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Chapter 6: Q13E (page 331)

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).

Short Answer

Expert verified

The distance between the vectors x and y is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) = 5\sqrt 5 \).

Step by step solution

01

Distance in \({\mathbb{R}^n}\)

Thelength of the vector is thedistance between u and v in \({\mathbb{R}^n}\), expressed as \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,v} \right)\). Therefore, \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,v} \right) = \left\| {{\mathop{\rm u}\nolimits} - v} \right\|\).

02

Find the distance between the vectors

Compute \({\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{10 + 1}\\{ - 3 + 5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{11}\\{ - 2}\end{aligned}} \right)\end{aligned}\)

Compute the distance between the vectors as shown below:

\(\begin{aligned}{c}{\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) &= \left\| {{\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} } \right\|\\ &= \sqrt {{{\left( {11} \right)}^2} + {{\left( { - 2} \right)}^2}} \\ &= \sqrt {121 + 4} \\ &= \sqrt {125} \\ &= 5\sqrt 5 \end{aligned}\)

Thus, the distance between the vectors x and y is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) = 5\sqrt 5 \).

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Most popular questions from this chapter

To measure the take-off performance of an airplane, the horizontal position of the plane was measured every second, from \(t = 0\) to \(t = 12\). The positions (in feet) were: 0, 8.8, 29.9, 62.0, 104.7, 159.1, 222.0, 294.5, 380.4, 471.1, 571.7, 686.8, 809.2.

a. Find the least-squares cubic curve \(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\) for these data.

b. Use the result of part (a) to estimate the velocity of the plane when \(t = 4.5\) seconds.

Suppose the x-coordinates of the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) are in mean deviation form, so that \(\sum {{x_i}} = 0\). Show that if \(X\) is the design matrix for the least-squares line in this case, then \({X^T}X\) is a diagonal matrix.

In Exercises 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

  1. \(\left( {1,0} \right),\left( {2,1} \right),\left( {4,2} \right),\left( {5,3} \right)\)

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram鈥揝chmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

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