JEE Main Class 11 Syllabus 2025 PDF for Paper 1 and 2

Equation Of A Line In Three Dimensions - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Equations for a Line in Space is considered one of the most asked concept.
15 Questions around this concept.

Solve by difficulty

EASY

The distance of the point (1, −5, 9) from the plane x − y + z = 5 measured along the

line x = y = z is :

$3\sqrt{10}$

$10\sqrt{3}$

$\frac{10}{\sqrt{3}}$

$\frac{20}{3}$

View Solution

A symmetrical form of the line of intersection of the planes $x=ay+b \: \text{ and } \: z=cy+d$ is:

$\frac{x-b}{a}=\frac{y-1}{1}=\frac{z-d}{c}$

$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$

$\frac{x-a}{b}=\frac{y-0}{1}=\frac{z-c}{d}$

$\frac{x-b-a}{b}=\frac{y-1}{0}=\frac{z-d-c}{d}$

View Solution

Concepts Covered - 1

Equations for a Line in Space

Equations for a Line in Space

A line is uniquely determined if

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

Equation of a line through a given point and parallel to a given vector

Let L be a line in space passing through point P(x₀ , y₀ , z₀). Let $\vec{\mathbf b}=a\hat{\mathbf i}+b\hat{\mathbf j}+c\hat{\mathbf k}$ be a vector parallel to L. Then, for any point on line Q(x, y, z), we know that vector PQ is parallel to vector b. Thus, there is a scalar, λ, such that $\overrightarrow{PQ} = \lambda\vec{\mathbf b}$ , which gives,

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\overrightarrow{PQ}=\lambda \vec{\mathbf b}}\\\mathrm{\;\;}\left ( x-x_0 \right )\hat i+\left ( y-y_0 \right )\hat j+\left ( z-z_0 \right )\hat k=\lambda\left ( a\hat i +b\hat j+c\hat k\right )\\\text{Using vector operations, we can rewrite,}\\\\\mathrm{\;\;}\left ( x\hat i+y\hat j+z\hat k \right )-\left ( x_0\hat i+y_0\hat j+z_0\hat k \right )=\lambda\left ( a\hat i +b\hat j+c\hat k\right )\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathrm{\;\;}\left ( x\hat i+y\hat j+z\hat k \right )=\left ( x_0\hat i+y_0\hat j+z_0\hat k \right )+\lambda\left ( a\hat i +b\hat j+c\hat k\right )$

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

Cartesian Form

Vector equation of a line $\vec{\mathbf r}=\vec{\mathbf r}_0+\lambda \vec{\mathbf b}$ 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 λ 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 λ assuming a, b, and c are non-zero, we get a different description of the same line:

$\frac{x-x_{0}}{a}=\lambda,\;\;\;\frac{y-y_{0}}{b}=\lambda,\;\;\;\frac{z-z_{0}}{c}=\lambda$

These are parametric equations of the line. Eliminating the parameter λ from 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 l, m, 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

Let P(x₀ , y₀ , z₀) and Q(x₁, y₁, z₁) be the points on a line, and let $\vec{\mathbf p}=x_0\hat i+y_0\hat j+z_0\hat k\;\text{and}\;\vec{\mathbf q}=x_1\hat i+y_1\hat j+z_1\hat k.$

Let $\vec{\mathbf r}$ 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 $\overrightarrow{PQ}=\left ( x_1-x_0 \right )\hat i+\left ( y_1-y_0 \right )\hat j+\left ( z_1-z_0 \right )\hat k$ as the direction vector, equation $\vec{\mathbf r}=\vec{\mathbf r}_0+\lambda \vec{\mathbf b}$ gives

$\vec{\mathbf r}=\vec{\mathbf p}+\lambda\left (\overrightarrow{PQ} \right )$

$\vec{\mathbf r}=\vec{\mathbf p}\;+\;\lambda\left ( \vec{\mathbf q}-\vec{\mathbf p} \right )$

Cartesian Form

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

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

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{\mathbf r}=\vec{\mathbf p}\;+\;\lambda\left ( \vec{\mathbf q}-\vec{\mathbf p} \right ) \\\hat{x} \hat{i}+y \hat{j}+z \hat{k}=x_{0} \hat{i}+y_{0} \hat{j}+z_{0} \hat{k}+\lambda\left[\left(x_{1}-x_{0}\right) \hat{i}+\left(y_{1}-y_{0}\right) \hat{j}+\left(z_{1}-z_{0}\right) \hat{k}\right]\\\text {Equating the like coefficients of } \hat{i}, \hat{j}, \hat{k}, \text { we get }\\x=x_{0}+\lambda\left(x_{1}-x_{0}\right) ; y=y_{0}+\lambda\left(y_{1}-y_{0}\right) ; z=z_{0}+\lambda\left(z_{1}-z_{0}\right)\\\text {On eliminating } \lambda, \text { we obtain }\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}\frac{x-x_{0}}{x_{1}-x_{0}}=\frac{y-y_{0}}{y_{1}-y_{0}}=\frac{z-z_{0}}{z_{1}-z_{0}}\\\\\text{which is the equation of the line in Cartesian form.}$