How do you normalize #(i+k)#?
1 Answer
Jan 3, 2016
Divide by the scalar
#sqrt(2)/2 i + sqrt(2)/2 k#
Explanation:
#abs(i+k) = sqrt(1^2+1^2) = sqrt(2)#
So multiplying by
#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#