Question

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

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

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

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