27.9 - Solved Examples on Normal Form

In the previous section, we saw equation of a plane in normal form. We saw some solved examples also. In this section, we will see a few more solved examples. Later in this section, we will see the method to find the coordinates of the foot of the perpendicular from the origin.

Solved example 27.35
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $\small{3\hat{i}+5\hat{j}-6\hat{k}}$
Solution:
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. Substituting the known values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{3^2 + 5^2 +(-6)^2}} \right)}    & {~=~}    &{7}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{70}} \right)}    & {~=~}    &{7}
\\ \end{array}}$

Solved example 27.36
Find the Cartesian equation of the following planes:
$\small{\text{(a)}~~\left(\vec{r} \right).\left(\hat{i}+\hat{j}-\hat{k}\right)=2~~~~\text{(b)}~~\left(\vec{r} \right).\left(2\hat{i}+3\hat{j}-4\hat{k}\right)=1}$
$\small{\text{(c)}~~\left(\vec{r} \right).\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]=15}$
Solution:
In each case, the scalar multiplication will give the Cartesian equation.
Part (a):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(\hat{i}+\hat{j}-\hat{k}\right)}    & {~=~}    &{2}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+\hat{j}-\hat{k}\right)}    & {~=~}    &{2}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{x+y-z}    & {~=~}    &{2}
\\ \end{array}}$

Part (b):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(2\hat{i}+3\hat{j}-4\hat{k}\right)}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}+3\hat{j}-4\hat{k}\right)}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{2x+3y-4z}    & {~=~}    &{1}
\\ \end{array}}$

Part (c):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]}    & {~=~}    &{15}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]}    & {~=~}    &{15}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{(s-2t)x+(3-t)y+(2s+t)z}    & {~=~}    &{15}
\\ \end{array}}$

---

Now we will see the method to find the coordinates of the foot of the perpendicular.
• Consider the fig.27.13 that we saw in the previous section. For convenience, it is shown again below:

Fig.27.13

• We know that, ON is the perpendicular drawn from the origin. We want a method to find the coordinates of N. It can be written in 5 steps:
1. Write the equation of the given plane in the vector form:
$\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
2. From the above equation, we get the unit vector:
$\small{\frac{\vec{n}}{\left|\vec{n} \right|}}$
• We get the distance $\small{d}$from the origin also.
3. Write down the coefficients of the above unit vector. These coefficients are the direction cosines of $\small{\vec{ON}}$
4. The magnitude of $\small{\vec{ON}}$ is $\small{d}$.Thus we get:
• x coordinate of N = $\small{d ~\times ~\text{first direction cosine}}$
• y coordinate of N = $\small{d ~\times ~\text{second direction cosine}}$
• z coordinate of N = $\small{d ~\times ~\text{third direction cosine}}$
5. Note that, $\small{\vec{ON}}$ is simply, the position vector of N. In the above steps, we are actually calculating the components of $\small{\vec{ON}}$.
• We know that:
    ♦ the coefficients of the components
    ♦ of the position vector of a point
    ♦ are the coordinates of that point 

---

Now we will see some solved examples:

Solved example 27.37
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
$\small{\text{(a)}~~2x+3y+4z-12=0~~~~\text{(b)}~~3y+4z-6=0}$
$\small{\text{(c)}~~x+y+z=1~~~~\text{(d)}~~5y+8=0}$
$\small{\text{(e)}~~2x-3y+4z-6=0}$
Solution:
Part (a):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2x+3y+4z-12}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x+3y+4z}    & {~=~}    &{12}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}+3\hat{j}+4\hat{k}\right)}    & {~=~}    &{12}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}+3\hat{j}+4\hat{k}}{\sqrt{2^2 + 3^2 + 4^2}}}    & {~=~}    &{\frac{12}{\sqrt{2^2 + 3^2 + 4^2}}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}+3\hat{j}+4\hat{k}}{\sqrt{29}}}    & {~=~}    &{\frac{12}{\sqrt{29}}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=\frac{2}{\sqrt{29}},~m=\frac{3}{\sqrt{29}},~n=\frac{4}{\sqrt{29}}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{12}{\sqrt{29}}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{12}{\sqrt{29}} \right)\left(\frac{2}{\sqrt{29}} \right) = \frac{24}{29}}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{12}{\sqrt{29}} \right)\left(\frac{3}{\sqrt{29}} \right) = \frac{36}{29}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{12}{\sqrt{29}} \right)\left(\frac{4}{\sqrt{29}} \right) = \frac{48}{29}}$

