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Q12E

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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

In Exercises 9-12, find a unit vector in the direction of the given vector.

12. \(\left( {\begin{array}{*{20}{c}}{\frac{8}{3}}\\2\end{array}} \right)\)

The unit vector \({\mathop{\rm u}\nolimits} \) in the direction of v is \({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}\\{\frac{3}{5}}\end{array}} \right)\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of a unit vector

A unit vector is a vector with a length of 1. When dividing a nonzero vector v by its length, namely, multiply by \(\frac{1}{{\left\| {\mathop{\rm v}\nolimits} \right\|}}\), we get a unit vector u since \(\left( {\frac{1}{{\left\| {\mathop{\rm v}\nolimits} \right\|}}} \right)\left\| {\mathop{\rm v}\nolimits} \right\|\) is the length of u. The process of producing u from v is known as the normalizing v, and we describe that u is in the same direction as v.

Step 2: Determine the unit vector in the direction

It is given that \({\mathop{\rm v}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{8}{3}}\\2\end{array}} \right)\).

Compute the length of \({\mathop{\rm v}\nolimits} \) as shown below:

\(\begin{array}{c}\left\| {\mathop{\rm v}\nolimits} \right\| &= \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \\ &= \sqrt {{{\left( {\frac{8}{3}} \right)}^2} + {2^2}} \\ &= \sqrt {\left( {\frac{{64}}{9}} \right) + 4} \\ &= \sqrt {\frac{{64 + 36}}{9}} \\ &= \sqrt {\frac{{100}}{9}} \end{array}\)

Multiply v by \(\frac{1}{{\left\| {\mathop{\rm v}\nolimits} \right\|}}\) to obtain the unit vector \({\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{array}{c}{\mathop{\rm u}\nolimits} = \frac{1}{{\left\| {\mathop{\rm v}\nolimits} \right\|}}{\mathop{\rm v}\nolimits} \\ = \frac{1}{{\sqrt {\frac{{100}}{9}} }}\left( {\begin{array}{*{20}{c}}{\frac{8}{3}}\\2\end{array}} \right)\\ = \frac{1}{{\frac{{10}}{3}}}\left( {\begin{array}{*{20}{c}}{\frac{8}{3}}\\2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{8}{{\frac{{30}}{3}}}}\\{\frac{2}{{\frac{{10}}{3}}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}\\{\frac{3}{5}}\end{array}} \right)\end{array}\)

Thus, the unit vector \({\mathop{\rm u}\nolimits} \) in the direction of v is \({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}\\{\frac{3}{5}}\end{array}} \right)\).

Most popular questions for Math Textbooks

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

a. If \(W = {\rm{span}}\left\{ {{x_1},{x_2},{x_3}} \right\}\) with \({x_1},{x_2},{x_3}\) linearly independent,

and if \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is an orthogonal set in \(W\) , then \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is a basis for \(W\) .

b. If \(x\) is not in a subspace \(W\) , then \(x - {\rm{pro}}{{\rm{j}}_W}x\) is not zero.

c. In a \(QR\) factorization, say \(A = QR\) (when \(A\) has linearly

independent columns), the columns of \(Q\) form an

orthonormal basis for the column space of \(A\).

Compute the least-squares error associated with the least square solution found in Exercise 4.

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

  1. \(\left( {\begin{aligned}{{}{}}3\\0\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}8\\5\\{ - 6}\end{aligned}} \right)\)

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

11. \(\left( {\begin{aligned}{{}{}}1&2&5\\{ - 1}&1&{ - 4}\\{ - 1}&4&{ - 3}\\1&{ - 4}&7\\1&2&1\end{aligned}} \right)\)

In Exercises 13 and 14, find the best approximation to \[{\bf{z}}\] by vectors of the form \[{c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2}\].

13. \[z = \left[ {\begin{aligned}3\\{ - 7}\\2\\3\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}2\\{ - 1}\\{ - 3}\\1\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\]
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