Find the value of (x+y+z)/(l+m+k)

[anemone]

$x,\,y,\,z,\,l,\,m,\,k$ are real numbers such that

$x^2+y^2+z^2=25$,

$l^2+m^2+k^2=36$, and

$xl+ym+zk=30$.

Evaluate $\dfrac{x+y+z}{l+m+k}$.

 

[kaliprasad]

Spoiler

Take the square root of each equation.

X+Y+Z=5

L+M+K=6

$\frac{X+Y+Z}{L+M+K}= 5/6$

Spoiler

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$ and not $x^2+y^2+z^2$

 

[ineedhelpnow]

I made an oopsie. Thanks for pointing it out :)

Spoiler

Add the two equations together first.

$x^2 + l^2 + y^2 + m^2 + z^2 + k^2 = 25 + 36$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(xl + ym + zk) = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(30) = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 60 = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 = 121$

1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100
4 + 36 + 81 = 121

So, x+l, y+m, and z+k are 2,6,9 in any order

From $x^2 + y^2 + z^2 = 25$
3 solution sets:
5,0,0
0,5,0
0,0,5

From $l^2 + m^2 + k^2 = 36$
3 solution sets:
6,0,0
0,6,0
0,0,6

It can only be :

x=5,y=0,z=0,l=6,m=0,n=0 or
x=0,y=5,z=0,l=0,m=6,n=0 or
x=0,y=0,z=5,l=0,m=0,k=6

In any of those cases, (x+y+z)/(l+m+k) would be :

(5+0+0)/(6+0+0) or (0+5+0)/(0+6+0) or (0+0+5)/(0+0+6)

sooooooooooo 5/6 :p

 

[kaliprasad]

I made an oopsie. Thanks for pointing it out :)

Spoiler

Add the two equations together first.

$x^2 + l^2 + y^2 + m^2 + z^2 + k^2 = 25 + 36$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(xl + ym + zk) = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(30) = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 60 = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 = 121$

1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100
4 + 36 + 81 = 121

So, x+l, y+m, and z+k are 2,6,9 in any order

From $x^2 + y^2 + z^2 = 25$
3 solution sets:
5,0,0
0,5,0
0,0,5

From $l^2 + m^2 + k^2 = 36$
3 solution sets:
6,0,0
0,6,0
0,0,6

It can only be :

x=5,y=0,z=0,l=6,m=0,n=0 or
x=0,y=5,z=0,l=0,m=6,n=0 or
x=0,y=0,z=5,l=0,m=0,k=6

In any of those cases, (x+y+z)/(l+m+k) would be :

(5+0+0)/(6+0+0) or (0+5+0)/(0+6+0) or (0+0+5)/(0+0+6)

sooooooooooo 5/6 :p

Spoiler

No one told you that they are integers

 

[ineedhelpnow]

Spoiler

No one told you that they are integers

I don't know what else to try. Someone else can have a go at it

 

[anemone]

I made an oopsie. Thanks for pointing it out :)

Spoiler

Add the two equations together first.

$x^2 + l^2 + y^2 + m^2 + z^2 + k^2 = 25 + 36$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(xl + ym + zk) = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(30) = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 - 60 = 61$

$(x + l)^2 + (y + m)^2 + (z + k)^2 = 121$

1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100
4 + 36 + 81 = 121

So, x+l, y+m, and z+k are 2,6,9 in any order

From $x^2 + y^2 + z^2 = 25$
3 solution sets:
5,0,0
0,5,0
0,0,5

From $l^2 + m^2 + k^2 = 36$
3 solution sets:
6,0,0
0,6,0
0,0,6

It can only be :

x=5,y=0,z=0,l=6,m=0,n=0 or
x=0,y=5,z=0,l=0,m=6,n=0 or
x=0,y=0,z=5,l=0,m=0,k=6

In any of those cases, (x+y+z)/(l+m+k) would be :

(5+0+0)/(6+0+0) or (0+5+0)/(0+6+0) or (0+0+5)/(0+0+6)

sooooooooooo 5/6 :p

Hi ineedhelpnow!

I applaud you for taking a stab at this challenge, but kaliprasad is right; we are not told that the 6 variables are integers. The good news is, your end result $\dfrac{5}{6}$ is correct, so, in an effort to encourage more members to participate in my challenges, I will give you 50% for the correct answer, hehehe...:P

 

[Greg]

Spoiler

We have two spheres centered at the origin. If \(\displaystyle x,y,z\) and \(\displaystyle l,m,k\) are treated as the components of
two collinear vectors the equation \(\displaystyle xl+ym+zk=30\) always holds:

\(\displaystyle <x,y,z>\cdot<l,m,k>\,=5\cdot6\cos(0)=30\)

and so

\(\displaystyle \frac{x+y+z}{l+m+k}=\frac{|<x,y,z>|}{|<l,m,k>|}=\frac56\)

as required.

 

[anemone]

Spoiler

We have two spheres centered at the origin. If \(\displaystyle x,y,z\) and \(\displaystyle l,k,m\) are treated as the components of
two collinear vectors the equation \(\displaystyle xl+ym+zk=30\) always holds:

\(\displaystyle <x,y,z>\cdot<l,m,k>\,=5\cdot6\cos(0)=30\)

and so

\(\displaystyle \frac{x+y+z}{l+m+k}=\frac{|<x,y,z>|}{|<l,m,k>|}=\frac56\)

as required.

Bravo, greg1313! Thanks for participating!

 

[anemone]

Another method of other to solve for this challenge:

Spoiler

$x^2+y^2+z^2=25$ gives us $\left(\dfrac{x}{5}\right)^2+\left(\dfrac{y}{5}\right)^2+\left(\dfrac{z}{5}\right)^2=1$,

$l^2+m^2+k^2=36$ gives us $\left(\dfrac{l}{6}\right)^2+\left(\dfrac{m}{6}\right)^2+\left(\dfrac{k}{6}\right)^2=1$ and

$\dfrac{xl}{30}+\dfrac{ym}{30}+\dfrac{zk}{30}=1$

So $\left(\dfrac{x}{5}\right)^2+\left(\dfrac{y}{5}\right)^2+\left(\dfrac{z}{5}\right)^2-2\left(\dfrac{xl}{30}+\dfrac{ym}{30}+\dfrac{zk}{30}\right)+\left(\dfrac{l}{6}\right)^2+\left(\dfrac{m}{6}\right)^2+\left(\dfrac{k}{6}\right)^2=1-2+1=0$

This implies $\left(\dfrac{x}{5}-\dfrac{l}{6}\right)^2+\left(\dfrac{y}{5}-\dfrac{m}{6}\right)^2+\left(\dfrac{z}{5}-\dfrac{k}{6}\right)^2=0$

$\therefore \dfrac{x}{5}-\dfrac{l}{6}=0,\,\,\dfrac{y}{5}-\dfrac{m}{6}=0,\,\,\dfrac{z}{5}-\dfrac{k}{6}=0$

thus we have $x=\dfrac{5l}{6}$, $y=\dfrac{5m}{6}$ and $z=\dfrac{5k}{6}$

Hence,

$\dfrac{x+y+z}{l+m+k}=\dfrac{\dfrac{5l}{6}+\dfrac{5m}{6}+\dfrac{5k}{6}}{l+m+k}=\dfrac{5}{6}$