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

through a parallel to b.

20. \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\)

Short Answer

Expert verified

The parametric equations are \({x_1} = 3 - 7t\) and \({x_2} = - 4 + 8t\).

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\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\) in the equation \({\bf{x}} = {\bf{a}} + t{\bf{b}}\)as shown below:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\)

Thus, \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\).

03

Equate the vectors

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

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\)

Simplify further.

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 7t}\\{8t}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 - 7t}\\{ - 4 + 8t}\end{array}} \right]\end{array}\)

04

Obtain the parametric equations of the line

Equate the vectors \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 - 7t}\\{ - 4 + 8t}\end{array}} \right]\) to obtain the parametric equations.

Thus, the parametric equations are \({x_1} = 3 - 7t\) and \({x_2} = - 4 + 8t\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose Ais an \(n \times n\) matrix with the property that the equation \(A{\mathop{\rm x}\nolimits} = 0\) has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

Find the general solutions of the systems whose augmented matrices are given

11. \(\left[ {\begin{array}{*{20}{c}}3&{ - 4}&2&0\\{ - 9}&{12}&{ - 6}&0\\{ - 6}&8&{ - 4}&0\end{array}} \right]\).

In Exercise 23 and 24, make each statement True or False. Justify each answer.

24.

a. Any list of five real numbers is a vector in \({\mathbb{R}^5}\).

b. The vector \({\mathop{\rm u}\nolimits} \) results when a vector \({\mathop{\rm u}\nolimits} - v\) is added to the vector \({\mathop{\rm v}\nolimits} \).

c. The weights \({{\mathop{\rm c}\nolimits} _1},...,{c_p}\) in a linear combination \({c_1}{v_1} + \cdot \cdot \cdot + {c_p}{v_p}\) cannot all be zero.

d. When are \({\mathop{\rm u}\nolimits} \) nonzero vectors, Span \(\left\{ {u,v} \right\}\) contains the line through \({\mathop{\rm u}\nolimits} \) and the origin.

e. Asking whether the linear system corresponding to an augmented matrix \(\left[ {\begin{array}{*{20}{c}}{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{2}}}}&{{{\rm{a}}_{\rm{3}}}}&{\rm{b}}\end{array}} \right]\) has a solution amounts to asking whether \({\mathop{\rm b}\nolimits} \) is in Span\(\left\{ {{a_1},{a_2},{a_3}} \right\}\).

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\(k\).

Solve the linear system of equations. You may use technology.

|3x+5y+3z=257X+9y+19z=654X+5y+11z=5|

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free