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### Section 11.2 : Vector Arithmetic

4. Find a vector that points in the same direction as $$\vec c = \left\langle { - 1,4} \right\rangle$$ with a magnitude of 10.

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At first glance this doesn’t appear to be all that similar to the previous problem. However, it’s actually very similar.

First, let’s check to see what the magnitude of this vector is.

$\left\| {\vec c} \right\| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 4 \right)}^2}} = \sqrt {17}$ Show Step 2

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!