Okay, oddly enough let’s determine a unit vector that points in the same direction. This doesn’t seem all that useful but it’s actually a very good thing to do in this case.

The unit vector is,

\[\vec u = \frac{{\vec c}}{{\left\| {\vec c} \right\|}} = \frac{1}{{\sqrt {17} }}\left\langle { - 1,4} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle { - \frac{1}{{\sqrt {17} }},\frac{4}{{\sqrt {17} }}} \right\rangle }}\]

Now, let’s think about what we did here. We took the original vector and multiplied it by a number, \(\frac{1}{{\sqrt {17} }}\) in this case, to change its magnitude. The result is a new vector, pointing in the same direction as the original vector, and has a new magnitude of one.

So, how can we use this new vector (and the process by which we found it) to get an answer for this problem?

Show Step 3

We know that scalar multiplication can change the magnitude of a vector. We’ve got a vector with magnitude of one that points in the correct direction. To convert this into a vector with magnitude of 10 all we need to do is multiply this new unit vector by 10 to get,

\[\vec v = 10\vec u = 10\left\langle { - \frac{1}{{\sqrt {17} }},\frac{4}{{\sqrt {17} }}} \right\rangle = \left\langle { - \frac{{10}}{{\sqrt {17} }},\frac{{40}}{{\sqrt {17} }}} \right\rangle \]

Now, let’s verify that this does what we want it to do with a quick magnitude computation.

\[\left\| {\vec v} \right\| = \sqrt {{{\left( { - \frac{{10}}{{\sqrt {17} }}} \right)}^2} + {{\left( {\frac{{40}}{{\sqrt {17} }}} \right)}^2}} = \sqrt {\frac{{1700}}{{17}}} = \sqrt {100} = 10\]

So, we do have a vector with magnitude 10 as predicted!