Part (b):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{3y+4z-6}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{3y+4z}    & {~=~}    &{6}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+3\hat{j}+4\hat{k}\right)}    & {~=~}    &{6}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}+3\hat{j}+4\hat{k}}{\sqrt{3^2 + 4^2}}}    & {~=~}    &{\frac{6}{\sqrt{3^2 + 4^2}}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}+3\hat{j}+4\hat{k}}{5}}    & {~=~}    &{\frac{6}{5}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=0,~m=\frac{3}{5},~n=\frac{4}{5}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{6}{5}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{6}{5} \right)\left(0 \right) = 0}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{6}{5} \right)\left(\frac{3}{5} \right) = \frac{18}{25}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{6}{5} \right)\left(\frac{4}{5} \right) = \frac{24}{25}}$

Part (c):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x+y+z}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+\hat{j}+\hat{k}\right)}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{1^2 + 1^2 + 1^2}}}    & {~=~}    &{\frac{1}{\sqrt{1^2 + 1^2 + 1^2}}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}}    & {~=~}    &{\frac{1}{\sqrt{3}}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=,\frac{1}{\sqrt{3}}~m=\frac{1}{\sqrt{3}},~n=\frac{1}{\sqrt{3}}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{1}{\sqrt{3}}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{1}{\sqrt{3}} \right)\left(\frac{1}{\sqrt{3}} \right) = \frac{1}{3}}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{1}{\sqrt{3}} \right)\left(\frac{1}{\sqrt{3}} \right) = \frac{1}{3}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{1}{\sqrt{3}} \right)\left(\frac{1}{\sqrt{3}} \right) = \frac{1}{3}}$

Part (d):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{5y+8}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{5y}    & {~=~}    &{-8}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+5\hat{j}+0\hat{k}\right)}    & {~=~}    &{-8}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}-5\hat{j}+0\hat{k}\right)}    & {~=~}    &{8}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}-5\hat{j}+0\hat{k}}{\sqrt{0^2 + 5^2 + 0^2}}}    & {~=~}    &{\frac{8}{\sqrt{0^2 + 5^2 + 0^2}}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}-5\hat{j}+0\hat{k}}{5}}    & {~=~}    &{\frac{8}{5}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(0\hat{i}-\hat{j}+0\hat{k} \right)}    & {~=~}    &{\frac{8}{5}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=0,~m=-1,~n=0}$
• Also, $\small{\left|\vec{ON} \right| = \frac{8}{5}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{8}{5} \right)\left(0 \right) = 0}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{8}{5} \right)\left(-1 \right) = \frac{-8}{5}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{8}{5} \right)\left(0 \right) = 0}$

Part (e):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2x-3y+4z-6}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x-3y+4z}    & {~=~}    &{6}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}-3\hat{j}+4\hat{k}\right)}    & {~=~}    &{6}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{2^2 + 3^2 + 4^2}}}    & {~=~}    &{\frac{6}{\sqrt{2^2 + 3^2 + 4^2}}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=\frac{2}{\sqrt{29}},~m=\frac{-3}{\sqrt{29}},~n=\frac{4}{\sqrt{29}}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{6}{\sqrt{29}}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{6}{\sqrt{29}} \right)\left(\frac{2}{\sqrt{29}} \right) = \frac{12}{29}}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{6}{\sqrt{29}} \right)\left(\frac{-3}{\sqrt{29}} \right) = \frac{-18}{29}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{6}{\sqrt{29}} \right)\left(\frac{4}{\sqrt{29}} \right) = \frac{24}{29}}$

Solved example 27.38
If O be the origin and the coordinates of P be (1,2,−3), then find the equation of the plane passing through P and perpendicular to OP.
Solution:
Part (a):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• So we want $\small{\hat{n}~\text{and}~d}$

2. The plane should be perpendicular to OP. That means, $\small{\vec{OP}}$ is normal to the plane
• $\small{\vec{OP}}$ is the position vector of P, which is: $\small{\hat{i}+2\hat{j}-3\hat{k}}$
• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{n}}    & {~=~}    &{\frac{\vec{OP}}{\left|\vec{OP} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{1^2 + 2^2 + (-3)^2}}}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{14}}}
\\ \end{array}}$

3. d= distance OP = $\small{\left|\vec{OP} \right|~=~\sqrt{14}}$

4. Substituting in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{14}} \right)}    & {~=~}    &{\sqrt{14}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+2\hat{j}-3\hat{k} \right)}    & {~=~}    &{14}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{x+2y-3z}    & {~=~}    &{14}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{x+2y-3z-14}    & {~=~}    &{0}
\\ \end{array}}$

---

In the next section, we will see plane perpendicular to a given vector and passing through a given point.

