i}+\hat{j}-\hat{k} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left[(x-1)\hat{i}+(y-1)\hat{j}+z\hat{k} \right].\left[-2\hat{i}-3\hat{j}+3\hat{k}\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(-2)(x-1)+(-3)(y-1)+3z}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{-2x + 2-3y+3+3z}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{-2x-3y+3z+5}    & {~=~}    &{0}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{2x+3y-3z}    & {~=~}    &{5}
\\ \end{array}}$

7. If we are asked to find only the Cartesian form, we can avoid all the above six steps and apply the determinant form directly:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-1     &{    y-1     }    &{    z-0      }    
\\ 1-1      &{    2-1     }    &{   1-0      }     
\\ -2-1      &{    2-1     }    &{ -1-0        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-1     &{    y-1     }    &{    z-0      }    
\\ 0      &{    1     }    &{   1      }     
\\ -3      &{    1     }    &{ -1        }     \\ \end{array}\right|~=~0$

$\small{\Rightarrow(x-1)(-1-1)-(y-1)(0+3)+z(0+3)=0}$

$\small{\Rightarrow(x-1)(-2)-(y-1)(3)+z(3)=0}$

$\small{\Rightarrow -2x+2-3y+3+3z=0}$

$\small{\Rightarrow 2x+3y-3z=5}$

---

In the next section, we will see intercept form.

Previous

Contents

Next

Copyright©2026 Higher secondary mathematics.blogspot.com