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