Question

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

\(\left[ {\begin{array}{*{20}{c}}3\\{-2}\\1\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{-1}\\3\\{-3}\\4\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\8\\7\\0\end{array}} \right]\)

Short answer

The given set is orthogonal.

Step-by-step solution

Definition of an orthogonal set

If \({u_i} \cdot {u_j}\) 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, \({u_1} = \left[ {\begin{align}3\\{ - 2}\\1\\3\end{align}} \right]\), \({u_2} = \left[ {\begin{align}{-1}\\{3}\\{-3}\\4\end{align}} \right]\) and \({u_3} = \left[ {\begin{align}3\\8\\7\\0\end{align}} \right]\).

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

\(\begin{align}{c}{u_1} \cdot {u_2} = \left( 3 \right)\left( { - 1} \right) + \left( { - 2} \right)\left( 3 \right) + \left( 1 \right)\left( { - 3} \right) + \left( 3 \right)\left( 4 \right)\\ = - 3 - 6 - 3 + 12\\ = 0\end{align}\)

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

\(\begin{align}{c}{u_2} \cdot {u_3} = \left( { - 1} \right)\left( 3 \right) + \left( 3 \right)\left( 8 \right) + \left( { - 3} \right)\left( 7 \right) + \left( 4 \right)\left( 0 \right)\\ = - 3 + 24 - 21 + 0\\ = 0\end{align}\)

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

\(\begin{align}{c}{u_1} \cdot {u_3} = \left( 3 \right)\left( 3 \right) + \left( { - 2} \right)\left( 8 \right) + \left( 1 \right)\left( 7 \right) + \left( 3 \right)\left( 0 \right)\\ = 9 - 16 + 7 + 0\\ = 0\end{align}\)

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