Question

What is the symmetric equation of a line in three dimensional space?

Answer 1

From the question we have been asked to find the equation of a line in three dimensional space. For solving this question we will use the concept of three dimensional geometry. We will use the formulae of symmetric equation of a line with the direction vector passing through a point which is $\dfrac{x-{{x}_{0}}}{a}=\dfrac{y-{{y}_{0}}}{b}=\dfrac{z-{{z}_{0}}}{c}$. Using this we will explain some examples and solve this question briefly. So, our solution will be as follows.

Complete step by step solution:
Generally in geometry that is in three dimensional geometry, the formulae of symmetric equation of a line with the direction vector $=\left( a,b,c \right)$ passing through a point $\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$ will be as follows.
$\Rightarrow \dfrac{x-{{x}_{0}}}{a}=\dfrac{y-{{y}_{0}}}{b}=\dfrac{z-{{z}_{0}}}{c}$
Here the directional vector points can’t be zero, that is $a,b,c$ can’t be zero.
If one of $a,b,c$ is zero; for example, $c=0$, then we can write as follows:
$\Rightarrow \dfrac{x-{{x}_{0}}}{a}=\dfrac{y-{{y}_{0}}}{b}$ and $z={{z}_{0}}$.
If two of $a,b,c$ are zero; for example, $b=c=0$, then we can write as follows.
$y={{y}_{0}},z={{z}_{0}}$
Here there is no restriction on x it can be any value that is it can be any real number.

Note: Students must be very careful in doing the calculations. Students must know the concept of three dimensional geometry very well to solve this question. We should know the formulae $\dfrac{x-{{x}_{0}}}{a}=\dfrac{y-{{y}_{0}}}{b}=\dfrac{z-{{z}_{0}}}{c}$ and the various conditions to solve this question briefly.