The vector U=2i +3j -2k, how do you find a unit vector parallel to U?

1 Answer
Jun 19, 2016

The unit vector is #v=(2/sqrt(17), 3/sqrt(17),-2/sqrt(17))#

Explanation:

If you have a vector #(a,b,c)# and multiply all the components by a same scalar #alpha#, you obtain the vector #(alpha a, alpha b, alpha c)#. This vector is parallel to the original one, and it is smaller than the original if #alpha<1#, and greater than the original if #alpha>1#.

Now, let's talk about norms. The norm of a vector is the quantity

#norm(a,b,c)=sqrt(a^2+b^2+c^2)#.

So, we want to multiply the vector #(2,3,-2)# by some scalar #alpha# such that the new vector #(2\alpha,3\alpha,-2\alpha)# has norm equal to one.

To do so, choose #alpha# as the inverse of the norm of the vector: we have

#norm(2,3,-2)=sqrt(4+9+4)=sqrt(17)#

and choosing #alpha=1/sqrt(17)# we have

#norm(2\alpha,3\alpha,-2\alpha)=norm(2/sqrt(17), 3/sqrt(17),-2/sqrt(17))#

#= (2/sqrt(17))^2+(3/sqrt(17))^2+(-2/sqrt(17))^2#

#=4/17+9/17+4/17=1#