Cartesian Equation of a Line

Here you will learn cartesian equation of line in 3d passing through a fixed point and passing through two points.

Let’s begin –

Cartesian Equation of a Line

The cartesian equations of a straight line passing through a fixed point $(x_1, y_1, z_1)$ having direction ratios proportional to a, b, c is given by

$x – x_1\over a$ = $y – y_1\over b$ = $z – z_1\over c$

Remark 1 : The above form of a line is known as the symmetrical form of a line.

Remark 2 : The parametric equations of the line $x – x_1\over a$ = $y – y_1\over b$ = $z – z_1\over c$ are

x = $x_1 + a\lambda$, y = $y_1 + b\lambda$, z = $z_1 + c\lambda$, where $\lambda$ is the parameter.

Remark 3 : The coordinates of any point on the line $x – x_1\over a$ = $y – y_1\over b$ = $z – z_1\over c$ are

($x_1 + a\lambda$,  $y_1 + b\lambda$, $z_1 + c\lambda$), where $\lambda$ $\in$ R.

Remark 4 : Since the direction cosines of a line are also its direcion ratios. Therefore, equations of a line passing through $(x_1, y_1, z_1)$ and having direction cosines l, m, n are

$x – x_1\over l$ = $y – y_1\over m$ = $z – z_1\over n$

Remark 5 : Since x, y and z-axes passes through the origin and have direction cosines 1, 0, 0; 0, 1, 0 and 0, 0, 1 respectively. Therefore, their equations are

x-axis : $x – 0\over 1$ = $y – 0\over 0$ = $z – 0\over 0$ or, y = 0 and z = 0

y-axis : $x – 0\over 0$ = $y – 0\over 1$ = $z – 0\over 0$ or, x = 0 and z = 0

z-axis : $x – 0\over 0$ = $y – 0\over 0$ = $z – 0\over 1$ or, y = 0 and y = 0

Cartesian Equation of a Line Passing Through Two Points

The Cartesian equation of aline passing through two given points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by

$x – x_1\over x_2 – x_1$ = $y – y_1\over y_2 – y_1$ = $z – z_1\over z – z_1$

Example : Find the cartesian equation of line passing through A(3, 4, -7) and B(1, -1, 6).

Solution : We have, A(3, 4, -7) and B(1, -1, 6)

The cartesian equation of line passing through two points is

$x – x_1\over x_2 – x_1$ = $y – y_1\over y_2 – y_1$ = $z – z_1\over z – z_1$

= $x – 3\over -2$ = $y – 4\over -5$ = $z + 7\over 13$