Question

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

8. \(\left\| {\mathop{\rm x}\nolimits} \right\|\)

Short answer

The length of x is \(\left\| {\mathop{\rm x}\nolimits} \right\| = 7\).

Step-by-step solution

The length of a vector

The nonnegative scalar \(\left\| {\mathop{\rm v}\nolimits} \right\|\) is defined aslength (ornorm) as shown below:

\(\left\| {\mathop{\rm v}\nolimits} \right\| = \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } = \sqrt {v_1^2 + v_2^2 + \cdots + v_n^2} ,\)and \({\left\| {\mathop{\rm v}\nolimits} \right\|^2} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} \)

Compute

\(\left\| {\mathop{\rm x}\nolimits} \right\|\)

It is given that \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\).

Compute the length of \({\mathop{\rm x}\nolimits} \) as shown below:

\(\begin{aligned}{c}\left\| {\mathop{\rm x}\nolimits} \right\| &= \sqrt {{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} } \\ &= \sqrt {{6^2} + {{\left( { - 2} \right)}^2} + {3^2}} \\ &= \sqrt {36 + 4 + 9} \\ &= \sqrt {49} \\ &= 7\end{aligned}\)

Thus, the length of x is \(\left\| {\mathop{\rm x}\nolimits} \right\| = 7\).