# Given a^2+b^2+c^2=25 ;x^2+y^2+z^2=36 and ax+by+cz=30 for a,b,c being real. How will you prove a/x=b/y=c/z=5/6 ?

May 31, 2016

See demo below

#### Explanation:

Suppose you have three points in ${R}^{3}$

${p}_{1} = \left(a , b , c\right)$
${p}_{2} = \left(x , y , z\right)$
${p}_{0} = \left(0 , 0 , 0\right)$

The respective lengths of sides ${p}_{1} - {p}_{0}$ and ${p}_{2} - {p}_{0}$
are

$\left\lVert {p}_{1} - {p}_{0} \right\rVert = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}$
$\left\lVert {p}_{2} - {p}_{0} \right\rVert = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

and their scalar product

$\left({p}_{1} - {p}_{0}\right) . \left({p}_{2} - {p}_{0}\right) = a x + b y + c z$

but (p_1-p_0).(p_2-p_0) = norm(p_1-p_0)norm(p_2-p_0) cos(hat(p_1p_0p_2))

so

$\cos \left(\hat{{p}_{1} {p}_{0} {p}_{2}}\right) = \frac{a x + b y + c z}{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}} = \frac{30}{5 \times 6} = 1$

Then the sides ${p}_{1} - {p}_{0}$ and ${p}_{2} - {p}_{0}$ are aligned so
$\left({p}_{1} - {p}_{0}\right) = \lambda \left({p}_{2} - {p}_{0}\right)$ and of course

$\lambda = \frac{5}{6}$

Jun 1, 2016

Another way

#### Explanation:

For those who know less

a^2+b^2+c^2=25….(1)
x^2+y^2+z^2=36…….(2)
ax+by+cz=30…..(3)

Dividing equation(1)by ${5}^{2}$ we have
(a/5)^2+(b/5)^2+(c/5)^2=1….(4)

Dividing equation(2)by ${6}^{2}$we have
(x/6)^2+(y/6)^2+(z/6)^2=1…(5)

Dividing equation(3)by 30 we have
(a/5*x/6)+(b/5*y/6)+(c/5*z/6)=1…..(6)

Adding equation (4) ;(5) and subtracting twice of equation(6) from their sum ,we get

${\left(\frac{a}{5} - \frac{x}{6}\right)}^{2} + {\left(\frac{b}{5} - \frac{y}{6}\right)}^{2} + {\left(\frac{c}{5} - \frac{z}{6}\right)}^{2} = 0$

a,b,c,x,y,z are real and sum of three squared exrpressions are zero,so each of them should be zero.

Hence $\frac{a}{5} - \frac{x}{6} = 0 \implies \frac{a}{x} = \frac{5}{6}$
Similarly, $\frac{b}{y} = \frac{5}{6} \mathmr{and} \frac{c}{z} = \frac{5}{6}$
$\therefore \frac{a}{x} = \frac{b}{y} = \frac{c}{z} = \frac{5}{6}$
Proved.