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.

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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.

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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.

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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.

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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:
$\small{\frac{-6}{7},~\frac{3}{7},~\frac{2}{7}}$

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In the next section, we will see a few more solved examples.

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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.

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

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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

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

Exercise 27.2

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In the next section, we will see Plane.

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27.6 - Distance Between Two Skew Lines

In the previous section, we saw the conditions for two lines to be parallel or perpendicular. In this section, we will see the shortest distance between two 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.

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Now we will see skew lines. It can be explained in 3 steps:
1. Consider two lines which satisfy the following two conditions:
(i) The two lines do not intersect at any point
(ii) The two lines are not parallel.
2. Two lines which satisfy both the conditions are called skew lines.
3. Skew lines will be lying in different planes. In other words, skew lines are non coplanar.

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Let us try to visualize a pair of skew lines. It can be done in 2 steps:
1. Fig.27.10 below shows a room of size:
    ♦ 2 units width along the x-axis
    ♦ 5 units length along the y-axis
    ♦ 3 units height along the z-axis.

Fig.27.10

2. Two lines are shown in the fig.
    ♦ The yellow line is aligned with the diagonal BD of the wall ABED
    ♦ The green line is aligned with the diagonal EG of the ceiling EFGD
• Those two lines do not intersect at any point. Also, they are not parallel. So the yellow and green lines form a pair of skew lines

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Now we will see the method to find the shortest distance between skew lines. This method make use of the fact that, the line of shortest distance between two skew lines, will be perpendicular to both the lines. The method can be explained in 8 steps:
1. In the fig.27.11 below, $\small{l_1~\text{and}~l_2}$ form a pair of skew lines.
    ♦ Vector equation of $\small{l_1~\text{is:}~\vec{r}=\vec{u_1}+\lambda \vec{v_1}}$
    ♦ Vector equation of $\small{l_2~\text{is:}~\vec{r}=\vec{u_2}+\mu \vec{v_2}}$

Fig.27.11

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 PQ. It is the shortest line between $\small{l_1~\text{and}~l_2}$
So we can write:
    ♦ PQ is perpendicular to $\small{l_1}$
    ♦ PQ is perpendicular to $\small{l_2}$ also

5. Imagine that the vector $\small{\vec{PQ}}$ is present between the points P and Q.
• This vector will be perpendicular to both $\small{l_1~\text{and}~l_2}$

6. Next we want a vector with the same direction as $\small{\vec{PQ}}$. It can be obtained in 4 steps:
(i) Based on the vector equations of $\small{l_1~\text{and}~l_2}$, we can write:
    ♦ $\small{l_1}$ is parallel to $\small{\vec{v_1}}$
    ♦ $\small{l_2}$ is parallel to $\small{\vec{v_2}}$
(ii) So $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will be a vector perpendicular to both $\small{\vec{v_1}~\text{and}~\vec{v_2}}$
(iii) Consequently, $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will be a vector perpendicular to both $\small{l_1~\text{and}~l_2}$
(iv) That is., $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will have the same direction as $\small{\vec{PQ}}$

7. Imagine that, $\small{\vec{ST}}$ is shifted in such a way that, the initial point S of $\small{\vec{ST}}$ coincides with the initial point P of $\small{\vec{PQ}}$
• In this situation, we can write:
The projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
= Length of the line $\small{PQ}$

8. We can easily calculate the projection. See section 26.9.
• We can write:
Shortest distance
= Length of the line $\small{PQ}$
= $\small{\vec{ST}.\hat{PQ}}$
    ♦ We can obtain $\small{\vec{ST}}$ from (3)
    ♦ $\small{\hat{PQ}}$ is the unit vector in the direction of $\small{\vec{PQ}}$. We can obtain it from 6(iv)

---

Let us see some solved examples:

Solved example 27.24
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+\hat{j}+\lambda\left(2\hat{i}-\hat{j}+\hat{k} \right)}$
and $\small{\vec{r} = 2\hat{i}+\hat{j}-\hat{k}+\mu\left(3\hat{i}-5\hat{j}+2\hat{k} \right)}$
Solution:
1. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$

2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u_1}}    & {~=~}    &{\hat{i}+\hat{j}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{u_2}}    & {~=~}    &{2\hat{i}+\hat{j}-\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{v_1}}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{\vec{v_2}}    & {~=~}    &{3\hat{i}-5\hat{j}+2\hat{k}}
\\ \end{array}}$

3. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = \hat{i}-\hat{k}}$

4. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{3\hat{i}-\hat{j}-7\hat{k}}{\sqrt{3^2 + (-1)^2 + 7^2}}}$
= $\small{\frac{3\hat{i}-\hat{j}+7\hat{k}}{\sqrt{59}}}$
• The reader may write all the steps related to the cross product

