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Question Number 181875 by mr W last updated on 01/Dec/22

$$ifa+b+c=0,findthemaximumof$$$$\frac{\mid a+2b+3c\mid }{\sqrt{a^2+b^2+c^2}}.$$

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

$$Q181318reposted,askingformore$$$$solutions.$$

Commented by SEKRET last updated on 01/Dec/22

$$\mathbf{what}\mathbf{happened}\mathbf{to}\mathbf{previos}\mathbf{solutions}$$

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

$$theyarethere,nothinghappened.$$

Answered by SEKRET last updated on 01/Dec/22

$$\mathbf{max}\left(\sqrt{2}\right)$$

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

$$pleaseshowyoursolution.$$

Answered by SEKRET last updated on 01/Dec/22

$$\mathbf{c}=-(\mathbf{b}+\mathbf{a})\mathbf{a}+2\mathbf{b}-3(\mathbf{b}+\mathbf{a})=-(\mathbf{b}+2\mathbf{a})$$$${\mathbf{a}}^2+{\mathbf{b}}^2+{(\mathbf{a}+\mathbf{b})}^2=2{\mathbf{a}}^2+2{\mathbf{b}}^2+2\mathbf{ab}$$$$\frac{\mid \mathbf{b}+2\mathbf{a}\mid }{\sqrt{2}\cdot \sqrt{{\mathbf{a}}^2+{\mathbf{b}}^2+\mathbf{ab}}}\mathbf{b}=\mathbf{ax}$$$$\frac{\mid \mathbf{a}\mid \cdot \mid \mathbf{x}+2\mid }{\sqrt{2}\cdot \mid \mathbf{a}\mid \cdot \sqrt{{\mathbf{x}}^2+\mathbf{x}+1}}=\frac{\mid \mathbf{x}+2\mid }{\sqrt{2}\cdot \sqrt{{\mathbf{x}}^2+\mathbf{x}+1}}=\mathbf{f}\left(\mathbf{x}\right)$$$$\mathbf{f}{}^{\prime }\left(\mathbf{x}\right)=0\frac{-3\mathbf{x}}{2\sqrt{2}\cdot \sqrt{{({\mathbf{x}}^2+\mathbf{x}+1)}^3}}=0$$$$\mathbf{x}=0\mathbf{f}\left(0\right)=\frac{2}{\sqrt{2}}=\sqrt{2}$$$$$$

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

$$thanks!$$

Answered by mr W last updated on 02/Dec/22

$$\text{Missing left or extra right}$$$$b=-(a+c)$$$$f=\frac{\mid a+2b+3c\mid }{\sqrt{a^2+b^2+c^2}}$$$$=\frac{\mid a-2a-2c+3c\mid }{\sqrt{a^2+{(a+c)}^2+c^2}}$$$$=\frac{\mid c-a\mid }{\sqrt{2(a^2+ac+c^2)}}$$$$=\frac{\mid k-1\mid }{\sqrt{2(1+k+k^2)}}withk=\frac{c}{a}$$$$=\frac{\mid k-1\mid }{\sqrt{2({(k-1)}^2+3(k-1)+3)}}$$$$=\frac{\mid u\mid }{\sqrt{2(u^2+3u+3)}}withu=k-1$$$$=\frac{1}{\sqrt{2(\frac{3}{u^2}+\frac{3}{u}+1)}}$$$$=\frac{1}{\sqrt{6[{(\frac{1}{u}+\frac{1}{2})}^2+\frac{1}{12}]}}$$$$⩽\frac{1}{\sqrt{6\times \frac{1}{12}}}=\sqrt{2}$$$$\Rightarrow maximum=\sqrt{2}$$$$atu=-2=k-1=\frac{c}{a}-1\Rightarrow c+a=0,b=0$$

Commented by SEKRET last updated on 01/Dec/22

$$\mathbf{super}$$

Answered by mr W last updated on 02/Dec/22

$$\text{Missing left or extra right}$$$$imagewehaveaplanewithfollowing$$$$equation:$$$$ax+by+cz=0$$$$thisplanepassesthroughtheorigin$$$$A(0,0,0).$$$$sincea+b+c=0,itmeanstheplane$$$$passesalsothroughthepointB(1,1,1).$$$$weknow\frac{\mid a+2b+3c\mid }{\sqrt{a^2+b^2+c^2}}representsthe$$$$distancefromthepointP(1,2,3)tothe$$$$planeax+by+cz=0.$$$$nowweonlyneedtofindthelargest$$$$distancefromthepointP(1,2,3)tothe$$$$planeax+by+cz=0.$$$$sincetheplanepassesthroughthe$$$$pointsA(0,0,0)andB(1,1,1),the$$$$largestdistancefrompointPtothe$$$$planeisthedistancefrompointPto$$$$thelineAB.$$$$\mathit{A}\mathit{B}=(1,1,1)$$$$\mathit{A}\mathit{P}=(1,2,3)$$$$\mathit{A}\mathit{P}\times \mathit{A}\mathit{B}=(1,2,3)\times (1,1,1)=(1,2,1)$$$$d=\frac{\mid \mathit{A}\mathit{P}\times \mathit{A}\mathit{B}\mid }{\mid \mathit{A}\mathit{B}\mid }=\frac{\sqrt{1^2+2^2+1^2}}{\sqrt{1^2+1^2+1^2}}=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}$$$$thatmeansmaximumof\frac{\mid a+2b+3c\mid }{\sqrt{a^2+b^2+c^2}}$$$$is\sqrt{2}.$$

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

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

$$alternative:$$$$\cos \theta =\frac{(1,2,3)\cdot (1,1,1)}{\sqrt{(1^2+1^2+1^2)(1^2+2^2+3^2)}}=\frac{\sqrt{6}}{\sqrt{7}}$$$$\sin \theta =\frac{1}{\sqrt{7}}$$$$d=AP\sin \theta =\sqrt{14}\times \frac{1}{\sqrt{7}}=\sqrt{2}$$

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