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Chapter 6: Q11E (page 331)

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

11. \(\left( {\begin{aligned}{*{20}{c}}{\frac{7}{4}}\\{\frac{1}{2}}\\1\end{aligned}} \right)\)

Short Answer

Expert verified

The unit vector \({\mathop{\rm u}\nolimits} \) in the direction of v is \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{7}{{\sqrt {69} }}}\\{\frac{2}{{\sqrt {69} }}}\\{\frac{4}{{\sqrt {69} }}}\end{aligned}} \right)\).

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 \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{7}{4}}\\{\frac{1}{2}}\\1\end{aligned}} \right)\).

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

\(\begin{aligned}{c}\left\| {\mathop{\rm v}\nolimits} \right\| &= \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \\ &= \sqrt {{{\left( {\frac{7}{4}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2} + {{\left( 1 \right)}^2}} \\ &= \sqrt {\left( {\frac{{49}}{{16}}} \right) + \left( {\frac{1}{4}} \right) + 1} \\ &= \sqrt {\frac{{49 + 4 + 16}}{{16}}} \\ &= \sqrt {\frac{{69}}{{16}}} \end{aligned}\)

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

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} &= \frac{1}{{\left\| {\mathop{\rm v}\nolimits} \right\|}}{\mathop{\rm v}\nolimits} \\ &= \frac{1}{{\sqrt {\frac{{69}}{{16}}} }}\left( {\begin{aligned}{*{20}{c}}{\frac{7}{4}}\\{\frac{1}{2}}\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{\frac{7}{{\frac{{4\sqrt {69} }}{4}}}}\\{\frac{1}{{\frac{{2\sqrt {69} }}{4}}}}\\{\frac{1}{{\frac{{\sqrt {69} }}{4}}}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{\frac{7}{{\sqrt {69} }}}\\{\frac{2}{{\sqrt {69} }}}\\{\frac{4}{{\sqrt {69} }}}\end{aligned}} \right)\end{aligned}\)

Thus, the unit vector \({\mathop{\rm u}\nolimits} \) in the direction of v is \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{7}{{\sqrt {69} }}}\\{\frac{2}{{\sqrt {69} }}}\\{\frac{4}{{\sqrt {69} }}}\end{aligned}} \right)\).

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Most popular questions from this chapter

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\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Find a \(QR\) factorization of the matrix in Exercise 12.

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a linear transformation that preserves lengths; that is, \(\left\| {T\left( {\bf{x}} \right)} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).

  1. Show that T also preserves orthogonality; that is, \(T\left( {\bf{x}} \right) \cdot T\left( {\bf{y}} \right) = 0\) whenever \({\bf{x}} \cdot {\bf{y}} = 0\).
  2. Show that the standard matrix of T is an orthogonal matrix.

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

10. \(\left( {\begin{aligned}{*{20}{c}}{ - 6}\\4\\{ - 3}\end{aligned}} \right)\)

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).
See all solutions

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