# Thread: finding unit vector orthagonal with vector

1. ## finding unit vector orthagonal with vector

Unsure how to do this.
Find two different unit vectors, in 2dimensions, that make an angle of 90degress with 4i+3j

2. Originally Posted by Craka
Unsure how to do this.
Find two different unit vectors, in 2dimensions, that make an angle of 90degress with 4i+3j
note that if two vectors are orthogonal, their dot product is zero.

so let one unit vector be <a,b> (note the other is <-a,-b>)

then $\displaystyle \left< a,b \right> \cdot \left< 4,3 \right> = 0$

then find the unit vector in the direction of <a,b>

3. Sorry but could please show the full solution. Thankyou

4. Originally Posted by Craka
Sorry but could please show the full solution.
You don't want to turn someone else's work do you?
Solve this system.
$\displaystyle \begin{gathered} 4a + 3b = 0 \hfill \\ a^2 + b^2 = 1 \hfill \\ \end{gathered}$

5. No i'm not trying to copy someone's work I am just trying to understand how to do the problem.
The answer is in the form of <3/5, -4/5> and <-3/5 , 4/5>

I can see where the 1/5 factor came in. It is from dividing by the magnitude of original vector ie the sqrt of [4^2 + 3^2], and yes I understand that if the dot product of two vector is zero than the two vectors must be orthagonal, however I still unsure how 4a+3b=0 and -4a-3b=0 and a^2+b^2=1 helps.

I would really like to understand what to do here. It is not copying to do some homework. It is for study. Thankyou

6. I can see how the answer works. ie the dot product of <4,3> . <-3/5, 4/5> and dot product <4,3> . <3/5, -4/5 > work both out to be zero.
What is the method of establishing this.

To get a unit vector is just a matter of dividing the vector by its length which is the sqrt of the squared sums of the vector components.

Guess what I'm also trying to ask here is in 2 dimensional vectors, to find a orthogonal vector is it just a matter of swapping say the x and y component of vector around and making one component negative , and how would you do so for 3 dimensional vectors?

7. Originally Posted by Craka
I can see how the answer works. ie the dot product of <4,3> . <-3/5, 4/5> and dot product <4,3> . <3/5, -4/5 > work both out to be zero.
What is the method of establishing this.

To get a unit vector is just a matter of dividing the vector by its length which is the sqrt of the squared sums of the vector components.

Guess what I'm also trying to ask here is in 2 dimensional vectors, to find a orthogonal vector is it just a matter of swapping say the x and y component of vector around and making one component negative , and how would you do so for 3 dimensional vectors?
<a,b>.<4,3> = 0

=> 4a + 3b = 0

pick a = 1. what does b have to be?

thus, you can find values a and b that work. infinitely many values will work, in fact. you just need 1 of them. once you find the unit vector in the direction <a,b>, take its negative to find the another unit vector

8. How would I go about doing this with a 3 dimensional vector ?

9. Originally Posted by Craka
How would I go about doing this with a 3 dimensional vector ?
same thing. pick random values for two of the components, and solve for the third

10. Originally Posted by Craka
How would I go about doing this with a 3 dimensional vector ?
One important point: in the plane there exist a single line perpendicular to a given line at a given point. That means that the set of vectors perpendicular to a given vector must all lie on that line. Given, for example, the vector <a, b> it is very easy to see that <a, b>.<-b, a>= 0 and so any vector perpendicular to <a, b> must b a multiple of <-b, a>. In particular, there are only two unit vectors perpendicular to a given vector, pointing in opposite directions.

In three dimensions, there exist an entire PLANE of line perpendicular to a given line at a given point. There are now an infinite number of unit vectors perpendicular to a given vector.