Question

In Exercises 3 and 4, display the following vectors using arrows

on an \(xy\)-graph: u, v, \( - {\bf{v}}\), \( - 2{\bf{v}}\), u + v , u - v, and u - 2v. Notice thatis the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

3. u and v as in Exercise 1

Short answer

The graph of the vectors is shown below:

Step-by-step solution

Write vectors u, v, \( - {\bf{v}}\), and\( - 2{\bf{v}}\)

From Exercise 1, the vectors are \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),and \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\).

Vector \( - v\)can be written as\(\left( { - 1} \right)v\).

Obtain the scalar multiple of vector \(v\)by scalar \(\left( { - 1} \right)\) to compute vector \( - v\).

\(\begin{aligned}{c}\left( { - 1} \right)v &= \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{\left( { - 1} \right)\left( { - 3} \right)}\\{\left( { - 1} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\end{aligned}\)

Thus, \( - v = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\).

Vector \( - 2v\)can be written as\(\left( { - 2} \right)v\).

Obtain the scalar multiple of vector \(v\)by scalar \(\left( { - 2} \right)\) to compute vector \( - 2v\).

\(\begin{aligned}{c}\left( { - 2} \right)v &= \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( { - 3} \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\end{aligned}\)

Thus, \( - 2v = \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\).

Write vectors u + v , and u - v

Obtain the vector \(u + v\)by using vectors \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),and \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\).

\(\begin{aligned}{c}u + v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + \left( { - 3} \right)}\\{2 + \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 - 3}\\{2 - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\end{aligned}\)

Thus, the vector is \(u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\).

Obtain the vector \(u - v\)by using vectors \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),and \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\).

\(\begin{aligned}{c}u - v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 - \left( { - 3} \right)}\\{2 - \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + 3}\\{2 + 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\end{aligned}\)

Thus, the vector is \(u - v = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\).

Compute vector \(u - 2v\)

Vector \(u - 2v\)can be written as\(u + \left( { - 2} \right)v\).

Obtain the scalar multiple of vector \(v\)by scalar \(\left( { - 2} \right)\), and then add the resultant vector with vector \(u\).

\(\begin{aligned}{c}u + \left( { - 2} \right)v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( { - 3} \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + 6}\\{2 + 2}\end{array}} \right]\end{aligned}\)

Solve further to get:

\(u + \left( { - 2} \right)v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\)

Thus, \(u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\).

Display the vectors on a graph

The graph of vectors \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\), \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\), \( - v = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\), \( - 2v = \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\), \(u - v = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\), \(u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\), and \(u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\) using arrows are shown below:

Thus, the graph is obtained.