Angle Between Two Lines

Q: What is the condition for perpendicularity?

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

Q: What is the condition for Lines to be parallel?

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

Q: What is the Cartesian equation of a line?

The Cartesian equation of a line is given by $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

Q: How do you calculate the angle between two lines?

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}

Q: What is the angle between two lines if there equation is $a_1 a_2+b_1 b_2+c_1 c_2=0$ ?

For perpendicular lines the condition is $a_1 a_2+b_1 b_2+c_1 c_2=0$ so the angle between two lines is $90^{\circ}$.

Angle Between Two Lines in 3D Space

Edited By Komal Miglani | Updated on Feb 14, 2025 11:48 PM IST

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.

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This Story also Contains

Angle Between Two Lines
Angle Between Two Lines in Vector Form
Angle Between Two Lines in Cartesian Form
Formulas for Angle Between Two Lines
Condition for Perpendicularly
Condition for parallelism
Ange between two lines in terms of direction cosines and direction ratios
Solved Examples Based on Angle Between Two Lines

Angle Between Two Lines in 3D Space

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.

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,
$\begin{aligned} \vec{r} = {\vec{r}}_{0} + λ \vec{b} \\ \vec{r} = {\vec{r^{'}}}_{0} + λ {\vec{b}}^{'} \end{aligned}$
As equation (i) and equation (ii) are straight lines in the directions of $\vec{b}$ and ${\vec{b}}^{'}$ , respectively.
Let $θ$ be the angle between the vectors $\vec{b}$ and ${\vec{b}}^{'}$
Using the dot product,

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$\begin{aligned} \vec{b} \cdot {\vec{b}}^{'} & = | \vec{b} | | {\vec{b}}^{'} | \cos θ \\ \Rightarrow \cos θ & = \frac{\vec{b} \cdot \vec{b^{'}}}{| \vec{b} | | \vec{b^{'}} |} \end{aligned}$

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,
$\vec{b} = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k} and {\vec{b}}^{'} = a_{2} \hat{i} + b_{2} \hat{j} + c_{2} \hat{k}$
So that,
$\vec{b} \cdot {\vec{b}}^{'} = a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$
$\begin{aligned} | \vec{b} | & = \sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}, and | {\vec{b}}^{'} | = \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}} \\ \cos θ & = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} \end{aligned}$

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

$θ = \tan^{- 1} \frac{- a}{b}$

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 $θ = \tan^{- 1} m$
3) The angle between two lines that are parallel to each other and have equal slopes $(m_{1} = m_{2})$ is $0^{\circ}$
4) The angle between two lines that are perpendicular to each other and have the product of their slopes equal to $- 1 (m_{1} m_{2} = - 1)$ is $90^{\circ}$
5) The angle between two lines having slopes $m_{1}$ and $m_{2}$ respectively is

$θ = \tan^{- 1} (\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}})$

Condition for Perpendicularly

The lines are perpendicular then $\cos θ = 90^{\circ}$
i.e.
$\begin{aligned} \vec{b} \cdot {\vec{b}}^{'} = 0 \\ [∵ \cos 90^{\circ} = 0] \\ \Rightarrow a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{aligned}$

Condition for parallelism

$\begin{aligned} The lines are parallel then \vec{b} = λ {\vec{b}}^{'} for some scalar λ . \\ \Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{aligned}$

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

If two lines having direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ then the angle between them is given by

$\cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
If two lines have direction ratios as $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ then the angle between them is given by $\cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$

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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 $l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ 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
$\begin{aligned} l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z - 2}{0} \\ l_{2} : \frac{x - 1}{1} = \frac{y + 3 / 2}{α / 2} = \frac{z + 5}{2} \\ l_{3} : \frac{x - 1}{- 3} = \frac{y - \frac{1}{2}}{- 2} = \frac{z - 0}{4} \end{aligned}$
$l_{1} perpendicular l_{2} :\Rightarrow \frac{| 3 - α + 0 |}{\sqrt{13} \sqrt{1 + \frac{α^{2}}{4}} + 4} = 0 \Rightarrow α = 3$
Angle between $l_{2}$ and $l_{3}$

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

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

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

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

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

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 $(α, 7, 1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$ , then $α$ is equal to
[JEE MAINS 2020]

Solution:
Since PQ is perpendicular to L , therefore

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

Hence, the answer is 4

Example 4: If the lines $x = a y + b, z = c y + d$ and $x = a^{'} z + b^{'}, y = c^{'} z + d^{'}$ 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

$\begin{aligned} l_{1} = l_{2}, m_{1} = m_{2}, n_{1} = n_{2} or \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{aligned}$

(ii) if two lines are perpendicular then

$\begin{aligned} l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 or 1 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{aligned}$

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

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

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

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

$\begin{aligned} \frac{x - 5}{2} = \frac{y - 2}{p / 7} = \frac{z - 13}{4} \\ \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} \\ 2 / 3 = \frac{4 + 2 p / 7 + 4}{3 \times \sqrt{(20 + \frac{p^{2}}{49})}} \\ \sqrt{20 + \frac{p^{2}}{49}} = 4 + p / 7 \\ p = 7 / 2 \end{aligned}$

Hence, the answer is 7/2

Frequently Asked Questions (FAQs)

1. What is the condition for perpendicularity?

The lines are perpendicular then $\cos θ = 90^{\circ}$
i.e.
$\vec{b} \cdot {\vec{b}}^{'} = 0$
$[∵ \cos 90^{\circ} = 0]$
$\Rightarrow a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

2. What is the condition for Lines to be parallel?

The lines are parallel then $\vec{b} = λ {\vec{b}}^{'}$ for some scalar $λ$ .

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

3. What is the Cartesian equation of a line?

The Cartesian equation of a line is given by $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$

4. How do you calculate the angle between two lines?

Let $θ$ be the angle between the vectors $\vec{b}$ and $\vec{b^{'}}$
Using the dot product,

$\begin{aligned} \vec{b} \cdot {\vec{b}}^{'} & = | \vec{b} | | {\vec{b}}^{'} | \cos θ \\ \Rightarrow \cos θ & = \frac{\vec{b} \cdot \vec{b^{'}}}{| \vec{b} | | \vec{b^{'}} |} \end{aligned}$