In the previous section, and in the section before that, we saw the basic details about the equation of a line in 3D space. In this section, we will see the relation between 3D and 2D. Later in this section, we will see angle between two lines in 3D.
First we will see the relation. It can be written in 4 steps:
1. We have seen one type of equation of a line:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}~=~\frac{z-z_1}{c}}$
• Here, the line passes through a point $\small{U\left(x_1,y_1,z_1 \right)}$
2. Suppose that, we are considering the XY-plane only. Then all the points will lie on that plane. There is no z-axis.
• In such a situation, the equation becomes:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}}$
• This can be rearranged as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{x-x_1}{a}} & {~=~} &{\frac{y-y_1}{b}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{\frac{y-y_1}{b}} & {~=~} &{\frac{x-x_1}{a}} \\
{~\color{magenta} 3 } &{\Rightarrow} &{y-y_1} & {~=~} &{\frac{b}{a}\left(x - x_1 \right)} \\
{~\color{magenta} 4 } &{\Rightarrow} &{y-y_1} & {~=~} &{\frac{b}{a}\left(x \right)~-~\frac{b}{a}\left(x_1 \right)} \\
{~\color{magenta} 5 } &{\Rightarrow} &{y} & {~=~} &{\frac{b}{a}\left(x \right)~-~\frac{b}{a}\left(x_1 \right)+y_1} \\
{~\color{magenta} 6 } &{\Rightarrow} &{y} & {~=~} &{\frac{b}{a}\left(x \right)~+~\left[y_1 - \frac{b}{a}\left(x_1 \right) \right]} \\
\end{array}}$
3. Let us examine the terms in the above result:
• There is one term in the L.H.S. It is a variable term.
• In the R.H.S, there are two terms. The first one is a variable term and the second one is a constant term.
4. We are familiar with this type of equations. It is of the form $\small{y = mx + c}$ It is the equation of the line with slope ‘m’ and y-intercept ‘c’.
• So we can write:
♦ In 3D, the real numbers $\small{a,~b~\text{and}~c}$ are the scalars of the parallel vector.
♦ In 2D, the real number $\small{\frac{b}{a}}$ is the slope of the line.
Let us see an example. It can be written in 3 steps:
1. Consider the equation of a line in 2D:
$\small{\frac{x+4}{3}~=~\frac{y-5}{2}}$
• It is clear that, the line passes through (−4,5).
• Also, it has a slope of $\small{\frac{b}{a} = \frac{2}{3}}$
• Then the line will make an angle of $\small{\tan^{-1}\left(\frac{2}{3} \right)~=~33.69^ \circ}$ with the x-axis
• The y-intercept of this line is:
$\small{y_1 - \frac{b}{a}\left(x_1 \right)~=~5 - \frac{2}{3}\left(-4 \right)~=~5 + \frac{8}{3} = 7.67}$
• This line is shown in yellow color in fig.27.8 below:
 |
| Fig.27.8 |
2. Consider another line with the same values of a and b:
$\small{\frac{x-1}{3}~=~\frac{y-4}{2}}$
• It is clear that, the line passes through (1,4).
• Also, it has a slope of $\small{\frac{b}{a} = \frac{2}{3}}$
• Then this line also will make an angle of $\small{\tan^{-1}\left(\frac{2}{3} \right)~=~33.69^ \circ}$ with the x-axis
• The y-intercept of this line is:
$\small{y_1 - \frac{b}{a}\left(x_1 \right)~=~4 - \frac{2}{3}\left(1 \right)~=~4 - \frac{2}{3} = 3.33}$
• This line is shown in red color in fig.27.8 above.
3. Let us form a 2D vector using the above values of a and b. We get:
$\small{\vec{a}=a\hat{i}+b\hat{j}=3\hat{i}+2\hat{j}}$
• It is shown in magenta color in fig.27.8 above. We see that, $\small{\vec{a}}$ is parallel to both the red and yellow lines.
The information obtained from the above discussion can be extended to 3D also. It can be written in 3 steps:
1. We are given a line in 3D:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}~=~\frac{z-z_1}{c}}$
• Then we can write a 3D vector: $\small{a\hat{i}+b\hat{j}+c\hat{k}}$
This vector will be parallel to the given line.
