How do you normalize # (2i+j+2k) #?

1 Answer
David G.
Jan 31, 2016

The normalized vector is #(2/3i+1/3j+2/3k)#.

This is a unit vector in the same direction as the original vector.

Explanation:

Normalizing a vector involves dividing the coefficient of each of its elements by the length of the vector, to yield a unit vector (length =1) in the same direction as the original vector.

To do this we first need to find the length of the vector. If the vector is in the form #(ai+bj+ck)# then its length is given by:

#l=sqrt(a^2 + b^2+c^2)#

In this case this is:

#l=sqrt(2^2 + 1^2+2^2) = sqrt(4+1+4) = sqrt9=3#

Now we divide #a#, #b# and #c# by #l#, which is #3#:

The normalized vector is #(2/3i+1/3j+2/3k)#.

This is a unit vector in the same direction as the original vector.

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