Question

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` ?

Answer 1

See demo below

Explanation:

Suppose you have three points in `R^3`

`p_1 = (a,b,c)`
`p_2 = (x,y,z)`
`p_0=(0,0,0)`

The respective lengths of sides `p_1-p_0` and `p_2-p_0`
are

`norm(p_1-p_0)=sqrt(a^2+b^2+c^2)`
`norm(p_2-p_0)=sqrt(x^2+y^2+z^2)`

and their scalar product

`(p_1-p_0).(p_2-p_0) = ax + by +cz`

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(hat(p_1p_0p_2))= (ax+by+cz)/(sqrt(a^2+b^2+c^2)sqrt(x^2+y^2+z^2)) = 30/(5times 6)=1`

Then the sides `p_1-p_0` and `p_2-p_0` are aligned so
`(p_1-p_0) = lambda(p_2-p_0)` and of course

`lambda = 5/6`

Answer 2

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

`(a/5-x/6)^2+(b/5-y/6)^2+(c/5-z/6)^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 `a/5-x/6=0=>a/x= 5/6`
Similarly, `b/y=5/6 and c/z =5/6`
`:.a/x=b/y=c/z= 5/6`
Proved.