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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.

  • 16 Questions around this concept.

Solve by difficulty

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

line x = y = z is :

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

Concepts Covered - 1

Equations for a Line in Space

Equations for a Line in Space

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 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.}

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Equations for a Line in Space

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