Question

Find the distance between \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right)\) and \({\mathop{\rm z}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\).

Short answer

The distance between the vectors u and z is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,{\mathop{\rm z}\nolimits} } \right) = 2\sqrt {17} \).

Step-by-step solution

Distance in \({\mathbb{R}^n}\)

Thelength of the vectoris thedistance between u and v in \({\mathbb{R}^n}\), expressed as \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,v} \right)\). Therefore, \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,v} \right) = \left\| {{\mathop{\rm u}\nolimits} - v} \right\|\).

Find the distance between the vectors

Compute \({\mathop{\rm u}\nolimits} - {\mathop{\rm z}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} - {\mathop{\rm z}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{0 + 4}\\{ - 5 + 1}\\{2 - 8}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}4\\{ - 4}\\{ - 6}\end{aligned}} \right)\end{aligned}\)

Compute the distance between the vectors as shown below:

\(\begin{aligned}{c}{\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,{\mathop{\rm z}\nolimits} } \right) &= \left\| {{\mathop{\rm u}\nolimits} - {\mathop{\rm z}\nolimits} } \right\|\\ &= \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2}} \\ &= \sqrt {16 + 16 + 36} \\ &= \sqrt {68} \\ &= 2\sqrt {17} \end{aligned}\)

Thus, the distance between the vectors u and z is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,{\mathop{\rm z}\nolimits} } \right) = 2\sqrt {17} \).