27.2 - Equation of A Line in Space

In the previous section, we saw the effect when direction ratios of two lines are proportional. In this section, we will see equation of line.

Equation of a line in space

In Class XI, we have seen the equation of line in two dimensional plane. Now we will see the equation in three dimensional space.
It can be explained in 10 steps:
1. Suppose that, we want a line to satisfy a condition:
The line must pass through a particular point $\small{U(x_1,y_1,z_1)}$.
• But here we encounter a problem. Infinite lines can pass through U. Because there are infinite directions possible through U. We cannot think about a unique line which satisfies the condition.

2. So we add one more condition. The two conditions are:
(i) The line must pass through a particular point $\small{U(x_1,y_1,z_1)}$.
(ii) The line must have a particular direction.
• There will be one and only one line which can satisfy the two conditions simultaneously. Our aim is to write the equation of such a line.

3. In fig.27.4 below, the magenta line $\small{L}$ satisfies two conditions.
(i) It passes through the given point $\small{U(x_1,y_1,z_1)}$
(ii) It is parallel to the given vector $\small{\vec{v}}$
• Our aim is to write the equation of this magenta line $\small{L}$.

Fig.27.4

4. A random point $\small{P(x,y,z)}$ is marked on the line $\small{L}$
• The position vector of P is $\small{\vec{r}}$

5. The position vector of U is $\small{\vec{u}}$

6. Imagine that, there is a vector $\small{\vec{UP}}$ between U and P
• $\small{\vec{UP}}$ is parallel to $\small{\vec{v}}$
• So we can write: $\small{\vec{UP}~=~\lambda \vec{v}}$
Where $\small{\lambda}$ is some real number.

7. Applying the triangle rule of vector addition, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u}+\vec{UP}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{\vec{u}+\lambda\vec{v}}    \\
\end{array}}$

8. In the above equation, we can input infinite values (real numbers) for $\small{\lambda}$.
• For each value of $\small{\lambda}$, we get a unique $\small{\vec{r}}$
• Each unique $\small{\vec{r}}$ gives a unique point on the line $\small{L}$
• So infinite values of $\small{\lambda}$ will give infinite points on $\small{L}$. Those infinite points together will give us the line $\small{L}$

9. Therefore, the vector equation of the line passing through U and parallel to $\small{\vec{v}}$ is: $\small{\vec{r}~=~\vec{u}+\lambda\vec{v}}$

10. Let us see an example:
• The vector equation of the line through (5,2,−4) and which is parallel to the vector $\small{3\hat{i}+2\hat{j}-8\hat{k}}$ can be obtained in 2 steps:
(i) $\small{\vec{u}}$ is the position vector of the given point (5,2,−4). So $\small{\vec{u} = 5\hat{i}+2\hat{j}-4\hat{k}}$
(ii) Then the required vector equation is:
$\small{5\hat{i}+2\hat{j}-4\hat{k}~+~\lambda \left(3\hat{i}+2\hat{j}-8\hat{k} \right)}$

Derivation of Cartesian form

This can be done in 7 steps:
1. In the fig.27.4 above, $\small{\vec{v}}$ is parallel to the magenta line $\small{L}$. Let the component form of the vector be: $\small{\vec{v}=v_1 \hat{i}+v_2\hat{j}+v_3\hat{k}}$.
• Based on the component form of $\small{\vec{v}}$, we can write:
Vector $\small{\vec{v}}$ has a set of direction ratios: $\small{v_1,~v_2,~v_3}$. This we proved in the previous section.

2. Suppose another line $\small{L_1}$ is also parallel to $\small{\vec{v}}$. Then $\small{L_1}$ will also have a set of direction ratios $\small{a_{L1},~b_{L1},~c_{L1}~~\text{as:}~~v_1,~v_2,~v_3}$.

3. $\small{L_1}$ will be parallel to $\small{L}$.
• Then the direction ratios of $\small{L~\text{and}~L_1}$ are proportional. This we proved in the previous section. We can write:
$\small{\frac{a_L}{a_{L1}}~=~\frac{b_L}{b_{L1}}~=~\frac{b_L}{b_{L1}}}$

4. So our next task is to find $\small{a_{L},~b_{L},~c_{L}}$
• We have two points $\small{U(x_1,y_1,z_1)~\text{and}~P(x,y,z)}$ on the line $\small{L}$
• Then a set of direction ratios of $\small{L}$ can be taken as: $\small{(x – x_1),~(y – y_1),~(z – z_1)}$

5. So from (3), we get:
$\small{\frac{x – x_1}{v_1}~=~\frac{y – y_1}{v_2}~=~\frac{z – z_1}{v_3}}$
• This is the equation of $\small{L}$ in Cartesian form.

6. Let us see an example:
The Cartesian equation of the line through (5,2,−4) and which is parallel to the vector $\small{3\hat{i}+2\hat{j}-8\hat{k}}$ can be obtained in 3 steps:
(i) The required line $\small{L}$, passes through (5,2,−4).
• Let (x,y,z) be any point on the line.
• Then (x−5), (y−2), (z+4) is a set of direction ratios of the required line.
(ii) Any line $\small{L_1}$ parallel to $\small{3\hat{i}+2\hat{j}-8\hat{k}}$, will have a set of direction ratios 3, 2, −8
(iii) The lines $\small{L ~\text{and}~ L_1}$ are parallel.
• So the direction ratios of the two lines will be proportional. Therefore, we can write:
$\small{\frac{x – 5}{3}~=~\frac{y – 2}{2}~=~\frac{z + 4}{-8}}$
• This is the equation of $\small{L}$ in Cartesian form.

