Question

Compute the quantities in Exercises 1-8 using the vectors

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

6. \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} \)

Short answer

The value is \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{{30}}{{49}}}\\{\frac{{ - 10}}{{49}}}\\{\frac{{15}}{{49}}}\end{aligned}} \right)\).

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

\(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} \)

It is given that \({\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\).

Evaluate \({\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} \) and \({\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right)\\ &= 6\left( 3 \right) + \left( { - 2} \right)\left( { - 1} \right) + 3\left( { - 5} \right)\\ &= 18 + 2 - 15\\ &= 5\end{aligned}\)

\(\begin{aligned}{c}{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\\ &= {\left( 6 \right)^2} + {\left( { - 2} \right)^2} + {3^2}\\ &= 36 + 4 + 9\\ &= 49\end{aligned}\)

Compute \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} \) as shown below:

\(\begin{aligned}{c}\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} = \frac{5}{{49}}\left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{\frac{{30}}{{49}}}\\{\frac{{ - 10}}{{49}}}\\{\frac{{15}}{{49}}}\end{aligned}} \right)\end{aligned}\)

Thus, the value is \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{{30}}{{49}}}\\{\frac{{ - 10}}{{49}}}\\{\frac{{15}}{{49}}}\end{aligned}} \right)\).