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### Section 11.2 : Vector Arithmetic

3. Find a unit vector that points in the same direction as $$\vec q = \vec i + 3\vec j + 9\vec k$$.

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Of course, the first step here really should be to check and see if we are lucky enough to actually have a unit vector already. It’s unlikely we do have a unit vector but you never know until you check!

$\left\| {\vec q} \right\| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 9 \right)}^2}} = \sqrt {91}$

Okay, as we pretty much had already guessed, this isn’t a unit vector (its magnitude isn’t one!) but we can use this to help find the answer.

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Recall that all we need to do to turn any vector into a unit vector is divide the vector by its magnitude. Doing that for this vector gives,

$\vec u = \frac{{\vec q}}{{\left\| {\vec q} \right\|}} = \frac{1}{{\sqrt {91} }}\left( {\vec i + 3\vec j + 9\vec k} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{\sqrt {91} }}\vec i + \frac{3}{{\sqrt {91} }}\vec j + \frac{9}{{\sqrt {91} }}\vec k}}$

As a quick check, not really required of course, we can compute a quick magnitude to verify that we do in fact have a unit vector.

$\left\| {\vec u} \right\| = \sqrt {{{\left( {\frac{1}{{\sqrt {91} }}} \right)}^2} + {{\left( {\frac{3}{{\sqrt {91} }}} \right)}^2} + {{\left( {\frac{9}{{\sqrt {91} }}} \right)}^2}} = \sqrt {\frac{{91}}{{91}}} = 1$

So, we do have a unit vector!