Find the direction angles of vector −8^i+3^j+2^k.
→a=−8^i+3^j+2^kcosα=−8√(−8)2+(3)2+(2)2=−8√77=−.91⇒α=cos−1(−.91)=156∘cosβ=3√(−8)2+(3)2+(2)2=3√77=.34⇒β=cos−1(.34)=70∘cosγ=2√(−8)2+(3)2+(2)2=2√77=.23⇒γ=cos−1(.23)=77∘
So the direction angles are 156∘,70∘,77∘
Option A is correct.