Find the unit vector along 2i + 2j+ 2k?

1 Answer
Apr 12, 2018

#hatv = sqrt3/3hati+sqrt3/3hatj+sqrt3/3hatk#

Explanation:

Let #vecv = 2hati+2hatj+2hatk#

The unit vector of any vector #vecv# is denoted by placing a hat, #hat#, (or caret) on top the letter #hatv#. The unit vector is obtained by dividing the vector by its magnitude:

#hatv = vecv/|vecv|" [1]"#

The magnitude of a vector is computed by taking the square root of the sum of the squares of its components:

#|vecv|= sqrt(v_x^2+v_y^2+v_z^2)" [2]"#

Substitute the components of vector, #vecv#, into equation [2]:

#|vecv|= sqrt(2^2+2^2+2^2)#

#|vecv|= sqrt12#

#|vecv|= 2sqrt3#

Substitute #vecv# and its magnitude into equation [1]:

#hatv = (2hati+2hatj+2hatk)/(2sqrt3)#

Divide each component:

#hatv = 1/sqrt3hati+1/sqrt3hatj+1/sqrt3hatk#

Rationalize the denominators:

#hatv = sqrt3/3hati+sqrt3/3hatj+sqrt3/3hatk#

To Check that this is a unit vector, we verify that its magnitude is 1:

#|hatv| = sqrt((sqrt3/3)^2+(sqrt3/3)^2+(sqrt3/3)^2)#

#|hatv| = sqrt(3/9+3/9+3/9)#

#|hatv| = sqrt(9/9)#

#|hatv| = sqrt(1) = 1#

This checks.