If a, b, c, x, y, z are real and `a^(2)+b^(2) + c^(2)=25, x^(2)+y^(2)+z^(2)=36 and ax+by+cz=30`, then `(a+b+c)/(x+y+z)` is equal to :

Options

A. 1
B. `(6)/(5)`
C. `(5)/(6)`
D. `(3)/(4)`

Answer

To solve the problem step by step, we need to find the value of \((a+b+c)/(x+y+z)\) given the conditions:

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

### Step 1: Analyze the equations
We have three equations to work with. The first two equations represent the squares of the magnitudes of vectors in 3D space, and the third equation represents the dot product of these vectors.

### Step 2: Choose values for \(b\) and \(c\)
To simplify our calculations, we can assume \(b = 0\) and \(c = 0\). This gives us:
- From \(a^2 + b^2 + c^2 = 25\), we have:
\[ a^2 = 25 \implies a = \pm 5 \]

### Step 3: Choose values for \(y\) and \(z\)
Similarly, we can assume \(y = 0\) and \(z = 0\). This gives us:
- From \(x^2 + y^2 + z^2 = 36\), we have:
\[ x^2 = 36 \implies x = \pm 6 \]

### Step 4: Check the dot product condition
Now we need to check if these values satisfy the dot product condition \(ax + by + cz = 30\):
- Substituting \(a = 5\) and \(x = 6\):
\[ ax + by + cz = 5 \cdot 6 + 0 + 0 = 30 \]
This condition is satisfied.

### Step 5: Calculate \((a+b+c)/(x+y+z)\)
Now we can substitute the values we have:
- \(a + b + c = 5 + 0 + 0 = 5\)
- \(x + y + z = 6 + 0 + 0 = 6\)

Thus,
\[ \frac{a+b+c}{x+y+z} = \frac{5}{6} \]

### Step 6: Consider negative values
If we consider the negative values \(a = -5\) and \(x = -6\), we still find:
\[ (-5) + 0 + 0 = -5 \]
\[ (-6) + 0 + 0 = -6 \]
Thus,
\[ \frac{-5}{-6} = \frac{5}{6} \]

### Final Answer
The value of \(\frac{a+b+c}{x+y+z}\) is \(\frac{5}{6}\).