# What Is the Length of a Vector?

The length of a vector $\stackrel{\to }{a}=\phantom{\rule{-0.17em}{0ex}}\left(x,y\right)$ is denoted by an absolute value sign $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}|$, and can be found by using this formula (notice the similarity to the Pythagorean theorem):

Formula

### TheLengthofaVector

 $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}|=\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(x,y\right)|=\sqrt{{x}^{2}+{y}^{2}}$

Example 1

Find the length of the vector $\stackrel{\to }{v}=\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)$.

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|& =\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)|=\sqrt{{3}^{2}+{4}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\sqrt{9+16}=\sqrt{25}=5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|=\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)|=\sqrt{{3}^{2}+{4}^{2}}=\sqrt{9+16}=\sqrt{25}=5$

Example 2

For what values of $t$ does the vector $\phantom{\rule{-0.17em}{0ex}}\left(t,{t}^{2}\right)$ have length $\sqrt{2}$?

In this case, you use the formula as an equation and solve it for $t$. This is how you do it:

$\begin{array}{llll}\hfill \sqrt{2}& =\sqrt{\phantom{\rule{-0.17em}{0ex}}{\left(t\right)}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left({t}^{2}\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\sqrt{{t}^{2}\phantom{\rule{-0.17em}{0ex}}\left(1+{t}^{2}\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =t\sqrt{1+{t}^{2}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ You divide each side of the equation by $t$ and get

You can now substitute back. Then you get this from the first factor: $\begin{array}{lll}\hfill u+2& =0\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\phantom{\rule{0.33em}{0ex}}\\ \hfill {t}^{2}+2& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {t}^{2}& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

From the second factor, you get $\begin{array}{lll}\hfill u-1& =0\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\phantom{\rule{0.33em}{0ex}}\\ \hfill {t}^{2}-1& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {t}^{2}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill t& =±1.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Since ${t}^{2}=-2$ from (1) has no solution, $t=±1$ from (2).