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Elementary Linear Algebra: An Active Learning Approach

Section 2.3 Vector Basics: Homework

Exercises Exercises

Exercise Group.

Let \(\u=\left(\begin{array}{r} 9\\-2\\6 \end{array}\right)\) and \(\v=\left(\begin{array}{r} -8\\8\\-4\end{array}\right)\text{.}\) Calculate the following:
1.
\(\u+\v\)
2.
\(\u-\v\)
3.
\(\|\u\|\)
4.
\(\|3\u\|\)
5.
\(\|\v\|\)
6.
\(\|2\v\|\)
7.
\(\|3\u-2\v\|\)

8.

Graph the vectors \(\left(\begin{array}{c}\theta\cos\theta\\\theta\sin\theta\end{array}\right)\) for \(\displaystyle{\theta=0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\frac{3\pi}{4},\frac{5\pi}{6},\pi}.\) Use a straightedge, and label the axes and the \(x,y-\)coordinates of the terminal point of each vector.

9.

Graph \(\left(\begin{array}{r}12\\-5\end{array}\right)\) and the unit vector parallel to it, labeling the coordinates of the terminal point for each. Use a straightedge and abel your axes.

Exercise Group.

Find a unit vector in the direction of the given vector (clearing all unnecessary radicals).
10.
\(\v=\left(\begin{array}{r}5\\-12 \end{array} \right)\)
11.
\(\v=\left(\begin{array}{r}3\\-4 \end{array} \right)\)
12.
\(\v=\left(\begin{array}{r}3\\4\\12 \end{array} \right)\)
13.
\(\v=\left(\begin{array}{r}-12\\4\\84\\3 \end{array} \right)\)

14.

Find two vectors of length \(5,\) one parallel to and one antiparallel to \(\left(\begin{array}{r}6\\-2\\3\end{array}\right).\)

Exercise Group.

Let \(\u= \left(\begin{array}{r} 3\\-1\end{array}\right)\) and \(\v= \left(\begin{array}{r} -2\\4\end{array}\right)\text{.}\)
15.
Find a linear combination of \(\u\) and \(\v\) which equals \(\left(\begin{array}{r} 7\\-9\end{array}\right)\) or explain why such a linear combination does not exist.
16.
Find a linear combination of \(\u\) and \(\v\) which equals \(\left(\begin{array}{r} 21\\-27\end{array}\right)\) or explain why such a linear combination does not exist.
17.
Find a linear combination of \(\u\) and \(\v\) which equals \(\left(\begin{array}{r} 14\\-9\end{array}\right)\) or explain why such a linear combination does not exist.

Exercise Group.

Let \(\u=\left(\begin{array}{r}-3\\9\end{array}\right)\) and \(\v=\left(\begin{array}{r}1\\-3\end{array}\right)\text{.}\)
18.
Find a linear combination of \(\u\) and \(\v\) which equals \(\left(\begin{array}{r}9\\-27\end{array}\right)\) or explain why such a linear combination does not exist.
19.
Find two distinct linear combinations of \(\u\) and \(\v\) each of which equals \(\left(\begin{array}{r}-5\\15\end{array}\right).\)
20.
Find a linear combination of \(\u\) and \(\v\) which equals \(\left(\begin{array}{r} -7\\2\end{array}\right)\) or explain why such a linear combination does not exist.

Exercise Group.

For each of the given \(\u,\v,\) and \(\w\) find a linear combination of \(\u\) and \(\v\) which equals \(\w\) or show that no such linear combination exists.
21.
\(\u=\left(\begin{array}{r} -2\\5\\-1 \end{array} \right),\,\v=\left(\begin{array}{r} 4\\1\\-6 \end{array} \right),\,\w=\left(\begin{array}{r}2\\17\\-15 \end{array} \right)\)
22.
\(\u=\left(\begin{array}{r} 1\\4\\1 \end{array} \right),\,\v=\left(\begin{array}{r} -3\\4\\2 \end{array} \right),\,\w=\left(\begin{array}{r}7\\12\\2 \end{array} \right)\)
23.
\(\u=\left(\begin{array}{r} -1\\4\\-5 \end{array} \right),\,\v=\left(\begin{array}{r} 3\\1\\-2 \end{array} \right),\,\w=\left(\begin{array}{r}13\\13\\0 \end{array} \right)\)
24.
\(\u=\left(\begin{array}{r} 1\\4\\1 \end{array} \right),\,\v=\left(\begin{array}{r} -3\\4\\2 \end{array} \right),\,\w=\left(\begin{array}{r}9\\4\\0 \end{array} \right)\)

25.

Let \(S=\left\{ \begin{pmatrix} 1\\0\\0\\ \vdots \\ 0 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \\ \vdots \\ 0 \end{pmatrix}, \cdots \begin{pmatrix} 0\\0\\ 0\\\vdots \\1 \end{pmatrix} \right\}\) be the set of canonical unit vectors in \(\R^n\text{.}\) Show that any vector, \(\u=\begin{pmatrix} u_1\\u_2\\ \vdots \\ u_n \end{pmatrix} \in \R^n\) can be written as a linear combination of the vectors in \(S\text{.}\)

28.

Prove that if \(\u,\v\) are parallel or antiparallel,
\begin{equation*} \left\{\left.t\u\,\right|\,t\in\R\right\}\,=\,\left\{\left.t\v\,\right|\,t\in\R\right\}. \end{equation*}
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