Question

Compute the quantities in Exercises 1-8 using the vectors

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

4. \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \)

Short answer

The value is \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 1}}{5}}\\{\frac{2}{5}}\end{aligned}} \right)\).

Step-by-step solution

Inner product

Consider \({\mathop{\rm u}\nolimits} ,v,\) and \({\mathop{\rm w}\nolimits} \) as the vectors in \({\mathbb{R}^n}\) and consider \(c\) as the scalar. Then

1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} \)
2. \(\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) \cdot {\mathop{\rm w}\nolimits} = {\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} + {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm w}\nolimits} \)
3. \(\left( {c{\mathop{\rm u}\nolimits} } \right) \cdot {\mathop{\rm v}\nolimits} = c\left( {{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \right) = {\mathop{\rm u}\nolimits} \cdot \left( {c{\mathop{\rm v}\nolimits} } \right)\)
4. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \ge 0\)and \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 0\) if and only if \({\mathop{\rm u}\nolimits} = 0\).

Compute

\(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \)

It is given that \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\).

Evaluate \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \) as shown below:

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

Compute \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{aligned}{c}\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} &= \frac{1}{5}\left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 1}}{5}}\\{\frac{2}{5}}\end{aligned}} \right)\end{aligned}\)

Thus, the value is \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 1}}{5}}\\{\frac{2}{5}}\end{aligned}} \right)\).