# What is the angle between <7 , 3 , 1 >  and  < 8 , -9 , 8 > ?

$\theta = {70.54}^{0} \left(2 \mathrm{dp}\right)$
Let $\vec{u} = < 7 , 3 , 1 > \mathmr{and} \vec{v} = < 8 , - 9 , 8 >$
We know the angle between two vectors as $\cos \theta = \frac{\vec{u} . \vec{v}}{| | \vec{u} | | . | | \vec{v} | |}$Now vecu.vecv= (7*8)+(3* -9)+ (1.8)=37 ; ||vecu||=sqrt(7^2+3^2+1^2)=sqrt59; ||vecv||=sqrt(8^2+(-9)^2+8^2)=sqrt209; :.cos theta= 37/sqrt(59*209)=37/sqrt 12331=0.3331:. theta = cos^-1(0.3331)=70.54^0(2dp)[Ans]