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.^i)
Now ^i.^i=^j.^j=1(|^i||^i| cos0=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