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

1 Answer
Jan 3, 2016

Answer:

Divide by the scalar #abs(i+k) = sqrt(2)# to get:

#sqrt(2)/2 i + sqrt(2)/2 k#

Explanation:

#abs(i+k) = sqrt(1^2+1^2) = sqrt(2)#

So multiplying by #1/sqrt(2) = sqrt(2)/2# we find:

#abs(sqrt(2)/2 i + sqrt(2)/2 k) = sqrt((sqrt(2)/2)^2 + (sqrt(2)/2)^2)#

#= sqrt(1/2+1/2) = sqrt(1) = 1#