Question

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

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

Short answer

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

Step-by-step solution

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.

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