Question

find the angle between the vectors a=i+2j+3k and b=2i-j+k​

Answer 1

Cos A = ( a1b1 +a2b2 + a3 b3 )/ square root of a1^2 + b1^2+c1^2 . Square root of a2^2+ b2^2 + c2^2
= 2-2+3/square root of (4+9+1)(4+1+1)
3/square root of 84