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Question-194648




Question Number 194648 by universe last updated on 12/Jul/23
Commented by Frix last updated on 12/Jul/23
(7/8)
$$\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Commented by universe last updated on 12/Jul/23
solution ?
$${solution}\:? \\ $$
Commented by Frix last updated on 12/Jul/23
a=b=c∧d=e=f
$${a}={b}={c}\wedge{d}={e}={f} \\ $$
Answered by witcher3 last updated on 12/Jul/23
(ad+be+cf)≤(√((a^2 +b^2 +c^2 )(d^2 +e^2 +f^2 )))  cachy shwartz   use when we have equality
$$\left(\mathrm{ad}+\mathrm{be}+\mathrm{cf}\right)\leqslant\sqrt{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{d}^{\mathrm{2}} +\mathrm{e}^{\mathrm{2}} +\mathrm{f}^{\mathrm{2}} \right)} \\ $$$$\mathrm{cachy}\:\mathrm{shwartz}\: \\ $$$$\mathrm{use}\:\mathrm{when}\:\mathrm{we}\:\mathrm{have}\:\mathrm{equality} \\ $$
Answered by AST last updated on 12/Jul/23
ad+be+cf≤(√(a^2 +b^2 +c^2 ))(√(d^2 +e^2 +f^2 ))=56  Equality holds when a=kd,b=ke,c=kf  ⇒((a+b+c)/(d+e+f))=k  a^2 +b^2 +c^2 =64=k^2 (d^2 +e^2 +f^2 )=49k^2   ⇒k=(8/7)=((a+b+c)/(d+e+f))⇒((d+ef)/(a+b+c))=(7/8)
$${ad}+{be}+{cf}\leqslant\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\sqrt{{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} }=\mathrm{56} \\ $$$${Equality}\:{holds}\:{when}\:{a}={kd},{b}={ke},{c}={kf} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{d}+{e}+{f}}={k} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{64}={k}^{\mathrm{2}} \left({d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)=\mathrm{49}{k}^{\mathrm{2}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{8}}{\mathrm{7}}=\frac{{a}+{b}+{c}}{{d}+{e}+{f}}\Rightarrow\frac{{d}+{ef}}{{a}+{b}+{c}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Answered by mr W last updated on 14/Jul/23
p=(a,b,c)  q=(d,e,f)  ∣p∣=(√(a^2 +b^2 +c^2 ))=(√(64))=8  ∣q∣=(√(d^2 +e^2 +f^2 ))=(√(49))=7  ad+be+cf=p∙q=∣p∣∣q∣cos θ  56=8×7 cos θ  ⇒cos θ=1 ⇒θ=0° ⇒p//q  ⇒(d/a)=(e/b)=(f/c)=k=±((∣q∣)/(∣p∣))=±(7/8)  ⇒((d+e+f)/(a+b+c))=k=±(7/8)
$$\boldsymbol{{p}}=\left({a},{b},{c}\right) \\ $$$$\boldsymbol{{q}}=\left({d},{e},{f}\right) \\ $$$$\mid\boldsymbol{{p}}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\sqrt{\mathrm{64}}=\mathrm{8} \\ $$$$\mid\boldsymbol{{q}}\mid=\sqrt{{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} }=\sqrt{\mathrm{49}}=\mathrm{7} \\ $$$${ad}+{be}+{cf}=\boldsymbol{{p}}\centerdot\boldsymbol{{q}}=\mid\boldsymbol{{p}}\mid\mid\boldsymbol{{q}}\mid\mathrm{cos}\:\theta \\ $$$$\mathrm{56}=\mathrm{8}×\mathrm{7}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\mathrm{1}\:\Rightarrow\theta=\mathrm{0}°\:\Rightarrow\boldsymbol{{p}}//\boldsymbol{{q}} \\ $$$$\Rightarrow\frac{{d}}{{a}}=\frac{{e}}{{b}}=\frac{{f}}{{c}}={k}=\pm\frac{\mid\boldsymbol{{q}}\mid}{\mid\boldsymbol{{p}}\mid}=\pm\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow\frac{{d}+{e}+{f}}{{a}+{b}+{c}}={k}=\pm\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Commented by AST last updated on 15/Jul/23
Observe that d^2 +e^2 +f^2 =k(ad+be+cf)=56k≥0  ⇒k is positive
$${Observe}\:{that}\:{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} ={k}\left({ad}+{be}+{cf}\right)=\mathrm{56}{k}\geqslant\mathrm{0} \\ $$$$\Rightarrow{k}\:{is}\:{positive} \\ $$
Commented by mr W last updated on 17/Jul/23
yes, you are right! thanks!
$${yes},\:{you}\:{are}\:{right}!\:{thanks}! \\ $$
Commented by kapoorshah last updated on 17/Jul/23
nice
$${nice} \\ $$

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