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
\(x – x_1\over x_2 – x_1\) = \(y – y_1\over y_2 – y_1\) = \(z – z_1\over z – z_1\)
= \(x – 3\over -2\) = \(y – 4\over -5\) = \(z + 7\over 13\)
