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Chapter 6: Q16E (page 331)

Determine which pairs of vectors in Exercises \(15 - 18\) are orthogonal.

16. \({\bf{u}} = \left( {\begin{aligned}{12}\\3\\{ - 5}\end{aligned}} \right),\,\,{\bf{v}} = \left( {\begin{aligned}{2}\\{ - 3}\\3\end{aligned}} \right)\)

Short Answer

Expert verified

The vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are orthogonal to each other.

Step by step solution

01

Definition of orthogonal vectors

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if \({\bf{u}} \cdot {\bf{v}} = 0\).

02

Checking Orthogonality for given vectors.

The given vectors are:

\({\bf{u}} = \left( {\begin{aligned}{*{20}{r}}{12}\\3\\{ - 5}\end{aligned}} \right),\,\,{\bf{v}} = \left( {\begin{aligned}{*{20}{r}}2\\{ - 3}\\3\end{aligned}} \right)\)

On having dot products, we get:

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{v}} &= \left( {\begin{aligned}{*{20}{r}}{12}\\3\\{ - 5}\end{aligned}} \right).\left( {\begin{aligned}{*{20}{r}}2\\{ - 3}\\3\end{aligned}} \right)\\ &= \left( {12} \right)\left( 2 \right) + \left( 3 \right)\left( { - 3} \right) + \left( { - 5} \right)\left( 3 \right)\\ &= 24 - 9 - 15\\ &= 0\end{aligned}\)

Since \({\bf{u}} \cdot {\bf{v}} = 0\).

Hence, the vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are orthogonal to each other.

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Most popular questions from this chapter

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\) may be written in the form

\(\begin{aligned}{c}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

Derive the normal equations (7) from the matrix form given in this section.

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set. Verify the following equality by induction, beginning with \(p = 2\). If \({\bf{x}} = {c_1}{{\bf{v}}_1} + \ldots + {c_p}{{\bf{v}}_p}\), then

\({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\)

A certain experiment produces the data \(\left( {1,1.8} \right),\left( {2,2.7} \right),\left( {3,3.4} \right),\left( {4,3.8} \right),\left( {5,3.9} \right)\). Describe the model that produces a least-squares fit of these points by a function of the form

\(y = {\beta _1}x + {\beta _2}{x^2}\)

Such a function might arise, for example, as the revenue from the sale of \(x\) units of a product, when the amount offered for sale affects the price to be set for the product.

a. Give the design matrix, the observation vector, and the unknown parameter vector.

b. Find the associated least-squares curve for the data.

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

3. \(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{ - {\bf{2}}}\\{ - {\bf{1}}}&{\bf{2}}\\{\bf{0}}&{\bf{3}}\\{\bf{2}}&{\bf{5}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{4}}}\\{\bf{2}}\end{aligned}} \right)\)

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)
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