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In Exercises 21 and 22, find a parametric equation of the line \(M\) through \({\mathop{\rm p}\nolimits} \) and \({\mathop{\rm q}\nolimits} \). [Hint: \(M\) is parallel to the vector \({\mathop{\rm q}\nolimits} - p\). See the figure below.]

22. \({\mathop{\rm p}\nolimits} = \left[ {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right]\), \(q = \left[ {\begin{array}{*{20}{c}}0\\{ - 4}\end{array}} \right]\)

Short Answer

Expert verified

The parametric equation of line \(M\) through \(p\) and \(q\) is \(x = \left[ {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}6\\{ - 7}\end{array}} \right].\)

Step by step solution

01

Observation from the given figure

Line \(M\) passing through \(p\) and \(q\) is parallel to the vector \({\mathop{\rm q}\nolimits} - {\mathop{\rm p}\nolimits} \).

02

Use the given part to solve the vector \(q - p\) 

It is given that\({\mathop{\rm p}\nolimits} = \left[ {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right]\)and\({\mathop{\rm q}\nolimits} = \left[ {\begin{array}{*{20}{c}}0\\{ - 4}\end{array}} \right]\).

\(\begin{array}{c}q - p = \left[ {\begin{array}{*{20}{c}}0\\{ - 4}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{0 + 6}\\{ - 4 - 3}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}6\\{ - 7}\end{array}} \right]\end{array}\)

03

Determine the parametric equation of line \(M\)

It is known that the equation of the linethrough \(p\) parallel to \({\mathop{\rm v}\nolimits} \) is represented by

\(x = {\mathop{\rm p}\nolimits} + t{\mathop{\rm v}\nolimits} \left( {t\,{\mathop{\rm in}\nolimits} \,\mathbb{R}} \right)\). The solution set of \(Ax = {\mathop{\rm b}\nolimits} \) is a line through \({\mathop{\rm p}\nolimits} \) parallel to the solution set \(Ax = 0\).

Write the parametric equation of line \(M\) parallel to vector \({\mathop{\rm q}\nolimits} - p\) as:

\(\begin{array}{c}x = p + t\left( {q - p} \right)\\ = \left[ {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}6\\{ - 7}\end{array}} \right]\end{array}\)

Thus, the parametric equation of line \(M\) through \(p\) and \(q\) is \(x = \left[ {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}6\\{ - 7}\end{array}} \right]\).

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Most popular questions from this chapter

In Exercises 31, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

31. \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\4&{ - 1}&3&{ - 6}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\0&7&{ - 1}&{ - 6}\end{array}} \right]\)

Question: Determine whether the statements that follow are true or false, and justify your answer.

19. There exits a matrix A such thatA[-12]=[357].

In Exercise 19 and 20, choose \(h\) and \(k\) such that the system has

a. no solution

b. unique solution

c. many solutions.

Give separate answers for each part.

19. \(\begin{array}{l}{x_1} + h{x_2} = 2\\4{x_1} + 8{x_2} = k\end{array}\)

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W.[Hint: Row operations are unnecessary.]

In Exercises 6, write a system of equations that is equivalent to the given vector equation.

6. \({x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\)

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