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-

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 a x + b y + c z = 20 then what is the

value of \frac{a + b + c}{x + y + z} ?

Commented by kapoorshah last updated on 11/Jun/23

l e t \vec{u} = (\begin{matrix} a \\ b \\ c \end{matrix}) a n d \vec{v} = (\begin{matrix} x \\ y \\ z \end{matrix})

∣ \vec{u} ∣= 4 a n d ∣ \vec{v} ∣ = 5

\cos θ = \frac{\vec{u} . \vec{v}}{∣ \vec{u} ∣ ∣ \vec{v} ∣} = 1

\vec{u} p a r a l l e l \vec{v} \Rightarrow \frac{a + b + c}{x + y + z} = \mp \frac{∣ \vec{u} ∣}{∣ \vec{v} ∣}

= \mp \frac{4}{5}

Answered by York12 last updated on 09/Jun/23

t r i v i a l y y o u c a n o b t a i n a = 0, b = 0 \Rightarrow c = \mp 4

x = 0, y = 0 \Rightarrow z = \mp 5

w h i c h s a t i f i e s t h e t h i r d e q u a t i o n

\Rightarrow

\frac{a + b + c}{x + y + z} = \frac{\mp 4}{5}

Answered by York12 last updated on 09/Jun/23

(a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) = {(a x + b y + c z)}^{2} = 400

\Rightarrow (a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) - {(a x + b y + c z)}^{2} = 0

\Rightarrow {(a y - b x)}^{2} + {(a x - c x)}^{2} + {(b z - c y)}^{2} = 0

\Rightarrow \frac{a}{x} = \frac{b}{y} = \frac{c}{z} = \frac{a + b + c}{x + y + z} = λ

\Rightarrow a = x λ, b = y λ, c = z λ

\Rightarrow λ^{2} (\underset{25}{\underset{⏟}{x^{2} + y^{2} + z^{2}}}) = 16 \Rightarrow λ = \frac{\mp 4}{5} ★

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