2. The converse can also be written:
We are given a 3D vector: $\small{a\hat{i}+b\hat{j}+c\hat{k}}$
Then we can write a 3D line:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}~=~\frac{z-z_1}{c}}$
• This line will be parallel to the given vector
3. The numbers $\small{x_1,~y_1,~z_1}$ can be calculated by equating all three fractions to any convenient real number.
The above discussion is applicable to the other type of the line also. It can be explained in 3 steps:
1. We have seen the other type of equation of a line:
$\small{\frac{x-x_1}{x_2 - x_1}~=~\frac{y-y_1}{y_2 - y_1}~=~\frac{z-z_1}{z_2 - z_1}}$
• Here, the line passes through two points:
$\small{U\left(x_1,y_1,z_1 \right)~\text{and}~V\left(x_2,y_2,z_2 \right)}$
2. Suppose that, we are considering the XY-plane only. Then all the points will lie on that plane. There is no z-axis.
• In such a situation, the equation becomes:
$\small{\frac{x-x_1}{x_2 - x_1}~=~\frac{y-y_1}{y_2 - y_1}}$
• This can be rearranged as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{x-x_1}{x_2 - x_1}} & {~=~} &{\frac{y-y_1}{y_2 - y_1}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{\frac{y-y_1}{y_2 - y_1}} & {~=~} &{\frac{x-x_1}{x_2 - x_1}} \\
{~\color{magenta} 3 } &{\Rightarrow} &{y-y_1} & {~=~} &{\frac{y_2 - y_1}{x_2 - x_1}\left(x - x_1 \right)} \\
\end{array}}$
3.
We are familiar with this type of equations. It is used in 2D when a line passes through two points $\small{U\left(x_1,y_1 \right)~\text{and}~V\left(x_2,y_2 \right)}$ .
• So we can write:
♦ In 3D, the real numbers $\small{\left(x_2 - x_1 \right)~\text{and}~\left(y_2 - y_1 \right)}$ are the scalars of the parallel vector.
♦ In 2D, the real number $\small{\frac{y_2 - y_1}{x_2 - x_1}}$ is the slope of the line.
Now we will see angle between two lines. It can be explained in steps:
1. In fig.27.9 below, two lines $\small{L_1~\text{and}~L_2}$ pass through the origin O.
 |
| Fig.27.9 |
2. The direction ratios of the two lines are given:
♦ $\small{L_1}$ has the direction ratios $\small{a_1,~b_1~\text{and}~c_1}$
♦ $\small{L_2}$ has the direction ratios $\small{a_2,~b_2~\text{and}~c_2}$
3. We want to find the acute angle $\small{\theta}$ between the two lines.
4. First we mark any two convenient points:
♦ Point $\small{P}$ on $\small{L_1}$
♦ Point $\small{Q}$ on $\small{L_2}$
5. Now we can imagine two vectors:
♦ $\small{\vec{OP}}$ aligned with $\small{L_1}$
♦ $\small{\vec{OQ}}$ aligned with $\small{L_2}$
6. Now we can think about parallel vectors
♦ $\small{\vec{p}=a_1 \hat{i}+b_1 \hat{j}+c_1\hat{k}}$ will be parallel to $\small{\vec{OP}}$
♦ $\small{\vec{q}=a_2 \hat{i}+b_2 \hat{j}+c_2\hat{k}}$ will be parallel to $\small{\vec{OQ}}$
7. So we can write about the angles:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\theta} & {~=~} &{\text{Angle between}~L_1~\text{and}~L_2} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\text{Angle between}~\vec{OP}~\text{and}~\vec{OQ}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\text{Angle between}~\vec{p}~\text{and}~\vec{q}} \\
\end{array}}$
8. We can easily find the angle between $\small{\vec{p}~\text{and}~\vec{q}}$ because, both are in the component form. We can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos\theta} & {~=~} &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{a_1 a_2 ~+~ b_1 b_2~+~c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}} \\
\end{array}}$
Suppose that, the two lines $\small{L_1~\text{and}~L_2}$ do not pass through the origin. In such a situation, we can use another method. It can be written in 4 steps:
1. Imagine a third line $\small{L_1’}$ such that:
♦ $\small{L_1’}$ is parallel to $\small{L_1}$
♦ $\small{L_1’}$ passes through the origin
• Since $\small{L_1’}$ is parallel to $\small{L_1}$, both will have the same direction ratios.
