Equation of a Line in Three Dimensions

What is the Equation of a Line?

The equation of a line is an algebraic way to express a line in terms of the coordinates of the points it joins. The equation of a line will always be a linear equation.

A line is uniquely determined if:

1. It passes through a given point and has given direction, or

2. It passes through two given points.

Equation of line in 2D

1) Point Slope Form

Using the Point Slope Form we represent a line using the slope $(m)$ and a point on the line $\left(x_1, y_1\right)$.

$
y-y_1=m\left(x-x_1\right)
$

2) Slope Intercept Form

In Slope Intercept Form we represent a line using the slope $(\mathrm{m})$ and the $y$ intercept $(b)$.

$
y=m x+b
$

3) Intercept Form

Using the Intercept Form we represent a line where it intersects the x -axis at $(a, 0)$ and the $y$-axis at $(0, b)$.

$
x / a+y / b=1
$
4) Normal Form

Represents a line using the angle $(\theta)$ the line makes with the positive $x$ axis and the perpendicular distance (p) from the origin to the line.

$
x \cos \theta+y \sin \theta=p
$

Equation of Line in 3D

The equation of the straight line in 3D requires two points that are located in space. The point in three coordinates is expressed as ( $x, y, z)$.

Equation of a line through a given point and parallel to a given vector in Vector Form

Let $L$ be $a$ line in space passing through point $P\left(x_0, y_0, z_0\right)$. Let $\overrightarrow{\mathbf{b}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$ be a vector parallel to $L$ . Then, for any point on line $\mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{z})$, we know that vector $P Q$ is parallel to vector $b$ . Thus, there is a scalar, $\lambda$, such that $\overrightarrow{P Q}=\lambda \overrightarrow{\mathbf{b}}$, which gives,

$
\begin{gathered}
\overrightarrow{P Q}=\lambda \tilde{\mathbf{b}} \\
\left(x-x_0\right) \hat{i}+\left(y-y_0\right) \hat{j}+\left(z-z_0\right) \hat{k}=\lambda(a \hat{i}+b \hat{j}+c \hat{k})
\end{gathered}
$

Using vector operations, we can rewrite,

$
\begin{aligned}
(x \hat{i}+y \hat{j}+z \hat{k})-\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right) & =\lambda(a \hat{i}+b \hat{j}+c \hat{k}) \\
(x \hat{i}+y \hat{j}+z \hat{k}) & =\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right)+\lambda(a \hat{i}+b \hat{j}+c \hat{k})
\end{aligned}
$

Setting $\overrightarrow{\mathbf{r}}=x \hat{i}+y \hat{j}+z \hat{k}$ and $\overrightarrow{\mathbf{r}}_0=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}$ we now have the vector equation of a line:

$
\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}
$

Equation of a line through a given point and parallel to a given vector in Cartesian Form

The Equation of a line through a given point and parallel to a given vector in Cartesian Form is

$
\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}
$

Derivation

The vector equation of a line $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}$ shows that the following equations are simultaneously true:

$
x=x_0+\lambda a ; y=y_0+\lambda b ; z=z_0+\lambda c
$

If we solve each of these equations for the component variables $x, y$, and $z$, we get a set of equations in which each variable is defined in terms of the parameter $\lambda$ and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:

If we solve each of the equations for $\lambda$ assuming $\mathrm{a}, \mathrm{b}$, and $c$ are non-zero, we get a different description of the same line:

$
\frac{x-x_0}{a}=\lambda, \quad \frac{y-y_0}{b}=\lambda, \quad \frac{z-z_0}{c}=\lambda
$

|These are parametric equations of the line. Eliminating the parameter $\lambda$ from the above equation, we get,

$
\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}
$

This is the Cartesian equation of the line.

NOTE:
If $\mathrm{I}, \mathrm{m}, \mathrm{n}$ are the direction cosines of the line, the equation of the line is

$
\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}
$

Equation of a line passing through two given points in vector form

Let $\overrightarrow{\mathbf{p}}=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k} \text { and } \overrightarrow{\mathbf{q}}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text {. }\\$
$\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})$

Derivation

Let $P\left(x_0, y_0, z_0\right)$ and $Q\left(x_1, y_1, z_1\right)$ be the points on a line, and let $\overrightarrow{\mathbf{p}}=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}$ and $\overrightarrow{\mathbf{q}}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}$

Let $\overrightarrow{\mathbf{r}}$ be the position vector of an arbitrary point $R(x, y, z)$ lying on this line We want to find a vector equation for the line PQ.

Using P as our known point on the line, and $\overrightarrow{P Q}=\left(x_1-x_0\right) \hat{i}+\left(y_1-y_0\right) \hat{j}+\left(z_1-z_0\right) \hat{k}$ as the direction vector, the equation $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}$ gives

$
\begin{aligned}
\overrightarrow{\mathbf{r}} & =\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{P Q}) \\
\overrightarrow{\mathbf{r}} & =\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})
\end{aligned}
$

Equation of a line passing through two given points in Cartesian Form

We can also find parametric equations for the line segment $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})$

As the position vector of point $\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is $\overrightarrow{\mathbf{r}}=(x \hat{i}+y \hat{j}+z \hat{k})$.

