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, 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-by-Step Solution:

**Step 1: Use Cauchy-Schwarz Inequality**

We can apply the Cauchy-Schwarz inequality, which states that:

\[ (ax + by + cz)^2 \leq (a^2 + b^2 + c^2)(x^2 + y^2 + z^2) \]

Substituting the given values into the inequality:

\[ 30^2 \leq (25)(36) \]

Calculating both sides:

\[ 900 \leq 900 \]

This shows that the equality holds, which implies that the vectors \((a, b, c)\) and \((x, y, z)\) are proportional.

**Step 2: Set Proportionality Constants**

Since the equality holds in the Cauchy-Schwarz inequality, we can say:

\[ \frac{a}{x} = \frac{b}{y} = \frac{c}{z} = k \]

for some constant \(k\).

**Step 3: Express \(a, b, c\) in terms of \(x, y, z\)**

From the proportionality, we can express \(a, b, c\) as:

\[ a = kx, \quad b = ky, \quad c = kz \]

**Step 4: Substitute into the first equation**

Now substitute these into the first equation \(a^2 + b^2 + c^2 = 25\):

\[ (kx)^2 + (ky)^2 + (kz)^2 = 25 \]

This simplifies to:

\[ k^2(x^2 + y^2 + z^2) = 25 \]

Using \(x^2 + y^2 + z^2 = 36\):

\[ k^2(36) = 25 \]

Thus,

\[ k^2 = \frac{25}{36} \implies k = \frac{5}{6} \quad (\text{since } k \text{ is positive}) \]

**Step 5: Find \(a + b + c\)**

Now we can find \(a + b + c\):

\[ a + b + c = kx + ky + kz = k(x + y + z) = \frac{5}{6}(x + y + z) \]

**Step 6: Find \((a + b + c)/(x + y + z)\)**

Now we can find the desired ratio:

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

### Final Answer:

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