Continuous equation of a straight line in the space

Consider the points $A = (2, 1,-2)$ and $B = (1,-2, 3)$, and find the continuous equations of the straight line that goes through $A$ anb $B$.

We will start computing a director vector of the straight line: $$\overrightarrow{AB}=B-A=(1,-2,3)-(2,1,-2)=(-1,-3,5)$$

Therefore, with the director vector and point $A$, we obtain the continuous equation: $$\dfrac{x-2}{-1}=\dfrac{y-1}{-3}=\dfrac{z+2}{5}$$

Continuous equation: $\dfrac{x-2}{-1}=\dfrac{y-1}{-3}=\dfrac{z+2}{5}$

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