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If-x-cy-bz-y-cx-az-and-z-bx-ay-then-prove-that-x-1-a-2-y-1-b-2-z-1-c-2-

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Question Number 214759 by MATHEMATICSAM last updated on 19/Dec/24
If x = cy + bz, y = cx + az and z = bx + ay then prove that (x/( (√(1 − a^2 )))) = (y/( (√(1 − b^2 )))) = (z/( (√(1 − c^2 )))) .
$$\mathrm{If}\:{x}\:=\:{cy}\:+\:{bz},\:{y}\:=\:{cx}\:+\:{az}\:\mathrm{and} \\ $$$${z}\:=\:{bx}\:+\:{ay}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{x}}{\:\sqrt{\mathrm{1}\:−\:{a}^{\mathrm{2}} }}\:=\:\frac{{y}}{\:\sqrt{\mathrm{1}\:−\:{b}^{\mathrm{2}} }}\:=\:\frac{{z}}{\:\sqrt{\mathrm{1}\:−\:{c}^{\mathrm{2}} }}\:. \\ $$
Answered by MathematicalUser2357 last updated on 19/Dec/24
It gives us x=y=z=0, and (0/( (√(1−a^2 ))))=(0/( (√(1−b^2 ))))=(0/( (√(1−c^2 )))) is true. Only for a,b,c∈(−1,1) ■
$$\mathrm{It}\:\mathrm{gives}\:\mathrm{us}\:{x}={y}={z}=\mathrm{0}, \\ $$$$\mathrm{and}\:\frac{\mathrm{0}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}=\frac{\mathrm{0}}{\:\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }}=\frac{\mathrm{0}}{\:\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }}\:\mathrm{is}\:\mathrm{true}.\:\mathrm{Only}\:\mathrm{for}\:{a},{b},{c}\in\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\blacksquare \\ $$
Answered by som(math1967) last updated on 19/Dec/24
x=cy+bz ⇒x=cy+b(bx+ay) ⇒x=cy+b^2 x+aby ⇒x(1−b^2 )=y(c+ab) ⇒(x/y)=(((c+ab))/((1−b^2 ))) .......(1) y=cx+az ⇒y=cx+a(bx+ay) ⇒y(1−a^2 )=x(c+ab) ⇒(x/y)=((1−a^2 )/(c+ab)) ........(2) (1)×(2) (x^2 /y^2 )=((c+ab)/(1−b^2 ))×((1−a^2 )/(c+ab)) ⇒(x/y)=((√(1−a^2 ))/( (√(1−b^2 )))) ∴ (x/( (√(1−a^2 ))))=(y/( (√(1−b^2 )))) same way we can show (y/( (√(1−b^2 ))))=(z/( (√(1−c^2 ))))
$$\:{x}={cy}+{bz} \\ $$$$\Rightarrow{x}={cy}+{b}\left({bx}+{ay}\right) \\ $$$$\Rightarrow{x}={cy}+{b}^{\mathrm{2}} {x}+{aby} \\ $$$$\Rightarrow{x}\left(\mathrm{1}−{b}^{\mathrm{2}} \right)={y}\left({c}+{ab}\right) \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\left({c}+{ab}\right)}{\left(\mathrm{1}−{b}^{\mathrm{2}} \right)}\:…….\left(\mathrm{1}\right) \\ $$$$\:{y}={cx}+{az} \\ $$$$\Rightarrow{y}={cx}+{a}\left({bx}+{ay}\right) \\ $$$$\Rightarrow{y}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)={x}\left({c}+{ab}\right) \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{1}−{a}^{\mathrm{2}} }{{c}+{ab}}\:\:\:……..\left(\mathrm{2}\right) \\ $$$$\:\left(\mathrm{1}\right)×\left(\mathrm{2}\right) \\ $$$$\:\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }=\frac{{c}+{ab}}{\mathrm{1}−{b}^{\mathrm{2}} }×\frac{\mathrm{1}−{a}^{\mathrm{2}} }{{c}+{ab}} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }} \\ $$$$\:\therefore\:\frac{{x}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}=\frac{{y}}{\:\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }} \\ $$$$\:{same}\:{way}\:{we}\:{can}\:{show} \\ $$$$\:\frac{{y}}{\:\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }}=\frac{{z}}{\:\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }} \\ $$$$ \\ $$

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