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Question Number 158081 by Eric002 last updated on 30/Oct/21
determine the angle between two vectors  A=4ax+ay−3az  and  B=2ax+4ay−3az
$${determine}\:{the}\:{angle}\:{between}\:{two}\:{vectors} \\ $$$${A}=\mathrm{4}{ax}+{ay}−\mathrm{3}{az}\:\:{and}\:\:{B}=\mathrm{2}{ax}+\mathrm{4}{ay}−\mathrm{3}{az} \\ $$
Answered by ajfour last updated on 30/Oct/21
cos θ=((4(2)+1(4)−3(−3))/( (√(4^2 +1^2 +(−3)^2 ))(√(2^2 +4^2 +(−3)^2 ))))    =((21)/( (√(26))(√(29))))  θ=cos^(−1) (((21)/( (√(26))(√(29))))).
$$\mathrm{cos}\:\theta=\frac{\mathrm{4}\left(\mathrm{2}\right)+\mathrm{1}\left(\mathrm{4}\right)−\mathrm{3}\left(−\mathrm{3}\right)}{\:\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }} \\ $$$$\:\:=\frac{\mathrm{21}}{\:\sqrt{\mathrm{26}}\sqrt{\mathrm{29}}} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{21}}{\:\sqrt{\mathrm{26}}\sqrt{\mathrm{29}}}\right). \\ $$

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