The symmetric form of the equation of a line in 3D is derived from the parametric equations of the line. Suppose the line passes through a point \( (x_1, y_1, z_1) \) and has a direction vector \( \mathbf{d} = (a, b, c) \). The parametric equations of the line are:
\[
x = x_1 + \lambda a, \quad y = y_1 + \lambda b, \quad z = z_1 + \lambda c
\]
where \( \lambda \) is a scalar parameter.
To express the line in symmetric form, we solve each of these parametric equations for \( \lambda \):
\[
\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}
\]
Thus, the symmetric form of the equation of a line is:
\[
\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}
\]
Example:
Let the line pass through the point \( (1, 2, 3) \) with the direction vector \( \mathbf{d} = (4, 5, 6) \). We can substitute these values into the symmetric form.
The parametric equations of the line are:
\[
x = 1 + \lambda \cdot 4, \quad y = 2 + \lambda \cdot 5, \quad z = 3 + \lambda \cdot 6
\]
Now, solving each equation for \( \lambda \):
\[
\frac{x – 1}{4} = \frac{y – 2}{5} = \frac{z – 3}{6}
\]
Thus, the symmetric form of the line is:
\[
\frac{x – 1}{4} = \frac{y – 2}{5} = \frac{z – 3}{6}
\]
This represents the equation of the line passing through \( (1, 2, 3) \) and having a direction vector \( \mathbf{d} = (4, 5, 6) \).
