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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

Expert verified

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

Step by step solution

01

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

02

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

03

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.

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

04

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 \({x_2} = 3t\).

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Most popular questions from this chapter

Let \(u = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\) and \(v = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\). Show that \(\left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) is in Span \(\left\{ {u,v} \right\}\) for all \(h\) and\(k\).

Suppose \(a,b,c,\) and \(d\) are constants such that \(a\) is not zero and the system below is consistent for all possible values of \(f\) and \(g\). What can you say about the numbers \(a,b,c,\) and \(d\)? Justify your answer.

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An important concern in the study of heat transfer is to determine the steady-state temperature distribution of a thin plate when the temperature around the boundary is known. Assume the plate shown in the figure represents a cross section of a metal beam, with negligible heat flow in the direction perpendicular to the plate. Let \({T_1},...,{T_4}\) denote the temperatures at the four interior nodes of the mesh in the figure. The temperature at a node is approximately equal to the average of the four nearest nodes—to the left, above, to the right, and below. For instance,

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\(T\left( x \right) = \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\)

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