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 that u - vis the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

4. u and v as in Exercise 2

Short answer

The graph of the vectors is shown below:

Step-by-step solution

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

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

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 \( - v\).

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

Thus, \( - v = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right]\).

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 \( - 2v\).

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

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

Write vectors u + v , and u - v

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

\(\begin{aligned}{c}u + v &= \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + \left( 2 \right)}\\{2 + \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + 2}\\{2 - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\end{aligned}\)

Thus, the vector is \(u + v = \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\).

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

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

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

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 \(u\).

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

Solve further to get:

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

Thus, \(u - 2v = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\).

Display the vectors on a graph

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

Thus, the graph of the vectors is obtained.