IF `veca,vecb,vecc` are three vectors such that each is inclined at an angle `pi/3` with the other two and `|veca|=1,|vecb|=2,|vecc|=3` then the scalar product of the vectors `2veca+3vecb-5vecc and 4veca-6vecb+10vecc` is equal to

Options

A. -334
B. 188
C. -522
D. -514

Answer

To solve the problem, we need to find the scalar product of the vectors \( \vec{u} = 2\vec{a} + 3\vec{b} - 5\vec{c} \) and \( \vec{v} = 4\vec{a} - 6\vec{b} + 10\vec{c} \).

### Step 1: Write the scalar product
The scalar product (dot product) of the vectors \( \vec{u} \) and \( \vec{v} \) can be expressed as:
\[ \vec{u} \cdot \vec{v} = (2\vec{a} + 3\vec{b} - 5\vec{c}) \cdot (4\vec{a} - 6\vec{b} + 10\vec{c}) \]

### Step 2: Expand the dot product
Using the distributive property of dot product, we expand:
\[ \vec{u} \cdot \vec{v} = 2\vec{a} \cdot 4\vec{a} + 2\vec{a} \cdot (-6\vec{b}) + 2\vec{a} \cdot 10\vec{c} + 3\vec{b} \cdot 4\vec{a} + 3\vec{b} \cdot (-6\vec{b}) + 3\vec{b} \cdot 10\vec{c} - 5\vec{c} \cdot 4\vec{a} - 5\vec{c} \cdot (-6\vec{b}) - 5\vec{c} \cdot 10\vec{c} \]

### Step 3: Simplify the terms
Calculating each term:
1. \( 2\vec{a} \cdot 4\vec{a} = 8|\vec{a}|^2 \)
2. \( 2\vec{a} \cdot (-6\vec{b}) = -12\vec{a} \cdot \vec{b} \)
3. \( 2\vec{a} \cdot 10\vec{c} = 20\vec{a} \cdot \vec{c} \)
4. \( 3\vec{b} \cdot 4\vec{a} = 12\vec{b} \cdot \vec{a} \)
5. \( 3\vec{b} \cdot (-6\vec{b}) = -18|\vec{b}|^2 \)
6. \( 3\vec{b} \cdot 10\vec{c} = 30\vec{b} \cdot \vec{c} \)
7. \( -5\vec{c} \cdot 4\vec{a} = -20\vec{c} \cdot \vec{a} \)
8. \( -5\vec{c} \cdot (-6\vec{b}) = 30\vec{c} \cdot \vec{b} \)
9. \( -5\vec{c} \cdot 10\vec{c} = -50|\vec{c}|^2 \)

Combining these, we get:
\[ \vec{u} \cdot \vec{v} = 8|\vec{a}|^2 + (-12 + 12)\vec{a} \cdot \vec{b} + (20 - 20)\vec{a} \cdot \vec{c} - 18|\vec{b}|^2 + 60\vec{b} \cdot \vec{c} - 50|\vec{c}|^2 \]

### Step 4: Substitute the magnitudes and angles
Given:
- \( |\vec{a}| = 1 \)
- \( |\vec{b}| = 2 \)
- \( |\vec{c}| = 3 \)
- The angle between each pair of vectors is \( \frac{\pi}{3} \), so \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).

Calculating the dot products:
- \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\left(\frac{\pi}{3}\right) = 1 \cdot 2 \cdot \frac{1}{2} = 1 \)
- \( \vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos\left(\frac{\pi}{3}\right) = 1 \cdot 3 \cdot \frac{1}{2} = \frac{3}{2} \)
- \( \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos\left(\frac{\pi}{3}\right) = 2 \cdot 3 \cdot \frac{1}{2} = 3 \)

### Step 5: Substitute these values into the equation
Now substituting back into the equation:
\[ \vec{u} \cdot \vec{v} = 8(1^2) - 18(2^2) + 60(3) - 50(3^2) \]
Calculating:
\[ = 8 - 18(4) + 180 - 50(9) \]
\[ = 8 - 72 + 180 - 450 \]
\[ = 8 - 72 + 180 - 450 = -334 \]

### Final Answer
Thus, the scalar product of the vectors \( \vec{u} \) and \( \vec{v} \) is:
\[ \boxed{-334} \]