7. Now we will see an interesting fact. It can be written in (ii) steps:
(i) Consider the Cartesian equation obtained in (6) above. The three fractions are equal. So we can write:
$\small{\frac{x – 5}{3}~=~\frac{y – 2}{2}~=~\frac{z + 4}{-8} = \lambda}$
(ii) $\small{\lambda}$ can be any real number.
• Let us put $\small{\lambda = -1}$. Then we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x – 5}{3}}    & {~=~}    &{-1}    \\
{~\color{magenta}     {}   }    &{\Rightarrow}    &{x}    & {~=~}    &{2}    \\
{~\color{magenta}    2    }    &{{}}    &{\frac{y – 2}{2}}    & {~=~}    &{-1}    \\
{~\color{magenta}     {}   }    &{\Rightarrow}    &{y}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{}}    &{\frac{z + 4}{-8}}    & {~=~}    &{-1}    \\
{~\color{magenta}     {}   }    &{\Rightarrow}    &{z}    & {~=~}    &{4}    \\
\end{array}}$

• So (2,0,4) is a point on the line. This is shown in the actual plot of $\small{L}$ in the fig.27.5 below:

Fig.27.5

• Note that in the fig. above, $\small{L}$ is parallel to $\small{\vec{v}}$, and is passing through U(5,2,−4)

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Alternate method to derive the Cartesian form

This can be written in 4 steps:
1. Based on fig.27.4 above, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}}    \\
{~\color{magenta}    2    }    &{{}}    &{\vec{v}}    & {~=~}    &{(v_1)\hat{i}+(v_2)\hat{j}+(v_3)\hat{k}}    \\
{~\color{magenta}    3    }    &{{}}    &{\vec{r}}    & {~=~}    &{x\hat{i}+y\hat{j}+z\hat{k}}    \\
\end{array}}$

2. We have the vector form:
$\small{\vec{r}~=~\vec{u}+\lambda\vec{v}}$
• Substituting the vectors, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u}+\lambda\vec{v}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x\hat{i}+y\hat{j}+z\hat{k}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}+\lambda\left[(v_1)\hat{i}+(v_2)\hat{j}+(v_3)\hat{k} \right]}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x\hat{i}+y\hat{j}+z\hat{k}}    & {~=~}    &{\left[x_1+ \lambda(v_1)\right]\hat{i}+\left[y_1+ \lambda(v_2)\right]\hat{j}+\left[z_1+ \lambda(v_3)\right]\hat{k}}    \\
\end{array}}$

3. Equating the coefficients of $\small{\hat{i},~\hat{j}~\text{and}~\hat{k}}$, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{x_1+ \lambda(v_1)}    \\
{~\color{magenta}    2    }    &{{}}    &{y}    & {~=~}    &{y_1+ \lambda(v_2)}    \\
{~\color{magenta}    3    }    &{{}}    &{z}    & {~=~}    &{z_1+ \lambda(v_3)}    \\
\end{array}}$

4. Eliminating $\small{\lambda}$, we get:
$\small{\frac{x - x_1}{v_1}~=~\frac{y - y_1}{v_2}~=~\frac{z - z_1}{v_3}~=~\lambda}$

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Now we will see some solved examples

Solved example 27.8
Find the equation of the line which passes through the point (1,2,3) and is parallel to the vector $\small{3\hat{i}+2\hat{j}-2\hat{k}}$.
Solution:
Part (i): Derivation of vector form
1. $\small{\vec{u}}$ is the position vector of the given point U(1,2,3). So $\small{\vec{u} = \hat{i}+2\hat{j}+3\hat{k}}$
2. Then the required vector equation is:
$\small{\vec{r}=\hat{i}+2\hat{j}+3\hat{k}~+~\lambda \left(3\hat{i}+2\hat{j}-2\hat{k} \right)}$

Part (ii): Derivation of Cartesian form
1. The required line $\small{L}$, passes through (1,2,3).
• Let (x,y,z) be any point on the line.
• Then (x−1), (y−2), (z−3) is a set of direction ratios of the required line.
(ii) Any line $\small{L_1}$ parallel to $\small{3\hat{i}+2\hat{j}-2\hat{k}}$, will have a set of direction ratios 3, 2, −2
(iii) The lines $\small{L ~\text{and}~ L_1}$ are parallel.
• So the direction ratios of the two lines will be proportional. Therefore, we can write:
$\small{\frac{x – 1}{3}~=~\frac{y – 2}{2}~=~\frac{z - 3}{-2}}$
• This is the equation of $\small{L}$ in Cartesian form. 

Solved example 27.9
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $\small{2\hat{i}-\hat{j}+4\hat{k}}$ and is in the direction $\small{\hat{i}+2\hat{j}-\hat{k}}$.
Solution:
Part (i): Derivation of vector form
1. $\small{\vec{u}}$ is the position vector of the given point. This vector is already given to us. So $\small{\vec{u} = 2\hat{i}-\hat{j}+4\hat{k}}$
2. Then the required vector equation is:
$\small{\vec{r}=2\hat{i}-\hat{j}+4\hat{k}~+~\lambda \left(\hat{i}+2\hat{j}-\hat{k} \right)}$

Part (ii): Derivation of Cartesian form
1. Based on the given position vector, we can write:
The required line $\small{L}$, passes through (2,−1,4).
• Let (x,y,z) be any point on the line.
• Then (x−2), (y+1), (z−4) is a set of direction ratios of the required line.
(ii) Any line $\small{L_1}$ parallel to $\small{\hat{i}+2\hat{j}-\hat{k}}$, will have a set of direction ratios 1, 2, −1
(iii) The lines $\small{L ~\text{and}~ L_1}$ are parallel.
• So the direction ratios of the two lines will be proportional. Therefore, we can write:
$\small{\frac{x – 2}{1}~=~\frac{y +1}{2}~=~\frac{z - 4}{-1}}$
• This is the equation of $\small{L}$ in Cartesian form. 

