Calculate the angle between the vectors: A → = 6.8 i ^ − 3.4 j ^ − 6.2 k ^ {\displaystyle {\vec {A}}=6.8{\hat {i}}-3.4{\hat {j}}-6.2{\hat {k}}} and B → = 8.2 i ^ + 2.3 j ^ − 7.0 k ^ {\displaystyle {\vec {B}}=8.2{\hat {i}}+2.3{\hat {j}}-7.0{\hat {k}}}
A → ⋅ B → = ( 6.8 ) ( 8.2 ) i ^ . i ^ + ( − 3.4 ) ( 2.3 ) j ^ . j ^ + ( − 6.2 ) ( − 7.0 ) k ^ . k ^ = 91.34 {\displaystyle {\vec {A}}\cdot {\vec {B}}=(6.8)(8.2){\hat {i}}.{\hat {i}}+(-3.4)(2.3){\hat {j}}.{\hat {j}}+(-6.2)(-7.0){\hat {k}}.{\hat {k}}=91.34}
A = A → ⋅ A → = ( 6.8 ) 2 + ( − 3.4 ) 2 + ( − 6.2 ) 2 = 9.81 {\displaystyle A={\sqrt {{\vec {A}}\cdot {\vec {A}}}}={\sqrt {(6.8)^{2}+(-3.4)^{2}+(-6.2)^{2}}}=9.81}
B = B → ⋅ B → = ( 8.2 ) 2 + ( 2.3 ) 2 + ( − 7.0 ) 2 = 11.0 {\displaystyle B={\sqrt {{\vec {B}}\cdot {\vec {B}}}}={\sqrt {(8.2)^{2}+(2.3)^{2}+(-7.0)^{2}}}=11.0}
A → ⋅ B → = A B cos θ → θ = cos − 1 A → ⋅ B → A B = 91.34 ( 9.81 ) ( 11.0 ) ) = 32 ∘ {\displaystyle {\vec {A}}\cdot {\vec {B}}=AB\cos \theta \rightarrow \theta =\cos ^{-1}{\frac {{\vec {A}}\cdot {\vec {B}}}{AB}}={\frac {91.34}{(9.81)(11.0)}})=32^{\circ }}
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