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Question Number 170832 by Shrinava last updated on 01/Jun/22

Commented by mr W last updated on 01/Jun/22

$$T=AM+BN⩾\sqrt{{AB}^2+{MN}^2}=\sqrt{1^2+2^2+5^2+2^2}=\sqrt{34}$$$$T_{min}=\sqrt{34}$$

Commented by Shrinava last updated on 01/Jun/22

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathrm{professor},$$$$\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{a}\mathrm{little}\mathrm{more},$$$$\mathrm{if}\mathrm{possible},\mathrm{please}$$

Commented by mr W last updated on 02/Jun/22

Commented by mr W last updated on 01/Jun/22

$$T=AM+BN=A^{\prime }P+{PB}^{\prime }⩾A^{\prime }{B}^{\prime }=\sqrt{{AB}^2+{MN}^2}$$

Commented by Shrinava last updated on 02/Jun/22

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{dear}\mathrm{ptofessor}$$

Commented by mr W last updated on 04/Jun/22

$$ithinkmysolutionaboveiswrong.$$

Commented by mr W last updated on 04/Jun/22

Commented by mr W last updated on 04/Jun/22

$$T=AM+BN=A^{\prime }N+NB⩾A^{\prime }B$$$${AB}^2={(2-1)}^2+{(3-1)}^2+{(4+1)}^2=30$$$${MN}^2=2^2=4$$$$A^{\prime }{B}^2=30+4-2\times \sqrt{30}\times 2\cos \theta$$$$\sin \theta =\cos (\frac{\pi }{2}-\theta )=\frac{5\times 1}{\sqrt{30}}=\frac{5}{\sqrt{30}}$$$$\cos \theta =\sqrt{1-\frac{25}{30}}=\frac{\sqrt{5}}{\sqrt{30}}$$$$A^{\prime }{B}^2=30+4-2\times \sqrt{30}\times 2\times \frac{\sqrt{5}}{\sqrt{30}}=34-4\sqrt{5}$$$$\Rightarrow T_{min}=\sqrt{34-4\sqrt{5}}$$

Commented by Shrinava last updated on 17/Jun/22

$$\mathrm{Perfect}-\mathrm{Amazing}\mathrm{Dear}\mathrm{Professor}$$

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