Previous

Contents

Next 

Copyright©2026 Higher secondary mathematics.blogspot.com

27.8 - Equation of A Plane in Normal Form

In the previous section, we completed a discussion on the shortest distance between two skew lines. In this section, we will see plane.

In 3D space, there are infinite planes in infinite orientations. But we can obtain a unique plane, if any one of the following three conditions are satisfied:
(i) Drop a perpendicular from the origin O, onto a plane. If the length of that perpendicular is fixed, then we get a unique plane.
(ii) A plane is passing through a given point $\small{\left(x_1,y_1,z_1 \right)}$. Also, the plane is perpendicular to a given vector or line. Then we get a unique plane.
(iii) The plane passes through three given non collinear points.

---

Now we will try to derive the vector equation of a plane. It can be done in 4 steps:
1. In fig.27.13 below, a plane passes through 3 points A, B and C.

Fig.27.13

• A perpendicular is dropped from O, onto the plane. The foot of the perpendicular is N. Let the length ON be $\small{d\,(d\ne0)}$.

• Suppose that, $\small{\hat{n}}$ is the unit vector perpendicular to the plane.
Then $\small{\vec{ON}=d\,\hat{n}}$

2. Mark any convenient point P on the plane.
• ON is perpendicular to the plane. So all lines on the plane will be perpendicular to ON. Obviously, NP will be perpendicular to ON.
• We can write: $\small{\vec{NP}~\text{and}~\vec{ON}}$ are perpendicular to each other.
• So we get: $\small{\vec{NP}.\vec{ON}=0}$

3. Let $\small{\vec{r}}$ be the position vector of $\small{P}$
Applying triangle law of vector addition, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{ON}+\vec{NP}}    & {~=~}    &{\vec{OP}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{NP}}    & {~=~}    &{\vec{OP}-\vec{ON}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\vec{NP}}    & {~=~}    &{\vec{r}-d\,\hat{n}}
\\ \end{array}}$

4. Substituting in (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{NP}.\vec{ON}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r}-d\,\hat{n} \right).d\,\hat{n}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r}-d\,\hat{n} \right).\hat{n}}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\vec{r}.\hat{n}-d\left(\hat{n}.\hat{n} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\vec{r}.\hat{n}-d}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ \end{array}}$
• This is the vector form of the equation of the plane.
◼ Remarks:
• 3 (magenta color): We are able to obtain this step from 2 (magenta colo) because, $\small{d}$ is the distance from origin. We wrote that, it is not zero.
• 5 (magenta color): Here we apply the fact that, $\small{\left(\hat{n}.\hat{n} \right)}$ is 1.

---

Now we will derive the Cartesian form. It can be done in 3 steps:
1. In fig.27.13 above, P is an arbitrary point. So we can write the component form of $\small{\vec{r}}$:
$\small{\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}}$

2. If $\small{l,~m~\text{and}~n}$ are the direction cosines of $\small{\hat{n}}$, then the component form of $\small{\hat{n}}$ can be written as:
$\small{\hat{n}=l\hat{i}+m\hat{j}+n\hat{k}}$

3. Substituting the above two results in the vector form, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(l\hat{i}+m\hat{j}+n\hat{k} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{lx + my + nz}    & {~=~}    &{d}
\\ \end{array}}$
• This is the Cartesian form.

---

Suppose that, instead of direction cosines $\small{l,~m~\text{and}~n}$ of $\small{\hat{n}}$, we are given the direction ratios $\small{a,~b~\text{and}~c}$ of the vector perpendicular to the plane. Then we can derive the Cartesian form of the plane in 4 steps:
1. We have the basic vector form:
$\small{\vec{r}.\hat{n}=d}$

2. In the present case,
    ♦ We do not have $\small{\hat{n}}$, which is the unit vector perpendicular to the plane
    ♦ But we do have $\small{\vec{n}}$, which is the vector perpendicular to the plane
• We have $\small{\vec{n}}$ because, we are given the direction ratios $\small{a,~b~\text{and}~c}$ of the vector perpendicular to the plane.
• We can write: $\small{\vec{n}=a\hat{i}+b\hat{j}+c\hat{k}}$

3. But we can obtain $\small{\hat{n}}$ easily:
$\small{\hat{n}=\frac{\vec{n}}{\left|\vec{n} \right|}}$

4. So substituting in the basic vector form, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\frac{a\hat{i}+b\hat{j}+c\hat{k}}{\sqrt{a^2 + b^2 + c^2}} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{ax + by + cz}    & {~=~}    &{d\,\sqrt{a^2 + b^2 + c^2}}
\\ \end{array}}$
• This is the Cartesian form when the direction ratios are given.