Solved example 27.10
Find the Cartesian equation of the line that passes through the point (−2,4,−5) and parallel to the line given by $\small{\frac{x+3}{3}~=~\frac{y-4}{5}~=~\frac{x+8}{6}}$.
Solution:
1. The Cartesian form of a line passing through $\small{U\left(x_1,y_1,z_1 \right)}$, and is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$ is:
$\small{\frac{x-x_1}{v_1}~=~\frac{y-y_1}{v_2}~=~\frac{z-z_1}{v_3}}$

2. $\small{v_2,~v_3,~v_4}$ can be taken as the direction ratios of any line parallel to $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$

3. In our present problem, the required line $\small{L}$ must be parallel to the line $\small{\frac{x+3}{3}~=~\frac{y-4}{5}~=~\frac{x+8}{6}}$
• This line has direction ratios 3, 5, 6
• So the required line $\small{L}$ must be parallel to the vector $\small{\vec{v}=3 \hat{i}+5 \hat{j}+6 \hat{k}}$  

4. The required line $\small{L}$ is said to pass through (−2,4,−5).
• So this line has the Cartesian equation:
$\small{\frac{x+2}{3}~=~\frac{y-4}{5}~=~\frac{z+5}{6}}$

Solved example 27.11
The Cartesian equation of a line is
$\small{\frac{x+3}{2}~=~\frac{y-5}{4}~=~\frac{x+6}{2}}$
Write its vector form
Solution:
1. The Cartesian form of a line passing through $\small{U\left(x_1,y_1,z_1 \right)}$, and is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$ is:
$\small{\frac{x-x_1}{v_1}~=~\frac{y-y_1}{v_2}~=~\frac{z-z_1}{v_3}}$

2. $\small{v_2,~v_3,~v_4}$ can be taken as the direction ratios of any line parallel to $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$

3. So in our present problem, the required line $\small{L}$ must be parallel to the vector
$\small{\vec{v}=2 \hat{i}+4 \hat{j}+2 \hat{k}}$

4. The given Cartesian form is:
$\small{\frac{x+3}{2}~=~\frac{y-5}{4}~=~\frac{x+6}{2}}$
• So the given line passes through $\small{U(-3,5,-6)}$
• So the position vector of $\small{U}$ is:
$\small{\vec{u}=-3\hat{i}+5\hat{j}-6\hat{k}}$

5. The general vector form is: $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$
• So for the present problem, we get:
$\small{\vec{r}=-3\hat{i}+5\hat{j}-6\hat{k}+\lambda \left(2 \hat{i}+4 \hat{j}+2 \hat{k} \right)}$

Easy method:
After becoming familiar with the above five steps, the reader may use the easy method. It can be written in three steps:
1. From the numerators of the given Cartesian form, we get the coordinates of U. These coordinates are the scalar components of $\small{\vec{u}}$.
• So we get: $\small{\vec{u}=-3\hat{i}+5\hat{j}-6\hat{k}}$
2. From the denominators of the given Cartesian form, we get the scalar components of $\small{\vec{v}}$.
• So we get: $\small{\vec{v}=2 \hat{i}+4 \hat{j}+2 \hat{k}}$
3. Now we can easily write $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$

Solved example 27.12
The Cartesian equation of a line is
$\small{\frac{x-5}{2}~=~\frac{y+4}{4}~=~\frac{x-6}{2}}$
Write its vector form
Solution:
1. The Cartesian form of a line passing through $\small{U\left(x_1,y_1,z_1 \right)}$, and is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$ is:
$\small{\frac{x-x_1}{v_1}~=~\frac{y-y_1}{v_2}~=~\frac{z-z_1}{v_3}}$

2. $\small{v_2,~v_3,~v_4}$ can be taken as the direction ratios of any line parallel to $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$

3. So in our present problem, the required line $\small{L}$ must be parallel to the vector
$\small{\vec{v}=3 \hat{i}+7 \hat{j}+2 \hat{k}}$

4. The given Cartesian form is:
$\small{\frac{x-5}{2}~=~\frac{y+4}{4}~=~\frac{x-6}{2}}$
• So the given line passes through $\small{U(5,-4,6)}$
• So the position vector of $\small{U}$ is:
$\small{\vec{u}=5\hat{i}-4\hat{j}+6\hat{k}}$

5. The general vector form is: $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$
• So for the present problem, we get:
$\small{\vec{r}=5\hat{i}-4\hat{j}+6\hat{k}+\lambda \left(3 \hat{i}+7 \hat{j}+2 \hat{k} \right)}$

Easy method:
After becoming familiar with the above five steps, the reader may use the easy method. It can be written in three steps:
1. From the numerators of the given Cartesian form, we get the coordinates of U. These coordinates are the scalar components of $\small{\vec{u}}$.
• So we get: $\small{\vec{u}=5\hat{i}-4\hat{j}+6\hat{k}}$
2. From the denominators of the given Cartesian form, we get the scalar components of $\small{\vec{v}}$.
• So we get: $\small{\vec{v}=3 \hat{i}+7 \hat{j}+2 \hat{k}}$
3. Now we can easily write $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$

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In the next section, we will see equation of a line passing through two given points.

