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Chapter 6: Q23E (page 331)

Let \({\bf{u}} = \left( {\begin{aligned}{2}\\{ - 5}\\{ - 1}\end{aligned}} \right)\) and \({\bf{v}} = \left( {\begin{aligned}{ - 7}\\{ - 4}\\6\end{aligned}} \right)\). Compute and compare \({\bf{u}}.{\bf{v}},\,\,{\left\| {\bf{u}} \right\|^2},\,\,{\left\| {\bf{v}} \right\|^2},{\rm{ and }}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\). Do not use the Pythagorean Theorem.

Short Answer

Expert verified

The required values are:

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{v}} &= 0\\{\left\| {\bf{u}} \right\|^2} &= 30\\{\left\| {\bf{v}} \right\|^2} &= 101\\{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} &= 131\end{aligned}\)

On comparing these, it can be observed that the sum of \({\left\| {\bf{u}} \right\|^2}\) and \({\left\| {\bf{v}} \right\|^2}\) is the same as \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\).

Step by step solution

01

 Compute \({\bf{u}} \cdot {\bf{v}},{\left\| {\bf{u}} \right\|^2},\,\,{\left\| {\bf{v}} \right\|^2},{\rm{ and }}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\)

The given vectors are:

\({\bf{u}} = \left( {\begin{aligned}{*{20}{c}}2\\{ - 5}\\{ - 1}\end{aligned}} \right){\rm{ and }}{\bf{v}} = \left( {\begin{aligned}{*{20}{c}}{ - 7}\\{ - 4}\\6\end{aligned}} \right)\)

Find \({\bf{u}} \cdot {\bf{v}}\).

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{v}} &= 2\left( { - 7} \right) + \left( { - 5} \right)\left( { - 4} \right) + \left( { - 1} \right)6\\ &= 0\end{aligned}\)

Find \({\left\| {\bf{u}} \right\|^2}\) and \({\left\| {\bf{v}} \right\|^2}\).

\(\begin{aligned}{c}{\left\| {\bf{u}} \right\|^2} &= {\bf{u}} \cdot {\bf{u}}\\ &= 2\left( 2 \right) + \left( { - 5} \right)\left( { - 5} \right) + \left( { - 1} \right)\left( { - 1} \right)\\ &= 30\end{aligned}\)

\(\begin{aligned}{c}{\left\| {\bf{v}} \right\|^2} &= {\bf{v}} \cdot {\bf{v}}\\ &= \left( { - 7} \right)\left( { - 7} \right) + \left( { - 4} \right)\left( { - 4} \right) + \left( 6 \right)\left( 6 \right)\\ &= 101\end{aligned}\)

Next, find \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\)

\(\begin{aligned}{c}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} &= \left( {{\bf{u}} + {\bf{v}}} \right) \cdot \left( {{\bf{u}} + {\bf{v}}} \right)\\ &= {\left( {2 + \left( { - 7} \right)} \right)^2} + {\left( { - 5 + \left( { - 4} \right)} \right)^2} + {\left( { - 1 + 6} \right)^2}\\ &= 131\end{aligned}\)

02

Make a comparison between obtained values

On comparing the obtained values, it can be observed that the sum of \({\left\| {\bf{u}} \right\|^2}\) and \({\left\| {\bf{v}} \right\|^2}\) is the same as \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\).

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Most popular questions from this chapter

Find an orthonormal basis of the subspace spanned by the vectors in Exercise 3.

A healthy child鈥檚 systolic blood pressure (in millimetres of mercury) and weight (in pounds) are approximately related by the equation

\({\beta _0} + {\beta _1}\ln w = p\)

Use the following experimental data to estimate the systolic blood pressure of healthy child weighing 100 pounds.

\(\begin{array} w&\\ & {44}&{61}&{81}&{113}&{131} \\ \hline {\ln w}&\\vline & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p&\\vline & {91}&{98}&{103}&{110}&{112} \end{array}\)

Use the Gram鈥揝chmidt process as in Example 2 to produce an orthogonal basis for the column space of

\(A = \left( {\begin{aligned}{{}{r}}{ - 10}&{13}&7&{ - 11}\\2&1&{ - 5}&3\\{ - 6}&3&{13}&{ - 3}\\{16}&{ - 16}&{ - 2}&5\\2&1&{ - 5}&{ - 7}\end{aligned}} \right)\)

Exercises 13 and 14, the columns of \(Q\) were obtained by applying the Gram Schmidt process to the columns of \(A\). Find anupper triangular matrix \(R\) such that \(A = QR\). Check your work.

14.\(A = \left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\), \(Q = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}\\{\frac{2}{7}}&{\frac{{ - 4}}{7}}\\{\frac{4}{7}}&{\frac{2}{7}}\end{aligned}} \right)\)

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