Question

Compute the quantities in Exercises 1-8 using the vectors

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

1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,{\rm{ }}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\)

Short answer

The values are \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 5\), \({\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 8\) and \(\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }} = \frac{8}{5}\).

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

\({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,{\rm{ }}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\)

It is given that, \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\).

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

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

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

\(\begin{aligned}{c}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\\ &4\left( { - 1} \right) + 6\left( 2 \right)\\ &= - 4 + 12\\ &= 8\end{aligned}\)

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

\(\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }} = \frac{8}{5}\)

Thus, the values are \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 5,{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 8\) and \(\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }} = \frac{8}{5}\).