Question

In Exercises 19 and 20, find the parametric equation of the line

through a parallel to b.

19. \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right]\)

Short answer

The parametric equations are \({x_1} = - 2 - 5t\) and \({x_2} = 3t\).

Step-by-step solution

Write the general parametric equation of the line

If a line passes through vector\({\bf{a}}\)and is parallel to vector b,then the parametric equation of the line is represented as\({\bf{x}} = {\bf{a}} + t{\bf{b}}\), where\(t\)is a parameter.

Here, \({\bf{x}}\) is represented as \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\).

Substitute the vectors in the parametric equations

Consider the parametric equation \({\bf{x}} = {\bf{a}} + t{\bf{b}}\).

Substitute the vectors\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right]\) in the equation \({\bf{x}} = {\bf{a}} + t{\bf{b}}\)as shown below:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right]\)

Thus, \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right]\).

Equate the vectors

Substitute \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\) in the above equation.

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right]\)

Simplify further.

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 5t}\\{3t}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2 - 5t}\\{3t}\end{array}} \right]\end{array}\)

Obtain the parametric equations of the line

Equate the vectors \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2 - 5t}\\{3t}\end{array}} \right]\) to obtain the parametric equations.

Thus, the parametric equations are \({x_1} = - 2 - 5t\) and \({x_2} = 3t\).