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27.1 - Direction Ratios of Two Parallel Lines

In the previous section, we saw the basic details about direction cosines and direction ratios. We saw the direction cosines of the line through two points P and Q. In this section, we will see more details about direction ratios.

Direction ratios of a line passing through two points

This can be explained in 6 steps:
1. Let $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ be two points in space.
2. We obtained the direction cosines of the line passing through P and Q:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l~=~\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{m~=~\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{n~=~\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
3. If we can find any three numbers a, b and c which satisfies the condition
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}}$, then we can say that a, b and c are the direction ratios of the line through P and Q
4. Let:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a}    & {~=~}    &{x_2 - x_1}
\\ {~\color{magenta}    2    }    &{}    &{b}    & {~=~}    &{y_2 - y_1}
\\ {~\color{magenta}    3    }    &{}    &{c}    & {~=~}    &{z_2 - z_1}
\\ \end{array}}$
5. We will check whether this a, b and c, satisfy the condition written in (3):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{l}{a}}& {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|\left(x_2 - x_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{m}{b}}& {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|\left(y2 - y_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{n}{c}}& {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|\left(z_2 - z_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
• The condition is satisfied.
• So $\small{\left(x_2 - x_1 \right),~\left(y_2 - y_1 \right),~\left(z_2 - z_1 \right)}$ are indeed direction ratios of the line through $\small{(x_1,x_2,x_3)~\text{and}~(x_1,x_2,x_3)}$
6. Note that, $\small{\left(x_1 - x_2 \right),~\left(y_1 - y_2 \right),~\left(z_1 - z_2 \right)}$ are also direction ratios of the line through $\small{(x_1,x_2,x_3)~\text{and}~(x_1,x_2,x_3)}$.
• This is shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{l}{a}}& {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|\left(x_1 - x_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{m}{b}}& {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|\left(y1 - y_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{n}{c}}& {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|\left(z_1 - z_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

Lines having proportional direction ratios

This can be explained in 10 steps:
1. Consider two directed lines L₁ and L₂
• L₁ has:
    ♦ direction cosines $\small{l_1,~m_1,~n_1}$
    ♦ direction ratios $\small{a_1,~b_1,~c_1}$
• L₂ has:
    ♦ direction cosines $\small{l_2,~m_2,~n_2}$
    ♦ direction ratios $\small{a_2,~b_2,~c_2}$

2. For L₁, we can write: $\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}}$

• Since the three fractions are equal, we can write:
$\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}~=~\lambda_1}$

3. For L₂, we can write: $\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}}$

• Since the three fractions are equal, we can write:
$\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}~=~\lambda_2}$

4. Suppose that, the direction ratios of the two lines satisfy the condition:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

• Since the three fractions are equal, we can write:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}~=~\lambda_3}$

5. Substituting from (4) into (2), we get:
$\small{\frac{l_1}{a_2 \lambda_3}~=~\frac{m_1}{b_2 \lambda_3}~=~\frac{n_1}{c_2 \lambda_3}~=~\lambda_1}$

6. Substituting from (3) into (5), we get:
$\small{\frac{l_1}{\left(\frac{l_2}{\lambda_2} \right) \lambda_3}~=~\frac{m_1}{\left(\frac{m_2}{\lambda_2} \right) \lambda_3}~=~\frac{n_1}{\left(\frac{n_2}{\lambda_2} \right) \lambda_3}~=~\lambda_1}$

$\small{\Rightarrow \frac{l_1 \lambda_2}{l_2 \lambda_3}~=~\frac{m_1 \lambda_2}{m_2 \lambda_3}~=~\frac{n_1 \lambda_2}{n_2 \lambda_3}~=~\lambda_1}$

• Multiplying throughout by $\small{\left(\frac{\lambda_3}{\lambda_2} \right)}$, we get:

$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}~=~\lambda_1\left(\frac{\lambda_3}{\lambda_2} \right)~=~\lambda_4}$

• Note that $\small{\lambda_1,~\lambda_2}$ . . . etc., are just some real numbers. Multiplying them will give new real numbers.

7. So we get an important result:
When direction ratios of two lines satisfy the condition $\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}~=~\lambda_3}$,
their direction ratios will satisfy the condition
$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}~=~\lambda_4}$  

We can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{b_1}{b_2}}    & {~=~}    &{\frac{c_1}{c_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{l_1}{l_2}}& {~=~}    &{\frac{m_1}{m_2}}    & {~=~}    &{\frac{n_1}{n_2}}
\\ \end{array}}$  

8. We have the direction cosines of L₁ and L₂. So we can write unit vectors parallel to them. We get:

$\small{\hat{L_1}= l_1\hat{i}+m_1\hat{j}+n_1\hat{k}}$

$\small{\hat{L_2}= l_2\hat{i}+m_2\hat{j}+n_2\hat{k}}$

9. Substituting from (7) into (8), we get:
$\small{\hat{L_1}= \left(l_2 \lambda_4 \right)\hat{i}+\left(m_2 \lambda_4 \right)\hat{j}+\left(n_2 \lambda_4 \right)\hat{k}}$

