Question

Find parametric equations for the line.

The line through \(\left( {{\rm{4, - 1,2}}} \right)\)and\(\left( {{\rm{1,1,5}}} \right)\)

Short answer

The parametric equations for the line is

\(\begin{aligned}{c}{\rm{x = 4 + 3t}}\;\;\;\\{\rm{y = - 1 - 2t}}\;\;\\{\rm{z = 2 - 3t}}\end{aligned}\)

Step-by-step solution

Find parametric equations for the line.

To solve for the line's parametric equation, a parallel vector and a point on the line are needed.

The parallel vector's solution is as follows:

\(\begin{aligned}{c}{\rm{v = }}\langle {\rm{4 - 1, - 1 - 1,2 - 5}}\rangle \\{\rm{ = }}\langle {\rm{3, - 2, - 3}}\rangle \end{aligned}\)

The plane's parametric equation is now: Using the initial point and the acquired parallel vector, the plane's parametric equation is:

\(\begin{aligned}{c}{\rm{x = 4 + 3t}}\;\;\;\\{\rm{y = - 1 - 2t}}\;\;\\{\rm{z = 2 - 3t}}\end{aligned}\)

Result.

Therefore, theparametric equations for the line is

\(\begin{aligned}{c}{\rm{x = 4 + 3t}}\;\;\;\\{\rm{y = - 1 - 2t}}\;\;\\{\rm{z = 2 - 3t}}\end{aligned}\)