---

Let us see some solved examples

Solved example 27.30
Find the distance of the plane $\small{2x-3y+4z-6=0}$ from the origin
Solution:
1. The given equation can be rearranged as:
$\small{2x-3y+4z=6}$

2. Writing this in vector form, we get:
$\small{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}-3\hat{j}+4\hat{k} \right)=6}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29}}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 6}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{6}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{6}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}~=~d}
\\ \end{array}}$

Solved example 27.31
Find the distance of the plane $\small{3x-4y+12z-3=0}$ from the origin
Solution:
1. The given equation can be rearranged as:
$\small{3x-4y+12z=3}$

2. Writing this in vector form, we get:
$\small{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(3\hat{i}-4\hat{j}+12\hat{k} \right)=3}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{169}=13}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 3}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{3}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{3}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{3}{13}~=~d}
\\ \end{array}}$

Solved example 27.32
Find the vector equation of the plane which is at a distance of $\small{\frac{6}{\sqrt{29}}}$ from the origin and its normal vector from the origin is $\small{2\hat{i}-3\hat{j}+4\hat{k}}$. Also find its Cartesian form.
Solution:
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$

2. In our present problem, we are given the normal vector: $\small{2\hat{i}-3\hat{j}+4\hat{k}}$
• This given normal vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• This given normal vector is $\small{\vec{n}}$, with magnitude $\small{\sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29}}$

3. We can obtain $\small{\hat{n}}$ as:
$\small{\hat{n}=\frac{\vec{n}}{\left|\vec{n} \right|}=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}}$

4. Substituting in (1), we get the vector equation of the plane as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{2}{\sqrt{29}}\,\hat{i}~-~\frac{3}{\sqrt{29}}\,\hat{j}~+~\frac{4}{\sqrt{29}}\,\hat{k} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ \end{array}}$

5. To write the Cartesian form, we must find the actual scalar multiplication:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(\frac{2}{\sqrt{29}}\,\hat{i}~-~\frac{3}{\sqrt{29}}\,\hat{j}~+~\frac{4}{\sqrt{29}}\,\hat{k} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\frac{2}{\sqrt{29}}\,\hat{i}~-~\frac{3}{\sqrt{29}}\,\hat{j}~+~\frac{4}{\sqrt{29}}\,\hat{k} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{2x}{\sqrt{29}}-\frac{3y}{\sqrt{29}}+\frac{4z}{\sqrt{29}}}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{2x - 3y + 4z - 6}    & {~=~}    &{0}
\\ \end{array}}$

Solved example 27.33
Find the direction cosines of the unit vector perpendicular to the plane $\small{\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)+1 = 0}$ passing through the origin.
Solution:
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given vector equation can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)+1}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)}    & {~=~}    &{-1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{(-1)\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)}    & {~=~}    &{(-1)(-1)}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(-6\hat{i}+3\hat{j}+2\hat{k} \right)}    & {~=~}    &{1}
\\ \end{array}}$

3.In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{(-6)^2 + (3)^2 + (2)^2} = \sqrt{49} = 7}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 1}$
• But the general form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{-1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{-1}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{-6\hat{i}+3\hat{j}+2\hat{k}}{7} \right)}    & {~=~}    &{\frac{-1}{7}~=~d}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{-6\hat{i}}{7}+\frac{3\hat{j}}{7}+\frac{2\hat{k}}{7} \right)}    & {~=~}    &{\frac{-1}{7}~=~d}
\\ \end{array}}$

6. From the above result, it is clear that:
$\small{\hat{n}=\frac{-6\hat{i}}{7}+\frac{3\hat{j}}{7}+\frac{2\hat{k}}{7}}$

7. For any unit vector, the coefficients are the direction cosines.
• So in our present case, the required direction cosines are:
$\small{\frac{-6}{7},~\frac{3}{7},~\frac{2}{7}}$

Solved example 27.34
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin
$\small{\text{(a)}~~z = 2~~\text{(b)}~~x+y+z=1}$
$\small{\text{(c)}~~2x+3y-z =5~~\text{(d)}~~5y+8=0}$
Solution:
Part (a):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{z}    & {~=~}    &{2}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+0\hat{j}+\hat{k} \right)}    & {~=~}    &{2}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(0\hat{i}+0\hat{j}+\hat{k} \right)}    & {~=~}    &{2}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is $\small{\hat{n}}$ because, its magnitude is one.