$\small{\Rightarrow \hat{L_1}= \lambda_4\left(l_2\hat{i}+m_2\hat{j}+n_2\hat{k} \right)}$

$\small{\Rightarrow \text{Unit vectors}~\hat{L_1}~\text{and}~\hat{L_2}~\text{are parallel}}$

$\small{\Rightarrow \text{Lines}~L_1~\text{and}~L_2~\text{are parallel}}$

10. So we can complete the result in (7). We get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{b_1}{b_2}}    & {~=~}    &{\frac{c_1}{c_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{l_1}{l_2}}& {~=~}    &{\frac{m_1}{m_2}}    & {~=~}    &{\frac{n_1}{n_2}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{L_1}& {~~\parallel~}    &{L_2}    & {{}}    &{{}}
\\ \end{array}}$  

◼ Remarks:
• 1 (magenta color): This line gives the condition. It indicates that, the direction ratios of the two lines are proportional
• 2 (magenta color): If the condition is satisfied, the direction cosines of the two lines will be proportional.
• 3 (magenta color): If the condition is satisfied, the two lines will be parallel to each other.

   

Solved example 27.6
Show that the points A(2,3,−4), B(1,−2,3) and C(3,8,−11) are collinear.
Solution:
1. First we find the direction ratios of AB:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(1-2) = -1}
\\ {~\color{magenta}    2    }    &{{}}    &{b_1}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(-2-3) = -5}
\\ {~\color{magenta}    3    }    &{{}}    &{c_1}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(3-(-4)) = 7}
\\ \end{array}}$  

2. Next we find the direction ratios of BC:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(3-1) = 2}
\\ {~\color{magenta}    2    }    &{{}}    &{b_2}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(8-(-2)) = 10}
\\ {~\color{magenta}    3    }    &{{}}    &{c_2}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(-11-3) = -14}
\\ \end{array}}$ 

3. Next we calculate the three ratios:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{-1}{2}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{b_1}{b_2}}& {~=~}    &{\frac{-5}{10}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{c_1}{c_2}}& {~=~}    &{\frac{7}{-14}}    & {~=~}    &{-\frac{1}{2}}
\\ \end{array}}$  

4. We see that:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$
• That means, the direction ratios of the two lines AB and BC, are proportional.
• That means, AB and BC are parallel.

5. The two parallel lines AB and BC have one point B in common. So the points A, B and C are collinear.

Solved example 27.7
Show that the points (2,3,4), (−1,−2,1) and (5,8,7) are collinear.
Solution:
1. Let the three points be: A(2,3,4), B(−1,−2,1) and C(5,8,7)
• First we find the direction ratios of AB:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(-1-2) = -3}
\\ {~\color{magenta}    2    }    &{{}}    &{b_1}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(-2-3) = -5}
\\ {~\color{magenta}    3    }    &{{}}    &{c_1}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(1-4) = -3}
\\ \end{array}}$  

2. Next we find the direction ratios of BC:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(5-(-1)) = 6}
\\ {~\color{magenta}    2    }    &{{}}    &{b_2}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(8-(-2)) = 10}
\\ {~\color{magenta}    3    }    &{{}}    &{c_2}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(7-1) = 6}
\\ \end{array}}$ 

3. Next we calculate the three ratios:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{-3}{6}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{b_1}{b_2}}& {~=~}    &{\frac{-5}{10}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{c_1}{c_2}}& {~=~}    &{\frac{-3}{6}}    & {~=~}    &{-\frac{1}{2}}
\\ \end{array}}$  

4. We see that:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$
• That means, the direction ratios of the two lines AB and BC, are proportional.
• That means, AB and BC are parallel.

5. The two parallel lines AB and BC have one point B in common. So the points A, B and C are collinear.

Direction ratios of a line, when the line is parallel to a vector

This can be explained in 5 steps:
1. Consider a line $\small{L}$, which is parallel to a vector:
$\small{\vec{v}=v_1 \hat{i}+v_2\hat{j}+v_3\hat{k}}$

2. We know that the direction cosines of $\small{\vec{v}}$ are:
$\small{\frac{v_1}{\left|\vec{v} \right|},~\frac{v_2}{\left|\vec{v} \right|},~\frac{v_3}{\left|\vec{v} \right|}}$

3. Let $\small{a_L,~b_L,~c_L}$ be the direction ratios of $\small{L}$.
• The direction cosines of $\small{L}$ will be same as in (2) above. This is because, $\small{L~\text{and}~\vec{v}}$ are parallel.
• Then we can write: $\small{\frac{v_1}{\left|\vec{v} \right|\,a_L}~=~\frac{v_2}{\left|\vec{v} \right|\,b_L}~=~\frac{v_3}{\left|\vec{v} \right|\,c_L}}$
• If we multiply all three fractions by $\small{\left|\vec{v} \right|}$, the equality will not change. So we get:
$\small{\frac{v_1}{a_L}~=~\frac{v_2}{b_L}~=~\frac{v_3}{c_L}}$   
• Since the three fractions are equal, we can write:
$\small{\frac{v_1}{a_L}~=~\frac{v_2}{b_L}~=~\frac{v_3}{c_L}~=~\lambda}$

4. Here $\small{\lambda}$ can be any real number.
• When $\small{\lambda = 1}$, we get:
$\small{a_L = v_1,~b_L = v_2,~c_L = v_3}$

5. So we can write:
When the line $\small{L}$ is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2\hat{j}+v_3\hat{k}}$, the numbers $\small{v_1,~v_2,~v_3}$ can be considered as a set of direction ratios of $\small{L}$
• This fact will become useful in the next section where we see lines parallel to a given vector.

