0
$\begingroup$

So I'm given the vector 5i-12j and I need to find a unit vector perpendicular to this line. I know I need to use the dot product in some way, shape, or form, but I just can't figure it out.

Thank you for your help.

$\endgroup$
  • $\begingroup$ If you find $a$ and $b$ such that $5a-12b=0$, then $5i-12j$ will be perpendicular to $ai+bj$. $\endgroup$ – angryavian Oct 13 '16 at 5:56
  • $\begingroup$ so I just have to pick random numbers till I find ones that satisfy that equation and that satisfy the definition of a unit vector? There's no precise math? $\endgroup$ – JeffreyWorley Oct 13 '16 at 6:01
  • $\begingroup$ How about rotating it by $90^\circ$? Try ${1 \over 13 } (12i+5j)$. Or you could try $-90^\circ$. Not much choice here... $\endgroup$ – copper.hat Oct 13 '16 at 6:01
  • $\begingroup$ @JeffreyWorley Ok so easy now $\endgroup$ – hamam_Abdallah Oct 13 '16 at 6:29
1
$\begingroup$

To figure it out do it that way: $$ \vec{v} = \left[\begin{matrix} 5 \\ -12 \end{matrix}\right] \\ \vec{w} = \left[\begin{matrix} a \\ b \end{matrix}\right]\\ \vec{u} = \frac{1}{\sqrt{a^2 + b^2}}\left[\begin{matrix} a \\ b \end{matrix}\right] $$ what you need to do is find $a$ and $b$ to get $\vec{w}$ and then compute that unit vector $\vec{u}$.

By dot product definition, you know that $cos(\theta) = 0$ when $\theta = \frac{\pi}{2}$. Hence, by it's algebric deffition, you'll get that it needs to be 0 when $\theta = \frac{\pi}{2}$.

By dot product computation, you'll get: $$ 5a -12b = 0\\ a = 12, b = 5 \rightarrow 60 - 60 - 0 $$

So your $\vec{w}$ is: $$ \vec{w} = \left[\begin{matrix} 12\\ 5 \end{matrix}\right] $$

Now all you need to do is normalize $\vec{w}$ to get $\vec{u}$, the unit vector perpendicular to $\vec{v}$:

$$ \sqrt{12^2 + 5^2} = \\ \sqrt{144+25} = 13\\ \vec{u} = \frac{1}{13}\vec{w}\\ \vec{u} = \left[\begin{matrix} \frac{12}{13}\\ \frac{5}{13} \end{matrix}\right] $$

$\endgroup$
1
$\begingroup$

given a non zero vector

$\vec{u}=(a,b)=a\vec{i}+b\vec{j}$

the vector $\vec{v}=(b,\color{red}{-}a)$ is perpendicular to $\vec{u}$.

the unit vector is then obtained by dividing by its norm

$||\vec{v}||=\sqrt{a^2+b^2}$.

so, the vector you seek is

$$\left(\frac{b}{||\vec{v}||},\frac{\color{red}{-}a}{||\vec{v}||}\right)$$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.