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Chapter 6: Q9E (page 331)

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

9. \(\left( {\begin{aligned}{*{20}{c}}{ - 30}\\{40}\end{aligned}} \right)\)

Short Answer

Expert verified

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

Step by step solution

01

Definition of a unit vector

Aunit vectoris 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 thenormalizingv, and we describe that u is in the same direction as v.

02

Determine the unit vector in the direction

It is given that \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 30}\\{40}\end{aligned}} \right)\).

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

\(\begin{aligned}{c}\left\| {\mathop{\rm v}\nolimits} \right\| &= \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \\ &= \sqrt {{{\left( { - 30} \right)}^2} + {{40}^2}} \\ &= \sqrt {900 + 1600} \\ &= \sqrt {2500} \\ &= 50\end{aligned}\)

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

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} &= \frac{1}{{\left\| {\mathop{\rm v}\nolimits} \right\|}}{\mathop{\rm v}\nolimits} \\ &= \frac{1}{{50}}\left( {\begin{aligned}{*{20}{c}}{ - 30}\\{40}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 30}}{{50}}}\\{\frac{{40}}{{50}}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 3}}{5}}\\{\frac{4}{5}}\end{aligned}} \right)\end{aligned}\)

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

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Most popular questions from this chapter

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

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

11. \(A = \left( {\begin{aligned}{{}{}}{\bf{4}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{ - {\bf{5}}}&{\bf{1}}\\{\bf{6}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{5}}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{9}}\\{\bf{0}}\\{\bf{0}}\\{\bf{0}}\end{aligned}} \right)\)

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

In Exercises 5 and 6, describe all least squares solutions of the equation \(A{\bf{x}} = {\bf{b}}\).

6.\(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\end{aligned}} \right)\),\({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{7}}\\{\bf{2}}\\{\bf{3}}\\{\bf{6}}\\{\bf{5}}\\{\bf{4}}\end{aligned}} \right)\)

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

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