Question

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

\(\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right]\), \(\left[ {\begin{align}0\\0\\0\end{align}} \right]\), \(\left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]\)

Short answer

The given set is orthogonal.

Step-by-step solution

Definition of an orthogonal set

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.

Check for orthogonality of vectors

Let the given vectors be, \({u_1} = \left[ {\begin{align}2\\{-5}\\{-3}\end{align}} \right]\), \({u_2} = \left[ {\begin{align}0\\0\\0\end{align}} \right]\) and \({u_3} = \left[ {\begin{align}4\\{ - 2}\\6\end{align}} \right]\).

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

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

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

\(\begin{align}{c}{u_2} \cdot {u_3} = \left( 0 \right)\left( 4 \right) + \left( 0 \right)\left( { - 2} \right) + \left( 0 \right)\left( 6 \right)\\ = 0 - 0 + 0\\ = 0\end{align}\)

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

\(\begin{align}{c}{u_1} \cdot {u_3} = \left( 2 \right)\left( 4 \right) + \left( { - 5} \right)\left( { - 2} \right) + \left( { - 3} \right)\left( 6 \right)\\ = 8 + 10 - 18\\ = 0\end{align}\)

Since all the pairs are orthogonal hence, the given set is orthogonal.