6. Always remember that, for a vector (or a directed line),
    ♦ There is only one set of direction cosines
    ♦ There are infinite sets of direction ratios
That is why in step (5) above, we wrote: "a set of direction ratios". We cannot write: "the set of direction ratios" 

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The link below gives a few more miscellaneous examples:

Exercise 27.1

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In the next section, we will see equation of a line in space.

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Chapter 27 - Three Dimensional Geometry

In the previous section, we completed a discussion on vector algebra. In this chapter, we will see Three dimensional geometry.

In class 11, we saw the basic details about three dimensional geometry. See chapter 12. In the present chapter, we will see some advanced topics related to three dimension. In the previous chapter, we saw vectors in three dimensional space. We can use those vectors for our present discussion. This will make the discussion more interesting.

Direction cosines of a line

This can be explained in 5 steps:
1. Consider a directed line L passing through the origin. It makes the following angles with the axes:
    ♦ $\small{\alpha}$ with the +ve direction of the x-axis
    ♦ $\small{\beta}$ with the +ve direction of the y-axis
    ♦ $\small{\gamma}$ with the +ve direction of the z-axis
• Then $\small{\cos \alpha,~\cos \beta~\text{and}~\cos \gamma}$ are the direction cosines of the line L

2. Suppose that, we reverse the direction of L. We will call the new line as L'.
• L' will make the following angles with the axes:
    ♦ $\small{\left( \pi - \alpha \right)}$ with the +ve direction of the x-axis
    ♦ $\small{\left( \pi - \beta \right)}$ with the +ve direction of the y-axis
    ♦ $\small{\left( \pi - \gamma \right)}$ with the +ve direction of the z-axis
• This can be easily proved if L lies in a two dimensional plane. In fig.27.1(a) below, $\small{\alpha~\text{and}~\beta}$ and are shown in white and green colors.

Fig.27.1

In fig.27.1(b), $\small{\alpha~\text{and}~\beta}$ are shown twice, by considering opposite angles. Based on those opposite angles, we can write:
• L' makes the following angles with the axes:
    ♦ $\small{\left( \pi - \alpha \right)}$ with the +ve direction of the x-axis
    ♦ $\small{\left( \pi - \beta \right)}$with the +ve direction of the y-axis
• So in the three dimensional space, $\small{\cos \left( \pi - \alpha \right),~\cos \left( \pi - \beta \right)~\text{and}~\cos \left( \pi - \gamma \right) }$ are the direction cosines of L'.
• We know that:
    ♦ $\small{\cos\left( \pi - \alpha \right)~=~-\cos \alpha}$
    ♦ $\small{\left(\cos \pi - \beta \right)~=~-\cos \beta}$
    ♦ $\small{\left(\cos \pi - \gamma \right)~=~-\cos \gamma}$
• That means, when a directed line is reversed, the direction cosines are also reversed.

3. If we are given a simple line in space, it can be extended in two opposite directions. So a simple line will have two sets of direction cosines.
• While doing problems, we must know which set to use. So we must specify the direction of the line.
• In other words, while doing problems in three dimensional geometry, we deal with directed lines only.

4. For a directed line, there will be a unique set of direction cosines. Those direction cosines are denoted by $\small{l,~m~\text{and}~n}$

5. Suppose that, the given line does not pass through the origin. Then we draw a line through the origin and parallel to the given line. The direction cosines of this new line will be same as those of the given line.

Relation between direction cosines of a line

This can be explained in 9 steps:
1. In fig.27.2 below, a line L is shown. This line does not pass through the origin. So we draw a line L₁ passing through the origin and parallel to L

Fig.27.2

2. Mark any convenient point P(x,y,z) on L₁
• Imagine a vector $\small{\vec{OP}}$ with initial point O and terminal point P. Since the coordinates of P are (x,y,z), we can write:
$\small{\vec{OP}~=~x\hat{i}+y\hat{j}+z\hat{k}}$

3. Based on the above component form, we can write:
    ♦ Projection of $\small{\vec{OP}}$ on the x-axis = x
    ♦ Projection of $\small{\vec{OP}}$ on the y-axis = y
    ♦ Projection of $\small{\vec{OP}}$ on the z-axis = z

4. The line L₁ makes an angle $\small{\alpha}$ with the +ve side of the x-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\alpha}$ with the +ve side of the x-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \alpha = x}$. Recall a similar fig., based on which, we discussed about vectors. (See fig.26.4 of the first section of the previous chapter 26)
• From this, we get:
$\small{\cos \alpha = \frac{x}{\left|\vec{OP} \right|} \Rightarrow l = \frac{x}{\left|\vec{OP} \right|}}$

5. The line L₁ makes an angle $\small{\beta}$ with the +ve side of the y-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\beta}$ with the +ve side of the y-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \beta = y}$.
• From this, we get:
$\small{\cos \beta= \frac{y}{\left|\vec{OP} \right|} \Rightarrow m = \frac{y}{\left|\vec{OP} \right|}}$
(See fig.26.5 of the first section of the previous chapter 26)

6. The line L₁ makes an angle $\small{\gamma}$ with the +ve side of the z-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\gamma}$ with the +ve side of the z-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \gamma = z}$.
• From this, we get:
$\small{\cos \gamma = \frac{z}{\left|\vec{OP} \right|} \Rightarrow n = \frac{z}{\left|\vec{OP} \right|}}$
(See fig.26.6 of the first section of the previous chapter 26)