2. Imagine a fourth line $\small{L_2’}$ such that:
♦ $\small{L_2’}$ is parallel to $\small{L_2}$
♦ $\small{L_2’}$ passes through the origin
• Since $\small{L_2’}$ is parallel to $\small{L_2}$, both will have the same direction ratios.
3. Now we can find the angle between $\small{L_1'~\text{and}~L_2'}$.
4. The required angle between $\small{L_1~\text{and}~L_2}$ will be same as the angle between $\small{L_1'~\text{and}~L_2'}$
Now we will see some solved examples
Solved example 27.15
Find the angle between the pair of lines given by
$\small{3\hat{i}+2\hat{j}-4\hat{k}~+~\lambda\left(\hat{i}+2\hat{j}+2\hat{k} \right)}$
and $\small{5\hat{i}-2\hat{j}~+~\mu\left(3\hat{i}+2\hat{j}+6\hat{k} \right)}$.
Solution:
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~\hat{i}+2\hat{j}+2\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~3\hat{i}+2\hat{j}+6\hat{k}}$
• Reason: In vector form, the vector inside braces is a vector parallel to the line.
2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos\theta} & {~=~} &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{(1)(3)~+~(2)(2)~+~(2)(6)}{\sqrt{1^2 + 2^2 + 2^2}\,\sqrt{3^2 + 2^2 + 6^2}}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{19}{\sqrt{9}\,\sqrt{49}}~=~\frac{19}{21}} \\
{~\color{magenta} 4 } &{\Rightarrow} &{\theta} & {~=~} &{\cos^{-1}\left(\frac{19}{21} \right)} \\
\end{array}}$
Solved example 27.16
Find the angle between the following pairs of lines
(i) $\small{2\hat{i}-5\hat{j}+\hat{k}~+~\lambda\left(3\hat{i}+2\hat{j}+6\hat{k} \right)}$ and
$\small{7\hat{i}-6\hat{j}~+~\mu\left(\hat{i}+2\hat{j}+2\hat{k} \right)}$
(ii) $\small{3\hat{i}+\hat{j}-2\hat{k}~+~\lambda\left(\hat{i}-\hat{j}-2\hat{k} \right)}$ and
$\small{2\hat{i}-\hat{j}-56\hat{k}~+~\mu\left(3\hat{i}-5\hat{j}-4\hat{k} \right)}$
Solution:
Part (i):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~3\hat{i}+2\hat{j}+6\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~\hat{i}+2\hat{j}+2\hat{k}}$
• Reason: In vector form, the vector inside braces is a vector parallel to the line.
2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos\theta} & {~=~} &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{(3)(1)~+~(2)(2)~+~(6)(2)}{\sqrt{3^2 + 2^2 + 6^2}\,\sqrt{1^2 + 2^2 + 2^2}}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{19}{\sqrt{49}\,\sqrt{9}}~=~\frac{19}{21}} \\
{~\color{magenta} 4 } &{\Rightarrow} &{\theta} & {~=~} &{\cos^{-1}\left(\frac{19}{21} \right)} \\
\end{array}}$
Part (ii):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~\hat{i}-\hat{j}-2\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~3\hat{i}-5\hat{j}-4\hat{k}}$
• Reason: In vector form, the vector inside braces is a vector parallel to the line.
