# Thread: another vector question

1. ## another vector question

Find a unit vector that is perpendicular to the vector 6i+8j

2. Originally Posted by hooke
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:

$\vec v = (6,8)~\implies~ \vec n = (-8,6)~\vee~ \vec n = (8, -6)$

The unit vector of $\vec n$ is calculated by: $\vec n^0 = \dfrac{\vec n}{|\vec n|}$

In 3-D space the normal unit vector is simply $\vec n = (0,0,1)$

3. Hello, hooke!

Find a unit vector that is perpendicular to the vector $6i+8j$
Two vectors are perpendicular if their dot product is zero.

Let the vector be: . $\vec v \:=\:ai+bj$

Then we have: . $(6i + 8j)\cdot(ai + bj) \:=\:0 \quad\Rightarrow\quad 6a + 8b \:=\:0 \quad\Rightarrow\quad b \:=\:-\frac{3a}{4}$

Let $a = 4\quad\hdots$ then: . $b = -3$

. . Hence, a perpendicular vector is: . $\vec v \:=\:4i - 3j$

For unit vector, divide by its magnitude: . $|\vec v| \:=\:\sqrt{4^2 + (\text{-}3)^2} \:=\:5$

Therefore: . $\vec u \:=\:\frac{\vec v}{|\vec v|} \:=\:\frac{4i-3j}{5} \:=\:\tfrac{4}{5}i - \tfrac{3}{5}j$