7. In the above steps (4), (5) and (6), we made use of our knowledge about vectors. Thereby, we obtained the direction cosines of L₁.
• We can write:
To obtain the direction cosines of L₁, we require three items:
    ♦ Any convenient point P on L₁
    ♦ The coordinates of P
    ♦ The length OP

8. Our next task is to find $\small{\left|\vec{OP} \right|}$
• This is the magnitude of the vector $\small{\vec{OP}}$.
• So it is simply, the distance between O and P.
• We have the coordinates of both O and P. So the distance between them is $\small{\sqrt{x^2 + y^2 + z^2}}$
That is., $\small{\left|\vec{OP} \right| = \sqrt{x^2 + y^2 + z^2}}$

9. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l^2 + m^2 + n^2}    & {~=~}    &{\frac{x^2}{\left|\vec{OP} \right|^2}~+~\frac{y^2}{\left|\vec{OP} \right|^2}~+~\frac{z^2}{\left|\vec{OP} \right|^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{x^2 ~+~y^2~+~z^2}{\left|\vec{OP} \right|^2}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{x^2 ~+~y^2~+~z^2}{x^2 ~+~y^2~+~z^2}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{1}
\\ \end{array}}$

Direction Ratios

This can be explained in 6 steps:
1. Consider a line with direction cosines l, m and n
2. Suppose that, there exist three numbers a, b and c which satisfy the following condition:
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}}$
3. If the above condition is satisfied, then we say that:
a, b and c are direction ratios of the line.

4. In step (2), there are three fractions. Since they are equal, we can write:
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}~=~k}$, where k is a real number.
• So we get:
$\small{l=ak,~m=bk~\text{and}~n=ck}$

5. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l^2 + m^2 + n^2}    & {~=~}    &{a^2 k^2~+~b^2 k^2~+~c^2 k^2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{1}    & {~=~}    &{k^2(a^2~+~b^2~+~c^2)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{k^2}    & {~=~}    &{\frac{1}{a^2 ~+~b^2~+~c^2}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{k}    & {~=~}    &{\pm\frac{1}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ \end{array}}$

◼ Remarks:
• 2 (magenta color): Here we use the fact that, $\small{l^2 + m^2 + n^2\,=\,1}$

6. So based on step (4), we get the direction cosines:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l}    & {~=~}    &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ {~\color{magenta}    2    }    &{{}}    &{m}    & {~=~}    &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ {~\color{magenta}    3    }    &{{}}    &{n}    & {~=~}    &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ \end{array}}$

• We must select the sign carefully. It depends on the direction of the line. This will become more clear when we do some solved examples

Solved example 27.1
If a line makes angles 90°, 60° and 30° with the positive direction of x, y and z axis respectively, find its direction cosines.
Solution:
1. The line makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$  

2. The line makes angle 60° with the positive direction of the y axis. So $\small{\beta = 60^{\circ}}$
• Then the direction cosine $\small{m = \cos 60 = \frac{1}{2}}$ 

3. The line makes angle 30° with the positive direction of the z axis. So $\small{\gamma = 30^{\circ}}$
• Then the direction cosine $\small{n = \cos 30 = \frac{\sqrt{3}}{2}}$

Solved example 27.2
If a line has direction ratios 2, −1, −2, determine its direction cosines.
Solution:
1. We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l}    & {~=~}    &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{2}{\sqrt{2^2 ~+~(-1)^2~+~(-2)^2}}~=~\pm\frac{2}{3}}
\\ {~\color{magenta}    2    }    &{{}}    &{m}    & {~=~}    &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-1}{3}}
\\ {~\color{magenta}    3    }    &{{}}    &{n}    & {~=~}    &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-2}{3}}
\\ \end{array}}$

2. So the two sets of direction cosines are:
$\small{\frac{2}{3},~\frac{-1}{3},~\frac{-2}{3}~\text{and}~\frac{-2}{3},~\frac{1}{3},~\frac{2}{3}}$

3. Depending on the direction of the directed line, we must choose the appropriate set.
• Let there be two points A and B on the line.
• If a person traveling from A to B has the direction cosines $\small{\frac{2}{3},~\frac{-1}{3},~\frac{-2}{3}}$, then a person traveling from B to A will have the direction cosines $\small{\frac{-2}{3},~\frac{1}{3},~\frac{2}{3}}$.  

Solved example 27.3
If a line has direction ratios −18, 12, −4, then what are its direction cosines.
Solution:
1. We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l}    & {~=~}    &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-18}{\sqrt{(-18)^2 ~+~12^2~+~(-4)^2}}~=~\pm\frac{-18}{22}~=~\pm\frac{-9}{11}}
\\ {~\color{magenta}    2    }    &{{}}    &{m}    & {~=~}    &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{12}{22}~=~\pm\frac{6}{11}}
\\ {~\color{magenta}    3    }    &{{}}    &{n}    & {~=~}    &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-4}{22}~=~\pm\frac{-2}{11}}
\\ \end{array}}$

2. So the two sets of direction cosines are:
$\small{\frac{-9}{11},~\frac{6}{11},~\frac{-2}{11}~\text{and}~\frac{9}{11},~\frac{-6}{11},~\frac{2}{11}}$

3. Depending on the direction of the directed line, we must choose the appropriate set.
• Let there be two points A and B on the line.
• If a person traveling from A to B has the direction cosines $\small{\frac{-9}{11},~\frac{6}{11},~\frac{-2}{11}}$, then a person traveling from B to A will have the direction cosines $\small{\frac{9}{11},~\frac{-6}{11},~\frac{2}{11}}$.

