Let θ \theta θ be the angle between the vectors g ⃗ \vec{g} g and h ⃗ \vec{h} h .
Using the transformed dot product equation,
θ = cos − 1 ( g ⃗ ⋅ h ⃗ ∣ g ⃗ ∣ ∣ h ⃗ ∣ ) \theta = \cos^{-1} \left( \frac{ \vec{g}\cdot \vec{h}}{|\vec{g}| |\vec{h}| } \right) θ = cos − 1 ( ∣ g ∣∣ h ∣ g ⋅ h )
where
g ⃗ ⋅ h ⃗ = ( 1 ) ( 0 ) + ( 2 ) ( − 3 ) + ( − 3 ) ( 4 ) = − 18 \vec{g}\cdot \vec{h} = (1)(0) + (2)(-3)+(-3)(4)=-18 g ⋅ h = ( 1 ) ( 0 ) + ( 2 ) ( − 3 ) + ( − 3 ) ( 4 ) = − 18 (the dot product of g ⃗ \vec{g} g and h ⃗ \vec{h} h )
and
∣ g ⃗ ∣ = ( 1 ) 2 + ( 2 ) 2 + ( − 3 ) 2 = 14 |\vec{g}| = \sqrt{(1)^2+(2)^2+(-3)^2} = \sqrt{14} ∣ g ∣ = ( 1 ) 2 + ( 2 ) 2 + ( − 3 ) 2 = 14 (magnitude of vector g ⃗ \vec{g} g )
∣ h ⃗ ∣ = ( 0 ) 2 + ( − 3 ) 2 + ( 4 ) 2 = 25 = 5 |\vec{h}| = \sqrt{(0)^2+(-3)^2+(4)^2} = \sqrt{25}=5 ∣ h ∣ = ( 0 ) 2 + ( − 3 ) 2 + ( 4 ) 2 = 25 = 5 (magnitude of vector h ⃗ \vec{h} h )
Therefore,
θ = cos − 1 ( − 18 14 ( 5 ) ) \theta = \cos^{-1} \left( \frac{-18}{\sqrt{14} (5)} \right) θ = cos − 1 ( 14 ( 5 ) − 18 )
θ = cos − 1 ( − 0.9621 ) \theta = \cos^{-1} \left( -0.9621 \right) θ = cos − 1 ( − 0.9621 )
θ = 164.1 8 0 \theta = 164.18^{0} θ = 164.1 8 0
Comments