What is the angle between the following pair vectors? →A=^i+^j+^k and →B=−2^i−2^j−2^k
180∘
→A.→B=AB cos θ=→A.→BABBut, →A.→B=(^i+^j+^k).(−2^i−2^j−2^k)→A.→B=−2−2−2=−6Again A=|→A|=√(1)2+(1)2+(1)2=√3B=|→B|=√(−2)2+(−2)2+(−2)2=√12=2√3Now, cos θ=−6√3×2√3=−1⇒θ=180∘