Problems from Continuous equation of a straight line in the space

Development:

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: $${\displaystyle \frac{x-2}{-1}}={\displaystyle \frac{y-1}{-3}}={\displaystyle \frac{z+2}{5}}$$

Solution:

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

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