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Q12E

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Found in: Page 331

### 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

# 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 $${\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}\\{\frac{3}{5}}\end{array}} \right)$$.

See the step by step solution

## 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 $${\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}\\{\frac{3}{5}}\end{array}} \right)$$.