Recommended Video Based on Equations for a Line in 3D Space

Solved Examples Based on Equations for a Line in 3D Space

Example 1: Consider the lines $L_1$ and $L_2$ given by

$
L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2} L_2: \frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}
$

line $L_3$ having direction ratios $1,-1,-2$, intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of the line segment $P Q$ is
[JEE MAINS 2023]

Solution

D R's of $P Q$ are $=(2 \lambda-\mu-1, \lambda-2 \mu+1,2 \lambda-3 \mu-1)$ given D.R's are $=(1,-1,-2)$

$
\begin{aligned}
& \frac{2 \lambda-\mu-1}{1}=\frac{\lambda-2 \mu+1}{-1}=\frac{2 \lambda-3 \mu-1}{-2} \\
& \lambda=\mu=3 \\
& \mathrm{P}=(7,6,8) \\
& \mathrm{Q}=(5,8,12) \\
& \mathrm{PQ}=2 \sqrt{6}
\end{aligned}
$

Hence, the answer is $2 \sqrt{6}$

Example 2: Let the line $L$ pass through the point $(0,1,2)$, intersect the line $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$ and be parallel to the plane $2 x+y-3 z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is :
[JEE MAINS 2023]

Solution

$\begin{aligned}
& \overrightarrow{\mathrm{PQ}}=(2 \lambda+1,3 \lambda+1,4 \lambda+1) \\
& \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{n}}=0 \quad \Rightarrow(2 \lambda+1) \cdot(2)+(3 \lambda+1)(1)+(4 \lambda+1)(-3)=0 \\
& \Rightarrow-5 \lambda=0 \\
& \Rightarrow \lambda=0 \\
& Q=(1,2,3) \\
& \mathrm{eq}^{\mathrm{n}} \text { of line } \\
& \frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu \\
& \text { distance of line from }(1,-9,2) \\
& \left(P^{\prime} \mathrm{Q}\right) \cdot(1,1,1)=0 \\
& \Rightarrow[\mu-1, \mu+10, \mu] \cdot[1,1,1]=0 \\
& \Rightarrow \mu-1+\mu+10+\mu=0 \\
& \mu=-3 \\
& Q^{\prime}=(-3,-2,1) \\
& P^{\prime} Q^{\prime}=\sqrt{16+49+9}=\sqrt{74}
\end{aligned}$

Hence, the answer is $\sqrt{74}$

Example 3: Let a line having direction ratios $1,-4,2$ intersect the lines $\frac{\mathrm{x}-7}{3}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}+2}{1}$ and $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}-7}{3}=\frac{\mathrm{z}}{1}$ at the points A and B . Then $(\mathrm{AB})^2$ is equal to $\qquad$ [JEE MAINS 2022]

Solution

Direction ratio of AB:

$
\frac{3 \mu-2 \lambda+7}{1}=\frac{-\mu-3 \lambda-6}{-4}=\frac{\mu-\lambda-2}{2}
$

On solving $\mu=-5$

$
\lambda=-3
$

$
\begin{aligned}
\therefore(\mathrm{AB})^2 & =(-15+6+7)^2+(5+9-6)^2+(-5+3-2)^2 \\
& =4+64+16 \\
& =84
\end{aligned}
$

Hence, the answer is 84

Example 4: The equation of the line through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{x+1}{3}=\frac{z-1}{-2}$ is: [JEE MAINS 2022]

Solution

$\begin{aligned}
& \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r \\
& \Rightarrow P(x, y, z)=(2 r+1,3 r-1,-2 r+1)
\end{aligned}$

Since, $\overrightarrow{Q P} \perp(2 \hat{i}+3 \hat{j}-2 \hat{k})$

$
\begin{aligned}
& \Rightarrow 4 r+2+9 r-6+4 r+2=0 \\
& \Rightarrow \mathrm{r}=\frac{2}{17} \\
& \Rightarrow \mathrm{P}\left(\frac{21}{17}, \frac{-11}{17}, \frac{13}{17}\right) \\
& \Rightarrow \overrightarrow{\mathrm{PQ}}=\frac{21 \hat{\mathrm{i}}-28 \hat{\mathrm{j}}-21 \hat{\mathrm{k}}}{17}
\end{aligned}
$

So, $\overrightarrow{\mathrm{QP}}: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-1}{4}=\frac{\mathrm{z}-2}{3}$
Hence, the answer is $\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

Example 5 : The equation of the line passing through $(-4,3,1)$, parallel to the plane $x+2 y-z-5=0$ and intersecting the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}$ is :

[JEE Mains 2021]
Solution: The normal vector of the plane containing two intersecting lines is parallel to the vector

$
\overrightarrow{v_1}=\left|\begin{array}{ccc}
\hat{i} & \hat{\hat{j}} & \hat{k} \\
3 & 0 & 1 \\
-3 & 2 & -1
\end{array}\right|=-2 \hat{i}+6 \hat{k}
$

$\therefore$ The required line is parallel to the vector

$
\left(\overrightarrow{v_2}\right)=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -1 \\
-2 & 0 & 6
\end{array}\right|=3 \hat{i}-\hat{j}+\hat{k}
$

$\therefore$ The required equation of the line is

$
\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}
$