Subject: Applied Physics 2

Topic: Electrodynamics

Difficulty: Medium

$$ \text {Formula a ⃗.b ⃗ = ab cosθ }$$

$(a_1 ) = a_1 i ̂ +a_2 j ̂ +a_3 k ̂ = 3i ̂ +4 j ̂+0\\ a_1=3 ,a_2=4 ,a_3=0 \\ $

$ (b_1 ) ⃗= b_1 i ̂ +b_2 j ̂ +b_3 k ̂ \\ = 6i ̂ +3j ̂+2k ̂ \\ 〖 b〗_1=6 ,b_2=3,b_3=2 \\ $

$ a ⃗.b ⃗ = 6 \\ |a ⃗ | = 5 \\ |b ⃗ | = 7 \\ $

$ \text {Therefore cosθ = (a ⃗.b ⃗)/(|a ⃗ |.|b ⃗ | ) = 6/35 } \\ $

$ \text{ θ=cos^{-1} 6/35 = 80029 } \\ $