Find a unit vector that is perpendicular to the vector 6i+8j
If this vector belongs to 2-D space then you only have to reverse the order of the components and make one component negantive:
$\displaystyle \vec v = (6,8)~\implies~ \vec n = (-8,6)~\vee~ \vec n = (8, -6)$
The unit vector of $\displaystyle \vec n$ is calculated by: $\displaystyle \vec n^0 = \dfrac{\vec n}{|\vec n|}$
In 3-D space the normal unit vector is simply $\displaystyle \vec n = (0,0,1)$
Hello, hooke!
Two vectors are perpendicular if their dot product is zero.Find a unit vector that is perpendicular to the vector $\displaystyle 6i+8j$
Let the vector be: .$\displaystyle \vec v \:=\:ai+bj$
Then we have: .$\displaystyle (6i + 8j)\cdot(ai + bj) \:=\:0 \quad\Rightarrow\quad 6a + 8b \:=\:0 \quad\Rightarrow\quad b \:=\:-\frac{3a}{4}$
Let $\displaystyle a = 4\quad\hdots$ then: .$\displaystyle b = -3$
. . Hence, a perpendicular vector is: .$\displaystyle \vec v \:=\:4i - 3j$
For unit vector, divide by its magnitude: .$\displaystyle |\vec v| \:=\:\sqrt{4^2 + (\text{-}3)^2} \:=\:5$
Therefore: .$\displaystyle \vec u \:=\:\frac{\vec v}{|\vec v|} \:=\:\frac{4i-3j}{5} \:=\:\tfrac{4}{5}i - \tfrac{3}{5}j $