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Chapter 6: Q15E (page 331)

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

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

Short Answer

Expert verified

The vectors \({\mathop{\rm a}\nolimits} \) and \({\mathop{\rm b}\nolimits} \) are not orthogonal.

Step by step solution

01

Definition of orthogonal vectors

Two vectors \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\) are said to beorthogonal to each other when \({\mathop{\rm u}\nolimits} \cdot v = 0\).

02

Determine which pairs of vectors are orthogonal

Determine whether the pairs of vectors are orthogonal as shown below:

\(\begin{aligned}{c}{\mathop{\rm a}\nolimits} \cdot {\mathop{\rm b}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\\ &= 8\left( { - 2} \right) + \left( { - 5} \right)\left( { - 3} \right)\\ &= - 16 + 15\\ &= - 1 \ne 0\end{aligned}\)

It is observed that the vectors \({\mathop{\rm a}\nolimits} \) and \({\mathop{\rm b}\nolimits} \) are not orthogonal.

Thus, the vectors \({\mathop{\rm a}\nolimits} \) and \({\mathop{\rm b}\nolimits} \) are not orthogonal.

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Most popular questions from this chapter

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

\(\left[ {\begin{align}{ 2}\\{ - 7}\\{-1}\end{align}} \right]\), \(\left[ {\begin{align}{ - 6}\\{ - 3}\\9\end{align}} \right]\), \(\left[ {\begin{align}{ 3}\\{ 1}\\{-1}\end{align}} \right]\)

In Exercises 7鈥�10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]鈥檚, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

10.\[y = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\],\[{{\bf{u}}_3} = \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

Suppose radioactive substance A and B have decay constants of \(.02\) and \(.07\), respectively. If a mixture of these two substances at a time \(t = 0\) contains \({M_A}\) grams of \(A\) and \({M_B}\) grams of \(B\), then a model for the total amount of mixture present at time \(t\) is

\(y = {M_A}{e^{ - .02t}} + {M_B}{e^{ - .07t}}\) (6)

Suppose the initial amounts \({M_A}\) and are unknown, but a scientist is able to measure the total amounts present at several times and records the following points \(\left( {{t_i},{y_i}} \right):\left( {10,21.34} \right),\left( {11,20.68} \right),\left( {12,20.05} \right),\left( {14,18.87} \right)\) and \(\left( {15,18.30} \right)\).

a.Describe a linear model that can be used to estimate \({M_A}\) and \({M_B}\).

b. Find the least-squares curved based on (6).

Suppose \(A = QR\) is a \(QR\) factorization of an \(m \times n\) matrix

A (with linearly independent columns). Partition \(A\) as \(\left[ {\begin{aligned}{{}{}}{{A_1}}&{{A_2}}\end{aligned}} \right]\), where \({A_1}\) has \(p\) columns. Show how to obtain a \(QR\) factorization of \({A_1}\), and explain why your factorization has the appropriate properties.

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 \(R = {Q^T}A\).
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