Question

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

\(\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]\)

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, \({u_1} = \left[ {\begin{align}{ 5}\\{-4}\\{ 0}\\3\end{align}} \right]\), \({u_2} = \left[ {\begin{align}{ - 4}\\1\\{ - 3}\\8\end{align}} \right]\) and \({u_3} = \left[ {\begin{align}3\\3\\5\\{ - 1}\end{align}} \right]\).

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

\(\begin{align}{c}{u_1} \cdot {u_2} = \left( 5 \right)\left( { - 4} \right) + \left( { - 4} \right)\left( 1 \right) + \left( 0 \right)\left( { - 3} \right) + \left( 3 \right)\left( 8 \right)\\ = - 20 - 4 + 0 + 24\\ = 0\end{align}\)

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

\(\begin{align}{c}{u_2} \cdot {u_3} = \left( { - 4} \right)\left( 3 \right) + \left( 1 \right)\left( 3 \right) + \left( { - 3} \right)\left( 5 \right) + \left( 8 \right)\left( { - 1} \right)\\ = - 12 + 3 - 15 - 8\\ = - 32\end{align}\)

Since \({u_2} \cdot {u_3} \ne 0\), hence, the given set is not orthogonal.