Question

Determine the unit vector in the direction of <-2, 9>.

Answer 1

Recall that the unit vector in the direction of a vector v≠<0,0> is:

$\vec{u}=\frac{v}{||v||}.$

Notice that:

$||<-2,9>||=\sqrt{(-2)^2+9^2}.$

Simplifying the above result we get:

$||<-2,9>||=\sqrt{4+81}=\sqrt{85}.$

Therefore the unite vector in the direction of <-2,9> is:

$\frac{<-2,9>}{\sqrt{85}}=<-\frac{2}{\sqrt{85}},\frac{9}{\sqrt{85}}>.$

Answer:

$\begin{equation*} <-\frac{2}{\sqrt{85}},\frac{9}{\sqrt{85}}> \end{equation*}$