Angle Between Two Lines in 3D Space

A straight line is a line that connects two points and extends to infinity in both directions. When two straight lines intersect, they form two sets of angles. The intersection results in two acute angles and two obtuse angles. The absolute value of angles is determined by the slopes of intersecting lines. Angle Between Two Lines helps us to find the relationship between two lines.

This Story also Contains

1. Angle Between Two Lines
2. Angle Between Two Lines in Vector Form
3. Angle Between Two Lines in Cartesian Form
4. Formulas for Angle Between Two Lines
5. Condition for Perpendicularly
6. Condition for parallelism
7. Ange between two lines in terms of direction cosines and direction ratios
8. Solved Examples Based on Angle Between Two Lines

In this article, we will cover the concept of Angle Between Two Lines. This topic falls under the broader category of Three Dimensional Geometry, which is a crucial chapter in Class 12 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of eight questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2018, one in 2019, two in 2020, one in 2021, and one in 2022.

Angle Between Two Lines

The intersection of two straight lines forms an angle. For two intersecting lines, there are two types of angles between the lines, the acute angle and the obtuse angle. The angle between two lines can be calculated by knowing the slopes of the two lines, or by knowing the equations of the two lines. The angle between two lines generally gives the acute angle between the two lines.

Angle Between Two Lines in Vector Form

Let the given lines be,
$
\begin{aligned}
& \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}} \\
& \overrightarrow{\mathbf{r}}={\overrightarrow{\mathbf{r}^{\prime}}}_0+\lambda \overrightarrow{\mathbf{b}}^{\prime}
\end{aligned}
$
As equation (i) and equation (ii) are straight lines in the directions of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{b}}^{\prime}$, respectively.
Let $\theta$ be the angle between the vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{b}}^{\prime}$
Using the dot product,

$
\begin{aligned}
\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}^{\prime} & =|\overrightarrow{\mathbf{b}}|\left|\overrightarrow{\mathbf{b}}^{\prime}\right| \cos \theta \\
\Rightarrow \quad \cos \theta & =\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}^{\prime}}}{|\overrightarrow{\mathbf{b}}|\left|\overrightarrow{\mathbf{b}^{\prime}}\right|}
\end{aligned}
$

Angle Between Two Lines in Cartesian Form

The equation of a straight line in cartesian form is

$
\begin{aligned}
& \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \\
& \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}
\end{aligned}
$

Then,
$
\overrightarrow{\mathbf{b}}=a_1 \hat{i}+b_1 \hat{j}+c_1 \hat{k} \quad \text { and } \quad \overrightarrow{\mathbf{b}}^{\prime}=a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k}
$
So that,
$
\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}^{\prime}=a_1 a_2+b_1 b_2+c_1 c_2
$
$
\begin{aligned}
|\overrightarrow{\mathbf{b}}| & =\sqrt{a_1^2+b_1^2+c_1^2}, \quad \text { and } \quad\left|\overrightarrow{\mathbf{b}}^{\prime}\right|=\sqrt{a_2^2+b_2^2+c_2^2} \\
\cos \theta & =\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{aligned}
$

Formulas for Angle Between Two Lines

1) The angle between two lines, of which, one of the lines is $a x+b y+c=0$, and the other line is the $x$-axis, is

$
\theta=\tan ^{-1} \frac{-a}{b}
$

2) The angle between two lines, of which one of the lines is $y=m x+c$ and the other line is the $x$ axis, is $\theta=\tan ^{-1} m$
3) The angle between two lines that are parallel to each other and have equal slopes $\left(m_1=m_2\right)$ is $0^{\circ}$
4) The angle between two lines that are perpendicular to each other and have the product of their slopes equal to $-1\left(m_1 m_2=-1\right)$ is $90^{\circ}$
5) The angle between two lines having slopes $m_1$ and $m_2$ respectively is

$
\theta=\tan ^{-1}\left(\frac{m_1-m_2}{1+m_1 m_2}\right)
$

Condition for Perpendicularly

The lines are perpendicular then $\cos \theta=90^{\circ}$
i.e.
$
\begin{aligned}
& \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}^{\prime}=0 \\
& {\left[\because \quad \cos 90^{\circ}=0\right]} \\
& \Rightarrow \quad a_1 a_2+b_1 b_2+c_1 c_2=0
\end{aligned}
$

Condition for parallelism

$\begin{aligned}
&\text { The lines are parallel then } \overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{b}}^{\prime} \text { for some scalar } \lambda \text {. }\\
&\Rightarrow \quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
\end{aligned}$

Ange between two lines in terms of direction cosines and direction ratios

If two lines having direction ratios $\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1$ and $\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2$ then the angle between them is given by

$
\cos \theta=\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}}
$
If two lines have direction ratios as $\mathrm{l}_1, \mathrm{~m}_1, \mathrm{n}_1$ and $\mathrm{l}_2, \mathrm{~m}_2, \mathrm{n}_2$ then the angle between them is given by$
\cos \theta=l_1 l_2+m_1 m_2+n_1 n_2
$

Recommendation Video Based on Angle Between Two Lines

Solved Examples Based on Angle Between Two Lines

Example 1: If the two lines $l_1: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_2: \frac{x-1}{1}=\frac{2 y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular, then an angle between the lines $l_2$ and $l_3: \frac{1-x}{3}=\frac{2 y-1}{-4}=\frac{z}{4}$ is.
[JEE MAINS 2022]

