Question

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

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

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

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