5. Substituting (3) and (4) in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Projection}}    & {~=~}    &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\left(\hat{i}-\hat{k} \right).\left(\frac{3\hat{i}-\hat{j}-7\hat{k}}{\sqrt{59}} \right)}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{(1)(3)+(0)(-1)+(-1)(-7)}{\sqrt{59}}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\frac{10}{\sqrt{59}}}
\\ \end{array}}$

• So the shortest distance = $\small{\frac{10}{\sqrt{59}}}$ units 

Solved example 27.25
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+2\hat{j}+\hat{k}+\lambda\left(\hat{i}-\hat{j}+\hat{k} \right)}$
and $\small{\vec{r} = 2\hat{i}-\hat{j}-\hat{k}+\mu\left(2\hat{i}+\hat{j}+2\hat{k} \right)}$
Solution:
1. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$

2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u_1}}    & {~=~}    &{\hat{i}+2\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{u_2}}    & {~=~}    &{2\hat{i}-\hat{j}-\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{v_1}}    & {~=~}    &{\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{\vec{v_2}}    & {~=~}    &{2\hat{i}+\hat{j}+2\hat{k}}
\\ \end{array}}$

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

4. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{-3\hat{i}+3\hat{k}}{\sqrt{(-3)^2  + 3^2}}}$
= $\small{\frac{-3\hat{i}+3\hat{k}}{\sqrt{18}}}$
• The reader may write all the steps related to the cross product

5. Substituting (3) and (4) in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Shortest distance}}    & {~=~}    &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\left(\hat{i}-3\hat{j}-2\hat{k} \right).\left(\frac{-3\hat{i}+3\hat{k}}{\sqrt{18}} \right)}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{(1)(-3)+(-3)(0)+(-2)(3)}{\sqrt{18}}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\frac{-9}{\sqrt{18}}=\frac{(-1)\sqrt{9}\,\sqrt{9}}{\sqrt{2}\,\sqrt{9}} = \frac{(-1)(3)}{\sqrt{2}}}
\\ {~\color{magenta}    5    }    &{{}}    &{}    & {~=~}    &{\frac{-3\sqrt{2}}{2}}
\\ \end{array}}$

• Projection is a distance. It cannot be −ve. So we need to take the absolute value.
• Therefore, the shortest distance = $\small{\left|\frac{-3\sqrt{2}}{2} \right|~=~\frac{3\sqrt{2}}{2}}$ units

Solved example 27.26
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+2\hat{j}+3\hat{k}+\lambda\left(\hat{i}-3\hat{j}+2\hat{k} \right)}$
and $\small{\vec{r} = 4\hat{i}+5\hat{j}+6\hat{k}+\mu\left(2\hat{i}+3\hat{j}+\hat{k} \right)}$
Solution:
1. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$

2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u_1}}    & {~=~}    &{\hat{i}+2\hat{j}+3\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{u_2}}    & {~=~}    &{4\hat{i}+5\hat{j}+6\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{v_1}}    & {~=~}    &{\hat{i}-3\hat{j}+2\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{\vec{v_2}}    & {~=~}    &{2\hat{i}+3\hat{j}+\hat{k}}
\\ \end{array}}$

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

4. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{-9\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{(-9)^2  + 3^2 + 9^2}}}$
= $\small{\frac{-9\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{171}}}$
• The reader may write all the steps related to the cross product

5. Substituting (3) and (4) in (1), we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Shortest distance}}    & {~=~}    &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\left(3\hat{i}-3\hat{j}+3\hat{k} \right).\left(\frac{-9\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{171}} \right)}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{(3)(-9)+(-3)(-3)+(3)(9)}{\sqrt{171}}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\frac{9}{\sqrt{171}}=\frac{9}{3\,\sqrt{19}} }
\\ {~\color{magenta}    5    }    &{{}}    &{}    & {~=~}    &{\frac{3}{\sqrt{19}}}
\\ \end{array}}$

• Therefore, the shortest distance = $\small{\frac{3}{\sqrt{19}}}$ units

Solved example 27.27
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = (1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}}$
and $\small{\vec{r} = (s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}}$
Solution:
1. Let us convert the given vector equations to standard form:
• The vector equation of $\small{l_1}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{(1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\hat{i}-t\hat{i}+t\hat{j}-2\hat{j}+3\hat{k}-2t\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\hat{i}-2\hat{j}+3\hat{k}-t\hat{i}+t\hat{j}-2t\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\hat{i}-2\hat{j}+3\hat{k}+t\left(-\hat{i}+\hat{j}-2\hat{k} \right)}
\\ \end{array}}$
• The vector equation of $\small{l_2}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{s\hat{i}+\hat{i}+2s\hat{j}-\hat{j}-2s\hat{k}-\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\hat{i}-\hat{j}-\hat{k}+s\hat{i}+2s\hat{j}-2s\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\hat{i}-\hat{j}-\hat{k}+s\left(\hat{i}+2\hat{j}-2\hat{k} \right)}
\\ \end{array}}$

2. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$

3. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u_1}}    & {~=~}    &{\hat{i}-2\hat{j}+3\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{u_2}}    & {~=~}    &{\hat{i}-\hat{j}-\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{v_1}}    & {~=~}    &{-\hat{i}+\hat{j}-2\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{\vec{v_2}}    & {~=~}    &{\hat{i}+2\hat{j}-2\hat{k}}
\\ \end{array}}$

4. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = \hat{j}-4\hat{k}}$

5. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{2\hat{i}-4\hat{j}-3\hat{k}}{\sqrt{2^2  + (-4)^2 + (-3)^2}}}$
= $\small{\frac{2\hat{i}-4\hat{j}-3\hat{k}}{\sqrt{29}}}$
• The reader may write all the steps related to the cross product

6. Substituting (4) and (5) in (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Shortest distance}}    & {~=~}    &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\left(\hat{j}-4\hat{k} \right).\left(\frac{2\hat{i}-4\hat{j}-3\hat{k}}{\sqrt{29}} \right)}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{(0)(2)+(1)(-4)+(-4)(-3)}{\sqrt{29}}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\frac{8}{\sqrt{29}} }
\\ {~\color{magenta}    5    }    &{{}}    &{}    & {~=~}    &{\frac{3}{\sqrt{19}}}
\\ \end{array}}$

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

---

• In the above discussion, we were given the equations of the lines in vector form. If the equations were given in the Cartesian form, we can quickly convert them into vector form. See the "easy method" mentioned in Solved examples 27.11 and 27.22 of section 27.2

Let us see a solved example

Solved example 27.28
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose Cartesian equations are
$\small{\frac{x+1}{7}~=~\frac{y+1}{-6}~=~\frac{z+1}{1}}$
and $\small{\frac{x-3}{1}~=~\frac{y-5}{-2}~=~\frac{z-7}{1}}$
Solution:
1. Let us convert the given Cartesian equations to vector equations:
• The vector equation of $\small{l_1}$ is:
$\small{\vec{r} = -\hat{i}-\hat{j}-\hat{k}+\lambda\left(7\hat{i}-6\hat{j}+\hat{k} \right)}$
• The vector equation of $\small{l_2}$ is:
$\small{\vec{r} = 3\hat{i}+5\hat{j}+7\hat{k}+\mu\left(\hat{i}-2\hat{j}+\hat{k} \right)}$

2. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$

3. Based on the given equations of the lines, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u_1}}    & {~=~}    &{-\hat{i}-\hat{j}-\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{u_2}}    & {~=~}    &{3\hat{i}+5\hat{j}+7\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{v_1}}    & {~=~}    &{7\hat{i}-6\hat{j}+\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{\vec{v_2}}    & {~=~}    &{\hat{i}-2\hat{j}+\hat{k}}
\\ \end{array}}$

4. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = 4\hat{i}+6\hat{j}+8\hat{k}}$

5. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{-4\hat{i}-6\hat{j}-8\hat{k}}{\sqrt{(-4)^2  + (-6)^2 + (-8)^2}}}$
= $\small{\frac{-4\hat{i}-6\hat{j}-8\hat{k}}{\sqrt{116}}}$
• The reader may write all the steps related to the cross product

6. Substituting (4) and (5) in (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Shortest distance}}    & {~=~}    &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\left(4\hat{i}+6\hat{j}+8\hat{k} \right).\left(\frac{-4\hat{i}-6\hat{j}-8\hat{k}}{\sqrt{116}} \right)}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{(4)(-4)+(6)(-6)+(8)(-8)}{\sqrt{116}}}
\\ {~\color{magenta}    4    }    &{{}}    &{}    & {~=~}    &{\frac{-116}{\sqrt{116}}=(-1)\sqrt{116}}
\\ {~\color{magenta}    5    }    &{{}}    &{}    & {~=~}    &{-2\sqrt{29}}
\\ \end{array}}$

• Projection is a distance. It cannot be −ve. So we need to take the absolute value.
• Therefore, the shortest distance = $\small{\left|-2\sqrt{29} \right|~=~2\sqrt{29}}$ units

---

In the next section, we will see distance between parallel lines.

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