Solution
$
\begin{aligned}
& l_1: \frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0} \\
& l_2: \frac{x-1}{1}=\frac{y+3 / 2}{\alpha / 2}=\frac{z+5}{2} \\
& l_3: \frac{x-1}{-3}=\frac{y-\frac{1}{2}}{-2}=\frac{z-0}{4}
\end{aligned}
$
$
l_1 \text { perpendicular } \mathrm{l}_2: \Rightarrow \frac{|3-\alpha+0|}{\sqrt{13} \sqrt{1+\frac{\alpha^2}{4}}+4}=0 \Rightarrow \alpha=3
$
Angle between $l_2$ and $l_3$

$
\begin{aligned}
& \cos \theta=\frac{|1 \times(-3)+(-2)(\alpha / 2)+2 \times 4|}{\sqrt{1+4+\frac{\alpha^2}{4}} \sqrt{9+16+4}} \\
& \cos \theta=\frac{|-3-\alpha+8|}{\sqrt{5+\frac{\alpha^2}{4}} \sqrt{29}} \quad \text { put } \alpha=3 \\
& \cos \theta=\frac{2}{\sqrt{\frac{29}{4}} \sqrt{29}}=\frac{4}{29} \\
& \theta=\cos ^{-1}\left(\frac{4}{29}\right) \Rightarrow \theta=\sec ^{-1}\left(\frac{29}{4}\right)
\end{aligned}
$
Hence, the answer is $
\sec ^{-1}\left(\frac{29}{4}\right)$

Example 2: For real numbers $\alpha$ and $\beta \neq 0$, if the point of intersection of the straight lines

$ \frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3} \text { and } \frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}, \\
x+2 y-z=8,$ lies on the plane $ \alpha-\beta$ is equal to
[JEE MAINS 2021]

Solution:
Let the point on the first line be $(p+\alpha, 2 p+1,3 p+1)$ and on the second line be $(q \beta+4,3 q+6,3 q+7)$
$
\begin{aligned}
& p+\alpha=q \beta+4 \\
& 2 p+1=3 q+6 \\
& 3 p+1=3 q+7
\end{aligned}
$
$
\begin{aligned}
& (i i i)-(i i) \Rightarrow p=1 \\
& (i i i) \Rightarrow 4=3 q+7 \Rightarrow q=-1 \\
& (i) \Rightarrow 1+\alpha=-\beta+4 \Rightarrow \alpha+\beta=3
\end{aligned}
$
So, the point of intersection is $(\alpha+1,3,4)$
It lies in a given plane, so

$
\begin{aligned}
& \alpha+1+2 \cdot 3-4=8 \\
\Rightarrow & \alpha=5 \\
\Rightarrow & \beta=-2 \\
\Rightarrow & \alpha-\beta=7
\end{aligned}
$

Hence, the answer is 7

Example 3: If the foot of the perpendicular drawn from the point $(1,0,3)$ on a line passing through $(\alpha, 7,1)$ is $\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$, then $\alpha$ is equal to $\qquad$
[JEE MAINS 2020]

Solution:
Since PQ is perpendicular to L , therefore

$\begin{aligned} & \left(1-\frac{5}{3}\right)\left(\alpha-\frac{5}{3}\right)+\left(-\frac{7}{3}\right)\left(7-\frac{7}{3}\right)+\left(3-\frac{17}{3}\right)\left(1-\frac{17}{3}\right)=0 \\ & \alpha=4\end{aligned}$

Hence, the answer is 4

Example 4: If the lines $x=a y+b, z=c y+d$ and $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$ are perpendicular, then:
[JEE MAINS 2019]

Solution:
Angle between two lines in terms of direction cosines and direction ratios -
(i) If two lines are parallel then

$
\begin{aligned}
& l_1=l_2, m_1=m_2, n_1=n_2 \text { or } \\
& \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
\end{aligned}
$

(ii) if two lines are perpendicular then

$
\begin{aligned}
& l_1 l_2+m_1 m_2+n_1 n_2=0 \text { or } 1 \\
& a_1 a_2+b_1 b_2+c_1 c_2=0
\end{aligned}
$

The equation of lines are

$
\begin{aligned}
& x=a y+b \\
& z=c y+d \\
& \Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}
\end{aligned}
$

and, lines

$
\begin{aligned}
& x=a^{\prime} y+b^{\prime} \\
& y=c^{\prime} z+d^{\prime} \\
& \Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=\frac{y-d^{\prime}}{c^{\prime}}=\frac{z}{1}
\end{aligned}
$
Given that both the lines are perpendicular to the concept

$
a a^{\prime}+c^{\prime}+c=0
$
Hence, the answer is $a a^{\prime}+c^{\prime}+c=0$

Example 5: If the angle between the lines, $x / 2=y / 2=z / 1$ and
$\frac{5-x}{-2}=\frac{7 y-14}{p}=\frac{z-3}{4} \cos ^{(-1)} \frac{2}{3}$, then $p$ is equal to :
[JEE MAINS
2018]

Solution:
$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$

$
\begin{aligned}
& \frac{x-5}{2}=\frac{y-2}{p / 7}=\frac{z-13}{4} \\
& \cos \theta=\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}} \\
& 2 / 3=\frac{4+2 p / 7+4}{3 \times \sqrt{\left(20+\frac{p^2}{49}\right)}} \\
& \sqrt{20+\frac{p^2}{49}}=4+p / 7 \\
& p=7 / 2
\end{aligned}
$

Hence, the answer is 7/2