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"
The link below gives a few more miscellaneous examples:
Exercise 27.1
In the next section, we will see equation of a line in space.
Previous
Contents
Copyright©2026 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment