If →A=2^i−3^j and →B=−4^i+2^j, then |→A.→B|=
→A=2^i−3^j→B=−4^i+2^j,→A⋅→B=(2^i−3^j).(−4^i+2^j)
=(2×−4)(^i⋅^i)+(2×2)(^i⋅^j)+(3×4)(^j⋅^i)−(3×2)(^j⋅^j)
Now ^i⋅^i=^j⋅^j=1(|^i||^i| cos 0=1)
^i⋅^j=^j⋅^i=0 (^i and ^j are perpendicular to each other)
So (2^i–3^j).(−4^i+2^j)
= -8 + 0 + 0 – 6
= -14
|→A.→B|=|−14|=14