# vectors problem

• Feb 25th 2007, 04:12 PM
faure72
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!
• Feb 25th 2007, 04:40 PM
Plato
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.
• Feb 25th 2007, 04:54 PM
faure72
Thanks, but what do you do if you are trying to find a perpendicular vector?
• Feb 25th 2007, 05:06 PM
Plato
Two vectors are perpendicular if their dot product is zero.
• Feb 25th 2007, 07:00 PM
AfterShock
Quote:

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)>
• Feb 25th 2007, 08:43 PM
Soroban
Hello, faure72!

Quote:

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.]