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Chapter 6: Q14E (page 331)

Find the distance between \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right)\) and \({\mathop{\rm z}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\).

Short Answer

Expert verified

The distance between the vectors u and z is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,{\mathop{\rm z}\nolimits} } \right) = 2\sqrt {17} \).

Step by step solution

01

Distance in \({\mathbb{R}^n}\)

Thelength of the vectoris 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 \({\mathop{\rm u}\nolimits} - {\mathop{\rm z}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} - {\mathop{\rm z}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{0 + 4}\\{ - 5 + 1}\\{2 - 8}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}4\\{ - 4}\\{ - 6}\end{aligned}} \right)\end{aligned}\)

Compute the distance between the vectors as shown below:

\(\begin{aligned}{c}{\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,{\mathop{\rm z}\nolimits} } \right) &= \left\| {{\mathop{\rm u}\nolimits} - {\mathop{\rm z}\nolimits} } \right\|\\ &= \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2}} \\ &= \sqrt {16 + 16 + 36} \\ &= \sqrt {68} \\ &= 2\sqrt {17} \end{aligned}\)

Thus, the distance between the vectors u and z is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,{\mathop{\rm z}\nolimits} } \right) = 2\sqrt {17} \).

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Most popular questions from this chapter

In Exercises 9-12 find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

12. \(A = \left[ {\begin{array}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{ - {\bf{1}}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{1}}&{ - {\bf{1}}}\end{array}} \right]\), \({\bf{b}} = \left( {\begin{array}{{}{}}{\bf{2}}\\{\bf{5}}\\{\bf{6}}\\{\bf{6}}\end{array}} \right)\)

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

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

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

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 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]\)
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