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Question-178942

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Question Number 178942 by cortano1 last updated on 23/Oct/22
Answered by mr W last updated on 23/Oct/22
Commented by mr W last updated on 23/Oct/22
A=(4,0,0)+λ(−1,0,1)=(4−λ,0,λ) B=(0,0,0)+μ(0,1,1)=(0,μ,μ) Φ=AB^2 =(4−λ)^2 +μ^2 +(λ−μ)^2 (∂Φ/∂λ)=−2(4−λ)+2(λ−μ)=0 ⇒2λ−μ=4 (∂Φ/∂μ)=2μ−2(λ−μ)=0 ⇒λ−2μ=0 ⇒μ=(4/3), λ=(8/3) (AB)_(min) ^2 =(4−(8/3))^2 +((4/3))^2 +((8/3)−(4/3))^2 =((16)/3) ⇒AB_(min) =(4/( (√3)))
$${A}=\left(\mathrm{4},\mathrm{0},\mathrm{0}\right)+\lambda\left(−\mathrm{1},\mathrm{0},\mathrm{1}\right)=\left(\mathrm{4}−\lambda,\mathrm{0},\lambda\right) \\ $$$${B}=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)+\mu\left(\mathrm{0},\mathrm{1},\mathrm{1}\right)=\left(\mathrm{0},\mu,\mu\right) \\ $$$$\Phi={AB}^{\mathrm{2}} =\left(\mathrm{4}−\lambda\right)^{\mathrm{2}} +\mu^{\mathrm{2}} +\left(\lambda−\mu\right)^{\mathrm{2}} \\ $$$$\frac{\partial\Phi}{\partial\lambda}=−\mathrm{2}\left(\mathrm{4}−\lambda\right)+\mathrm{2}\left(\lambda−\mu\right)=\mathrm{0}\:\Rightarrow\mathrm{2}\lambda−\mu=\mathrm{4} \\ $$$$\frac{\partial\Phi}{\partial\mu}=\mathrm{2}\mu−\mathrm{2}\left(\lambda−\mu\right)=\mathrm{0}\:\Rightarrow\lambda−\mathrm{2}\mu=\mathrm{0} \\ $$$$\Rightarrow\mu=\frac{\mathrm{4}}{\mathrm{3}},\:\lambda=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left({AB}\right)_{{min}} ^{\mathrm{2}} =\left(\mathrm{4}−\frac{\mathrm{8}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\Rightarrow{AB}_{{min}} =\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by Tawa11 last updated on 23/Oct/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 23/Oct/22
alternative: OP^(→) =(4,0,0) n^→ =(−1,0,1)×(0,1,1)=(−1,1,−1) AB=d=∣(((−1,1,−1)∙(4,0,0))/( (√((−1)^2 +1^2 +(−1)^2 ))))∣=(4/( (√3))) ✓
$${alternative}: \\ $$$$\overset{\rightarrow} {{OP}}=\left(\mathrm{4},\mathrm{0},\mathrm{0}\right) \\ $$$$\overset{\rightarrow} {{n}}=\left(−\mathrm{1},\mathrm{0},\mathrm{1}\right)×\left(\mathrm{0},\mathrm{1},\mathrm{1}\right)=\left(−\mathrm{1},\mathrm{1},−\mathrm{1}\right) \\ $$$${AB}={d}=\mid\frac{\left(−\mathrm{1},\mathrm{1},−\mathrm{1}\right)\centerdot\left(\mathrm{4},\mathrm{0},\mathrm{0}\right)}{\:\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}\mid=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\:\checkmark \\ $$

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