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Question Number 9060 by tawakalitu last updated on 16/Nov/16

Answered by mrW last updated on 17/Nov/16

b) 60^2 =40^2 +92^2 −2×40×92×cos α cos α=((40^2 +92^2 −60^2 )/(2×40×92)) x^2 =40^2 +46^2 −2×40×46×cos α =40^2 +46^2 −2×40×46×((40^2 +92^2 −60^2 )/(2×40×92)) =40^2 +46^2 −((40^2 +92^2 −60^2 )/2)=484 x=22=∣A^→ ∣

$$\left.{b}\right) \\ $$$$\mathrm{60}^{\mathrm{2}} =\mathrm{40}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} −\mathrm{2}×\mathrm{40}×\mathrm{92}×\mathrm{cos}\:\alpha \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{40}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} −\mathrm{60}^{\mathrm{2}} }{\mathrm{2}×\mathrm{40}×\mathrm{92}} \\ $$$${x}^{\mathrm{2}} =\mathrm{40}^{\mathrm{2}} +\mathrm{46}^{\mathrm{2}} −\mathrm{2}×\mathrm{40}×\mathrm{46}×\mathrm{cos}\:\alpha \\ $$$$=\mathrm{40}^{\mathrm{2}} +\mathrm{46}^{\mathrm{2}} −\mathrm{2}×\mathrm{40}×\mathrm{46}×\frac{\mathrm{40}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} −\mathrm{60}^{\mathrm{2}} }{\mathrm{2}×\mathrm{40}×\mathrm{92}} \\ $$$$=\mathrm{40}^{\mathrm{2}} +\mathrm{46}^{\mathrm{2}} −\frac{\mathrm{40}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} −\mathrm{60}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{484} \\ $$$${x}=\mathrm{22}=\mid\overset{\rightarrow} {{A}}\mid \\ $$

Commented by mrW last updated on 17/Nov/16

Commented by tawakalitu last updated on 17/Nov/16

Great. God bless you.

$$\mathrm{Great}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Answered by mrW last updated on 17/Nov/16

c) point A(−4,2,−2) point B(−1,3,2) point C(5,λ,μ) AB^(→) =(3,1,4) AC^(→) =(9,λ−2,μ+2) (9/3)=((λ−2)/1) =((μ+2)/4) λ−2=3 λ=5 μ+2=3×4=12 μ=10

$$\left.{c}\right) \\ $$$${point}\:{A}\left(−\mathrm{4},\mathrm{2},−\mathrm{2}\right) \\ $$$${point}\:{B}\left(−\mathrm{1},\mathrm{3},\mathrm{2}\right) \\ $$$${point}\:{C}\left(\mathrm{5},\lambda,\mu\right) \\ $$$$\overset{\rightarrow} {{AB}}=\left(\mathrm{3},\mathrm{1},\mathrm{4}\right) \\ $$$$\overset{\rightarrow} {{AC}}=\left(\mathrm{9},\lambda−\mathrm{2},\mu+\mathrm{2}\right) \\ $$$$\frac{\mathrm{9}}{\mathrm{3}}=\frac{\lambda−\mathrm{2}}{\mathrm{1}}\:=\frac{\mu+\mathrm{2}}{\mathrm{4}} \\ $$$$\lambda−\mathrm{2}=\mathrm{3} \\ $$$$\lambda=\mathrm{5} \\ $$$$\mu+\mathrm{2}=\mathrm{3}×\mathrm{4}=\mathrm{12} \\ $$$$\mu=\mathrm{10} \\ $$

Commented by tawakalitu last updated on 17/Nov/16

Wow. thank you very much.

$$\mathrm{Wow}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}. \\ $$

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