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If-a-2-b-2-c-2-16-x-2-y-2-z-2-25-and-ax-by-cz-20-then-what-is-the-value-of-a-b-c-x-y-z-

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Question Number 193296 by MATHEMATICSAM last updated on 09/Jun/23
If a^2 + b^2 + c^2 = 16, x^2 + y^2 + z^2 = 25 and ax + by + cz = 20 then what is the value of ((a + b + c)/(x + y + z)) ?
$$\mathrm{If}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:=\:\mathrm{16},\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$$\mathrm{and}\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{20}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:? \\ $$
Commented by kapoorshah last updated on 11/Jun/23
let u^→ = ((a),(b),(c) ) and v^→ = ((x),(y),(z) ) ∣ u^→ ∣= 4 and ∣v^→ ∣ = 5 cos θ = ((u^→ .v^→ )/(∣u^→ ∣ ∣v^→ ∣)) = 1 u^→ parallel v^→ ⇒ ((a + b + c)/(x + y + z)) = ∓ ((∣u^→ ∣)/(∣v^→ ∣)) =∓ (4/5)
$${let}\:\overset{\rightarrow} {{u}}\:=\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}\:\:{and}\:\:\overset{\rightarrow} {{v}}\:=\:\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix} \\ $$$$\mid\:\overset{\rightarrow} {{u}}\mid=\:\mathrm{4}\:\:\:{and}\:\:\mid\overset{\rightarrow} {{v}}\mid\:=\:\mathrm{5} \\ $$$$\mathrm{cos}\:\theta\:=\:\frac{\overset{\rightarrow} {{u}}\:.\overset{\rightarrow} {{v}}}{\mid\overset{\rightarrow} {{u}}\mid\:\mid\overset{\rightarrow} {{v}}\mid}\:=\:\mathrm{1} \\ $$$$\overset{\rightarrow} {{u}}\:{parallel}\:\overset{\rightarrow} {{v}}\:\Rightarrow\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\mp\:\frac{\mid\overset{\rightarrow} {{u}}\mid}{\mid\overset{\rightarrow} {{v}}\mid}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mp\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$
Answered by York12 last updated on 09/Jun/23
trivialy you can obtain a=0 , b=0 ⇒c=∓4 x=0 , y=0 ⇒ z =∓5 which satifies the third equation ⇒ ((a+b+c)/(x+y+z))= ((∓4)/5)
$${trivialy}\:{you}\:{can}\:{obtain}\:{a}=\mathrm{0}\:,\:{b}=\mathrm{0}\:\Rightarrow{c}=\mp\mathrm{4} \\ $$$${x}=\mathrm{0}\:,\:{y}=\mathrm{0}\:\Rightarrow\:{z}\:=\mp\mathrm{5} \\ $$$${which}\:{satifies}\:{the}\:{third}\:{equation} \\ $$$$\Rightarrow \\ $$$$\:\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\:\frac{\mp\mathrm{4}}{\mathrm{5}} \\ $$$$ \\ $$
Answered by York12 last updated on 09/Jun/23
(a^2 +b^2 +c^2 )(x^2 +y^2 +z^2 )=(ax+by+cz)^2 =400 ⇒(a^2 +b^2 +c^2 )(x^2 +y^2 +z^2 )−(ax+by+cz)^2 =0 ⇒(ay−bx)^2 +(ax−cx)^2 +(bz−cy)^2 =0 ⇒ (a/x)=(b/y)=(c/z)=((a+b+c)/(x+y+z))=λ ⇒a=xλ , b=yλ , c = zλ ⇒λ^2 (x^2 +y^2 +z^2 _(25) )=16 ⇒ λ=((∓4)/5)★
$$ \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)=\left({ax}+{by}+{cz}\right)^{\mathrm{2}} =\mathrm{400} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\left({ax}+{by}+{cz}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({ay}−{bx}\right)^{\mathrm{2}} +\left({ax}−{cx}\right)^{\mathrm{2}} +\left({bz}−{cy}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\lambda \\ $$$$\Rightarrow{a}={x}\lambda\:,\:{b}={y}\lambda\:,\:{c}\:=\:{z}\lambda \\ $$$$\Rightarrow\lambda^{\mathrm{2}} \left(\underset{\mathrm{25}} {\underbrace{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\right)=\mathrm{16}\:\Rightarrow\:\lambda=\frac{\mp\mathrm{4}}{\mathrm{5}}\bigstar \\ $$$$ \\ $$$$ \\ $$

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