Find a unit vector in the direction of the given vector.
v=(1,-1)
for any unit vector in the direction of (a,b)
....
find the length of the vector
= √(a^2+b^2)
now divide each component by that magnitude
to get (a/√(a^2+b^2) , b/√(a^2+b^2)
or you could write it as
(1/√(a^2+b^2) (a,b)
apply this to your vector, let me know what you got
apply this to your vector
I got (1/Square root of 2,-1/square root of 2)
Thanks
correct
notice you can check that the magnitude is 1
magnitude
= √( (1/√2)^2 + (-1/√2)^2 )
=√( 1/2 + 1/2)
= √1
= 1
To find a unit vector in the direction of a given vector, we need to normalize the given vector.
The formula for normalizing a vector is to divide each component of the vector by its magnitude.
Given vector v = (1, -1), we first need to compute the magnitude (length) of the vector:
|v| = √(1^2 + (-1)^2)
= √(1 + 1)
= √2
Now, we can normalize the vector by dividing each component by the magnitude:
v_unit = (1/√2, -1/√2)
So, the unit vector in the direction of the given vector v = (1, -1) is (1/√2, -1/√2).