# Math Help - vectors problem

1. ## vectors problem

i know that the definition of a unit vector is one such that if you square all the vectors, add them together, and take the square root, you get 1.

i also know that in order for a vector to be perpendicular to another, the dot product of the two vectors has to equal 0.

however, i am having trouble finding a unit vector that is perpendicular to the vector (1, -2, 3) so any help would be appreciated!

2. First, I think that you have very confused ideas.
A unit vector is a vector of length one.
The length of a vector is the square root of the sum of the squares of its components. Another way: it is the square root of the dot product of the vector with itself.

To convert a vector to an equivalent( parallel) unit vector simply divide each component of the vector by the length of the vector.

3. Thanks, but what do you do if you are trying to find a perpendicular vector?

4. Two vectors are perpendicular if their dot product is zero.

5. Originally Posted by faure72
i know that the definition of a unit vector is one such that if you square all the vectors, add them together, and take the square root, you get 1.

i also know that in order for a vector to be perpendicular to another, the dot product of the two vectors has to equal 0.

however, i am having trouble finding a unit vector that is perpendicular to the vector (1, -2, 3) so any help would be appreciated!
<1, 5, 3>; the unit vector, then, is:

sqrt(1^2 + 5^2 + 3^2) = sqrt(35)

<1/sqrt(35), 5/sqrt(35), 3/sqrt(35)>

6. Hello, faure72!

i am having trouble finding a unit vector that is perpendicular to the vector (1, -2, 3)
Let v = (a,b,c) be perpendicular to (1,-2,3)

Then: .(a,b,c)·(1,-2,3) .= .0

And we have: .a - 2b + 3c .= .0

Obviously, there is no unique answer,
. . so select any set of numbers that satisfy the equation.

For example: .(2,1,0)

Now convert it to a unit vector.
. . . . . . . . . . . . . . ___________ - - - _
Its magnitude is: .√2² + 1² + 0² .= .√5
. . . . . . . . . . . . . . . . . . . . . . _ . . . _
Divide by the magnitude: . (2/√5, 1/√5, 0) . . . . there!

[Note: This is but one of a brizillion possible perpendicular vectors.]