4. So the vector equation in (2) is comparable to the general form:
$\small{\vec{r}.\hat{n}=d}$

5. Thus we get:
$\small{\hat{n}=0\hat{i}+0\hat{j}+\hat{k}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{0,~0,~1}$

7. The vector equation obtained in (2) is comparable to the general form. That means, the quantity in the R.H.S is d.
• Therefore, the distance of the plane from the origin is 2 units.

Part (b):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x+y+z}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+\hat{j}+\hat{k} \right)}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\hat{i}+\hat{j}+\hat{k} \right)}    & {~=~}    &{1}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 1}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{1}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \right)}    & {~=~}    &{\frac{1}{\sqrt{3}}~=~d}
\\ \end{array}}$
• Thus we get:
$\small{\hat{n}=\frac{\hat{i}}{\sqrt{3}}+\frac{\hat{j}}{\sqrt{3}}+\frac{\hat{k}}{\sqrt{3}}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{\frac{1}{\sqrt{3}},~\frac{1}{\sqrt{3}},~\frac{1}{\sqrt{3}}}$

7. From (5) we get d. We can write:
Distance of the plane from the origin is $\small{\frac{1}{\sqrt{3}}}$ units.

Part (c):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2x+3y-z}    & {~=~}    &{5}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}+3\hat{j}-\hat{k} \right)}    & {~=~}    &{5}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(2\hat{i}+3\hat{j}-\hat{k} \right)}    & {~=~}    &{5}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14}}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 5}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{5}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{5}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{2\hat{i}+3\hat{j}-\hat{k}}{\sqrt{14}} \right)}    & {~=~}    &{\frac{5}{\sqrt{14}}~=~d}
\\ \end{array}}$
• Thus we get:
$\small{\hat{n}=\frac{2\hat{i}}{\sqrt{14}}+\frac{3\hat{j}}{\sqrt{14}}-\frac{\hat{k}}{\sqrt{14}}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{\frac{2}{\sqrt{14}},~\frac{3}{\sqrt{14}},~\frac{-1}{\sqrt{14}}}$

7. From (5) we get d. We can write:
Distance of the plane from the origin is $\small{\frac{5}{\sqrt{14}}}$ units.

Part (d):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{5y}    & {~=~}    &{-8}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+5\hat{j}+0\hat{k} \right)}    & {~=~}    &{-8}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(5\hat{j} \right)}    & {~=~}    &{-8}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(-5\hat{j} \right)}    & {~=~}    &{8}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is 5

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 8}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{8}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{8}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{-5\hat{j}}{5} \right)}    & {~=~}    &{\frac{8}{5}~=~d}
\\ \end{array}}$
• Thus we get:
$\small{\hat{n}=\frac{-5\hat{j}}{5}=-\hat{j}=0\hat{i}-\hat{j}+0\hat{k}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{0,~-1,~0}$

7. From (5) we get d. We can write:
Distance of the plane from the origin is $\small{\frac{8}{5}}$ units.

---

In the next section, we will see a few more solved examples. We will see coordinates of the foot of the perpendicular from origin also.

Previous

Contents

Next

Copyright©2026 Higher secondary mathematics.blogspot.com

27.7 - Distance Between Two Parallel Lines

In the previous section, we saw the shortest distance between two skew lines. In this section, we will see the distance between two parallel lines.

• If two lines in space intersect at a point, we can say that the shortest distance between the two lines is zero.
• If two lines in space are parallel, the shortest distance can be obtained in 3 steps:
(i) Mark any convenient point on one line.
(ii) Drop a perpendicular from that point onto the other line.
(iii) Length of that perpendicular is the shortest distance.

