In exercises 1-6, determine which sets of vectors are orthogonal.

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

If \[{{\bf{u}}_i} \cdot {{\bf{u}}_j} = 0\] for \[i \ne j\], then the set of vectors \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}\] is said to be orthogonal.

Let the given vectors are, \({u_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\\{ - 3}\end{array}} \right]\), \({u_2} = \left[ {\begin{array}{*{20}{c}}5\\2\\1\end{array}} \right]\) and \({u_3} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\\{ - 7}\end{array}} \right]\).

\(\begin{array}{c}{u_1} \cdot {u_2} = \left( { - 1} \right)\left( 5 \right) + \left( 4 \right)\left( 2 \right) + \left( { - 3} \right)\left( 1 \right)\\ = - 5 + 8 - 3\\ = 0\end{array}\)

First, find \({u_1} \cdot {u_2}\):

Now, find \({u_2} \cdot {u_3}\):

\(\begin{array}{c}{u_2} \cdot {u_3} = \left( 5 \right)\left( 3 \right) + \left( 2 \right)\left( { - 4} \right) + \left( 1 \right)\left( { - 7} \right)\\ = 15 - 8 - 7\\ = 0\end{array}\)

And find \({u_1} \cdot {u_3}\):

\(\begin{array}{c}{u_1} \cdot {u_3} = \left( { - 1} \right)\left( 3 \right) + \left( 4 \right)\left( { - 4} \right) + \left( { - 3} \right)\left( { - 7} \right)\\ = - 3 - 16 + 21\\ = 2\end{array}\)

Since, \({u_1} \cdot {u_3} \ne 0\). Hence, the given set is not orthogonal.