2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos\theta} & {~=~} &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{(1)(3)~+~(-1)(-5)~+~(-2)(-4)}{\sqrt{1^2 + (-1)^2 + (-2)^2}\,\sqrt{3^2 + (-5)^2 + (-4)^2}}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{16}{\sqrt{6}\,\sqrt{50}}~=~\frac{16}{\sqrt{6}\,(5)\sqrt{2}}~=~\frac{8\sqrt{2} \sqrt{2}}{\sqrt{3}\sqrt{2}\,(5)\sqrt{2}}~=~\frac{8}{5\sqrt{3}}} \\
{~\color{magenta} 4 } &{\Rightarrow} &{\theta} & {~=~} &{\cos^{-1}\left(\frac{8}{5\sqrt{3}} \right)} \\
\end{array}}$
Solved example 27.17
Find the angle between the pair of lines given by
$\small{\frac{x+3}{3}~=~\frac{y-1}{5}~=~\frac{z+3}{4}}$
and $\small{\frac{x+1}{1}~=~\frac{y-4}{1}~=~\frac{z-5}{2}}$.
Solution:
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~3\hat{i}+5\hat{j}+4\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~\hat{i}+\hat{j}+2\hat{k}}$
• Reason: The direction ratios in Cartesian form, can be used as the scalars of a vector parallel to the line.
2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos\theta} & {~=~} &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{(3)(1)~+~(5)(1)~+~(4)(2)}{\sqrt{3^2 + (5)^2 + (4)^2}\,\sqrt{1^2 + (1)^2 + (2)^2}}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{16}{\sqrt{50}\,\sqrt{6}}~=~\frac{16}{(5)\sqrt{2}\,\sqrt{6}}~=~\frac{8\sqrt{2} \sqrt{2}}{(5)\sqrt{2}\,\sqrt{3}\sqrt{2}}~=~\frac{8}{5\sqrt{3}}} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{\frac{8\sqrt{3}}{5\sqrt{3}\,\sqrt{3}}~=~\frac{8\sqrt{3}}{15}} \\
{~\color{magenta} 5 } &{\Rightarrow} &{\theta} & {~=~} &{\cos^{-1}\left(\frac{8}{5\sqrt{3}} \right)} \\
\end{array}}$
Solved example 27.18
Find the angle between the following pairs of lines
(i) $\small{\frac{x-2}{2}~=~\frac{y-1}{5}~=~\frac{z+3}{-3}}$ and
$\small{\frac{x+2}{-1}~=~\frac{y-4}{8}~=~\frac{z-5}{4}}$
(ii) $\small{\frac{x}{2}~=~\frac{y}{2}~=~\frac{z}{1}}$ and
$\small{\frac{x-5}{4}~=~\frac{y-2}{1}~=~\frac{z-3}{8}}$
Solution:
Part (i):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~2\hat{i}+5\hat{j}-3\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~-\hat{i}+8\hat{j}+4\hat{k}}$
• Reason: The direction ratios in Cartesian form, can be used as the scalars of a vector parallel to the line.
2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos\theta} & {~=~} &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{(2)(-1)~+~(5)(8)~+~(-3)(4)}{\sqrt{2^2 + (5)^2 + (-3)^2}\,\sqrt{(-1)^2 + (8)^2 + (4)^2}}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{26}{\sqrt{38}\,\sqrt{81}}~=~\frac{26}{9\sqrt{38}}} \\
{~\color{magenta} 4 } &{\Rightarrow} &{\theta} & {~=~} &{\cos^{-1}\left(\frac{26}{9\sqrt{38}} \right)} \\
\end{array}}$
Part (ii):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~2\hat{i}+2\hat{j}+\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~4\hat{i}+\hat{j}+8\hat{k}}$
• Reason: The direction ratios in Cartesian form, can be used as the scalars of a vector parallel to the line.
2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos\theta} & {~=~} &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}} \\
{~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{(2)(4)~+~(2)(1)~+~(1)(8)}{\sqrt{2^2 + (2)^2 + (1)^2}\,\sqrt{(4)^2 + (1)^2 + (8)^2}}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{18}{\sqrt{9}\,\sqrt{81}}~=~\frac{18}{(3)\,(9)}~=~\frac{2}{3}} \\
{~\color{magenta} 4 } &{\Rightarrow} &{\theta} & {~=~} &{\cos^{-1}\left(\frac{2}{3} \right)} \\
\end{array}}$
In the next section, we will see condition for two lines to be parallel or perpendicular.
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