---

Now we will see the method to find the distance between parallel lines. The method can be explained in 10 steps:
1. In the fig.27.12 below, $\small{l_1~\text{and}~l_2}$ form a pair of parallel lines.
    ♦ Vector equation of $\small{l_1~\text{is:}~\vec{r}=\vec{u_1}+\lambda \vec{v}}$
    ♦ Vector equation of $\small{l_2~\text{is:}~\vec{r}=\vec{u_2}+\mu \vec{v}}$
• Note that both lines are parallel to $\small{\vec{v}}$. So the lines themselves are parallel to each other

Fig.27.12

2. Next, we want a point on each line.
• Mark the point S on $\small{l_1}$ such that, the position vector of S is $\small{\vec{u_1}}$
• Mark the point T on $\small{l_2}$ such that, the position vector of T is $\small{\vec{u_2}}$

3. Based on the above two position vectors, we can write:
$\small{\vec{ST}=\vec{u_2} - \vec{u_1}}$

4. Next we concentrate on PT. It is obtained by dropping a perpendicular from T onto $\small{l_1}$. Point P is the foot of the perpendicular. PT is the shortest line between $\small{l_1~\text{and}~l_2}$
So we can write:
    ♦ PT is perpendicular to $\small{l_1}$
    ♦ PT is perpendicular to $\small{l_2}$ also

5. We want the distance TP.
• From the triangle STP,we get: $\small{TP = ST \sin \theta}$.
• If we use vectors, the above calculation can be easily done.

6. The lines $\small{l_1,~l_2,~ST~\text{and}~TP}$ lie in the same plane.
The cross product $\small{\left(\vec{v}\times\vec{ST} \right)}$ will be a vector which is perpendicular to that plane.

7. We have:
$\small{\vec{v}\times\vec{ST} = \left(\left|\vec{v} \right|\,\left|\vec{ST} \right|\,\sin\theta \right)\hat{n}}$
• Where $\small{\hat{n}}$ is the unit vector perpendicular to the plane.

8. Substituting for $\small{\vec{ST}}$ from (3), we get:
$\small{\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) = \left(\left|\vec{v} \right|\,\left|\vec{ST} \right|\,\sin\theta \right)\hat{n}}$
• $\small{\left|\vec{ST} \right|}$ can be written as the length ST. So we get:
$\small{\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) = \left(\left|\vec{v} \right|\,ST\,\sin\theta \right)\hat{n}}$
• $\small{ST\,\sin\theta = PT}$. So we get:
$\small{\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) = \left(\left|\vec{v} \right|PT \right)\hat{n}}$

9. The above result is the equality of two vectors. Their magnitudes will be the same. So we can write:
$\small{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right| = \left|\vec{v} \right|\,PT}$

10. Thus we get:
$\small{PT = \frac{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right|}{\left|\vec{v} \right|}}$

---

Let us see a solved example

Solved example 27.29
Find the distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+2\hat{j}-4\hat{k}+\lambda\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
and $\small{\vec{r} = 3\hat{i}+3\hat{j}-5\hat{k}+\mu\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
Solution:
The given two lines are parallel because, both the lines are parallel to the vector $\small{2\hat{i}+3\hat{j}+6\hat{k} }$
1. We have:
Shortest distance
= Length of the line $\small{PT}$
= $\small{\frac{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right|}{\left|\vec{v} \right|}}$

2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u_1}}    & {~=~}    &{\hat{i}+2\hat{j}-4\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{u_2}}    & {~=~}    &{3\hat{i}+3\hat{j}-5\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{v}}    & {~=~}    &{2\hat{i}+3\hat{j}+6\hat{k}}
\\ \end{array}}$

3. $\small{\vec{u_2} - \vec{u_1} = 2\hat{i}+\hat{j}-\hat{k}}$

4. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{PT}    & {~=~}    &{\frac{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right|}{\left|\vec{v} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\frac{\left|\left(2\hat{i}+3\hat{j}+6\hat{k} \right)\times\left(2\hat{i}+\hat{j}-\hat{k} \right) \right|}{\left|2\hat{i}+3\hat{j}+6\hat{k} \right|}}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{\left|-9\hat{i}+14\hat{j}-4\hat{k}  \right|}{\left|2\hat{i}+3\hat{j}+6\hat{k} \right|}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\frac{\sqrt{293}}{\sqrt{49}}~=~\frac{\sqrt{293}}{7}}
\\ \end{array}}$

• So the shortest distance = $\small{\frac{\sqrt{293}}{7}}$ units
• Therefore, the shortest distance = $\small{\frac{8}{\sqrt{29}}}$ units

---

The link below gives the exercise questions. All those exercise questions are already answered in our discussions.

Exercise 27.2

---

In the next section, we will see Plane.

Previous

Contents

Next

Copyright©2026 Higher secondary mathematics.blogspot.com