Question

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).

Short answer

The distance between the vectors x and y is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) = 5\sqrt 5 \).

Step-by-step solution

Distance in \({\mathbb{R}^n}\)

Thelength of the vector is 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 \({\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{10 + 1}\\{ - 3 + 5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{11}\\{ - 2}\end{aligned}} \right)\end{aligned}\)

Compute the distance between the vectors as shown below:

\(\begin{aligned}{c}{\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) &= \left\| {{\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} } \right\|\\ &= \sqrt {{{\left( {11} \right)}^2} + {{\left( { - 2} \right)}^2}} \\ &= \sqrt {121 + 4} \\ &= \sqrt {125} \\ &= 5\sqrt 5 \end{aligned}\)

Thus, the distance between the vectors x and y is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) = 5\sqrt 5 \).