Direction cosines of a line passing through two points

This can be explained in 9 steps:
1. In fig.27.3 below, $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ are two points in space.

Fig.27.3

• There can be one and only one line through both P and Q. That line is shown in magenta color. We want the direction cosines of this magenta line.
2. Imagine that, a vector $\small{\vec{PQ}}$ is present between P and Q. Based on our knowledge in vector algebra, we can write the component form of $\small{\vec{PQ}}$:
$\small{\vec{PQ}~=~\left(x_2 - x_1 \right)\hat{i}+\left(y_2 - y_1 \right)\hat{j}+\left(z_2 - z_1 \right)\hat{k}}$
3. Let $\small{\vec{PQ}}$ make angles $\small{\alpha,~\beta~\text{and}~\gamma}$ with the positive side of the x, y and z-axis respectively. Then the magenta line will also be making the same angles with the three axes.
4. $\small{\left(x_2 - x_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the x-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x_2 - x_1}    & {~=~}    &{\left|\vec{PQ} \right|\cos \alpha}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

5. $\small{\left(y_2 - y_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the y-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y_2 - y_1}    & {~=~}    &{\left|\vec{PQ} \right|\cos \beta}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

6. $\small{\left(z_2 - z_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the z-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{z_2 - z_1}    & {~=~}    &{\left|\vec{PQ} \right|\cos \gamma}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

7. So we obtained the three direction cosines of $\small{\vec{PQ}}$. Note that, $\small{\left|\vec{PQ} \right|}$ can be easily determined because, we have the coordinates of both P and Q. We get:
$\small{\left|\vec{PQ} \right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}$

8. The magenta line has the same direction cosines as that of $\small{\vec{PQ}}$. So we can compile the result.
• The direction cosines of a line through two points $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ are:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    3    }    &{}    &{\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ \end{array}}$

9. The direction cosines written in (8) above, are the ones for a person traveling from P to Q. If he travels from Q to P, he must use the reversed direction cosines:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \alpha}    & {~=~}    &{\frac{x_1 - x_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{\cos \beta}    & {~=~}    &{\frac{y_1 - y_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    3    }    &{}    &{\cos \gamma}    & {~=~}    &{\frac{z_1 - z_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ \end{array}}$
   

Solved example 27.4
Find the direction cosines of the line passing through the two points (−2,4,−5) and (1,2,3).
Solution:
1. Let the two points be: P(−2,4,−5) and Q(1,2,3).
2. So for a person traveling in the direction P to Q, the direction cosines are:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{1-(-2)}{\sqrt{\left(1-(-2) \right)^2 + \left(2 - 4 \right)^2 + \left(3 - (-5) \right)^2}}~=~\frac{3}{\sqrt{\left(3 \right)^2 + \left(-2 \right)^2 + \left(8 \right)^2}}~=~\frac{3}{\sqrt{77}}}
\\ {~\color{magenta}    2    }    &{}    &{\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{2-4}{\sqrt{77}}~=~\frac{-2}{\sqrt{77}}}
\\ {~\color{magenta}    3    }    &{}    &{\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{3-(-5)}{\sqrt{77}}~=~\frac{8}{\sqrt{77}}}
\\ \end{array}}$

3. Also, for a person traveling in the direction Q to P, the direction cosines are:
$\small{\frac{-3}{\sqrt{77}},~\frac{2}{\sqrt{77}},~\frac{-8}{\sqrt{77}}}$

Solved example 27.5
Find the direction cosines of x, y and z-axis.
Solution:
Part (a): Direction cosines of x-axis
1. The x-axis makes angle 0° with the positive direction of the x axis. So $\small{\alpha = 0^{\circ}}$
• Then the direction cosine $\small{l = \cos 0 = 1}$ 
2. The x-axis makes angle 90° with the positive direction of the y axis. So $\small{\beta = 90^{\circ}}$
• Then the direction cosine $\small{m = \cos 90 = 0}$
3. The x-axis makes angle 90° with the positive direction of the z axis. So $\small{\gamma = 90^{\circ}}$
• Then the direction cosine $\small{n = \cos 90 = 0}$
4. So the direction cosines of the x-axis are: 1,0,0

Part (b): Direction cosines of y-axis
1. The y-axis makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$ 
2. The y-axis makes angle 0° with the positive direction of the y axis. So $\small{\beta = 0^{\circ}}$
• Then the direction cosine $\small{m = \cos 0 = 1}$
3. The y-axis makes angle 90° with the positive direction of the z axis. So $\small{\gamma = 90^{\circ}}$
• Then the direction cosine $\small{n = \cos 90 = 0}$
4. So the direction cosines of the y-axis are: 0,1,0

Part (b): Direction cosines of z-axis
1. The z-axis makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$ 
2. The z-axis makes angle 90° with the positive direction of the y axis. So $\small{\beta = 90^{\circ}}$
• Then the direction cosine $\small{m = \cos 90 = 0}$
3. The y-axis makes angle 0° with the positive direction of the z axis. So $\small{\gamma = 0^{\circ}}$
• Then the direction cosine $\small{n = \cos 0 = 1}$
4. So the direction cosines of the z-axis are: 0,0,1

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In the next section, we will see more details about direction ratios.

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