Question

Determine which pairs of vectors in Exercises \(15 - 18\) are orthogonal.

18. \(y = \left( {\begin{aligned}{ - 3}\\7\\4\\0\end{aligned}} \right),\,\,z = \left( {\begin{aligned}{1}\\{ - 8}\\{15}\\{ - 7}\end{aligned}} \right)\)

Short answer

The vectors \({\bf{y}}{\rm{ and }}{\bf{z}}\) are not orthogonal to each other.

Step-by-step solution

Definition of orthogonal vectors

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if \({\bf{u}} \cdot {\bf{v}} = 0\).

Checking Orthogonality for given vectors.

The given vectors are:

\({\bf{y}} = \left( {\begin{aligned}{*{20}{r}}{ - 3}\\7\\4\\0\end{aligned}} \right),\,\,{\bf{z}} = \left( {\begin{aligned}{*{20}{r}}1\\{ - 8}\\{15}\\{ - 7}\end{aligned}} \right)\)

On having dot products, we get:

\(\begin{aligned}{c}{\bf{y}} \cdot {\bf{z}} = \left( {\begin{aligned}{*{20}{r}}{ - 3}\\7\\4\\0\end{aligned}} \right) \cdot \left( {\begin{aligned}{*{20}{r}}1\\{ - 8}\\{15}\\{ - 7}\end{aligned}} \right)\\ = \left( { - 3} \right)\left( 1 \right) + \left( 7 \right)\left( { - 8} \right) + \left( 4 \right)\left( {15} \right) + \left( 0 \right)\left( { - 7} \right)\\ = - 3 - 56 + 60 + 0\\ = 1\end{aligned}\)

Since \({\bf{y}} \cdot {\bf{z}} \ne 0\).

Hence, the vectors \(y{\rm{ and }}z\) are not orthogonal to each other.