IF `bara,barb,barc` are three vectors such that each is inclined at an angle `pi/3` with the other two and `|bara|=1,|barb|=2,|barc|=3` then the scalar product of the vectors `2bara+3barb-5barc and 4bara-6barb+10barc` 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 \(2\mathbf{a} + 3\mathbf{b} - 5\mathbf{c}\) and \(4\mathbf{a} - 6\mathbf{b} + 10\mathbf{c}\), given that the vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are inclined at an angle of \(\frac{\pi}{3}\) with each other and have magnitudes \(|\mathbf{a}| = 1\), \(|\mathbf{b}| = 2\), and \(|\mathbf{c}| = 3\).

### Step-by-Step Solution:

1. **Identify the Magnitudes and Angles:**
- \(|\mathbf{a}| = 1\)
- \(|\mathbf{b}| = 2\)
- \(|\mathbf{c}| = 3\)
- The angle between any two vectors is \(\frac{\pi}{3}\).

2. **Calculate the Scalar Product:**
The scalar product of two vectors \(\mathbf{x}\) and \(\mathbf{y}\) can be calculated using the formula:
\[ \mathbf{x} \cdot \mathbf{y} = |\mathbf{x}| |\mathbf{y}| \cos(\theta) \]
where \(\theta\) is the angle between the vectors.

3. **Expand the Scalar Product:**
We need to compute:
\[ (2\mathbf{a} + 3\mathbf{b} - 5\mathbf{c}) \cdot (4\mathbf{a} - 6\mathbf{b} + 10\mathbf{c}) \]

Using the distributive property of the dot product, we have:
\[ = 2\mathbf{a} \cdot 4\mathbf{a} + 2\mathbf{a} \cdot (-6\mathbf{b}) + 2\mathbf{a} \cdot 10\mathbf{c} + 3\mathbf{b} \cdot 4\mathbf{a} + 3\mathbf{b} \cdot (-6\mathbf{b}) + 3\mathbf{b} \cdot 10\mathbf{c} - 5\mathbf{c} \cdot 4\mathbf{a} - 5\mathbf{c} \cdot (-6\mathbf{b}) - 5\mathbf{c} \cdot 10\mathbf{c} \]

4. **Calculate Each Term:**
- \(2\mathbf{a} \cdot 4\mathbf{a} = 8|\mathbf{a}|^2 = 8 \cdot 1^2 = 8\)
- \(2\mathbf{a} \cdot (-6\mathbf{b}) = -12\mathbf{a} \cdot \mathbf{b} = -12 \cdot |\mathbf{a}||\mathbf{b}|\cos\left(\frac{\pi}{3}\right) = -12 \cdot 1 \cdot 2 \cdot \frac{1}{2} = -12\)
- \(2\mathbf{a} \cdot 10\mathbf{c} = 20\mathbf{a} \cdot \mathbf{c} = 20 \cdot |\mathbf{a}||\mathbf{c}|\cos\left(\frac{\pi}{3}\right) = 20 \cdot 1 \cdot 3 \cdot \frac{1}{2} = 30\)
- \(3\mathbf{b} \cdot 4\mathbf{a} = 12\mathbf{b} \cdot \mathbf{a} = 12 \cdot |\mathbf{b}||\mathbf{a}|\cos\left(\frac{\pi}{3}\right) = 12 \cdot 2 \cdot 1 \cdot \frac{1}{2} = 12\)
- \(3\mathbf{b} \cdot (-6\mathbf{b}) = -18|\mathbf{b}|^2 = -18 \cdot 2^2 = -72\)
- \(3\mathbf{b} \cdot 10\mathbf{c} = 30\mathbf{b} \cdot \mathbf{c} = 30 \cdot |\mathbf{b}||\mathbf{c}|\cos\left(\frac{\pi}{3}\right) = 30 \cdot 2 \cdot 3 \cdot \frac{1}{2} = 90\)
- \(-5\mathbf{c} \cdot 4\mathbf{a} = -20\mathbf{c} \cdot \mathbf{a} = -20 \cdot |\mathbf{c}||\mathbf{a}|\cos\left(\frac{\pi}{3}\right) = -20 \cdot 3 \cdot 1 \cdot \frac{1}{2} = -30\)
- \(-5\mathbf{c} \cdot (-6\mathbf{b}) = 30\mathbf{c} \cdot \mathbf{b} = 30 \cdot |\mathbf{c}||\mathbf{b}|\cos\left(\frac{\pi}{3}\right) = 30 \cdot 3 \cdot 2 \cdot \frac{1}{2} = 90\)
- \(-5\mathbf{c} \cdot 10\mathbf{c} = -50|\mathbf{c}|^2 = -50 \cdot 3^2 = -450\)

5. **Combine All Terms:**
Now, we can combine all these results:
\[ 8 - 12 + 30 + 12 - 72 + 90 - 30 + 90 - 450 \]

Simplifying this:
\[ = 8 - 12 + 30 + 12 - 72 + 90 - 30 + 90 - 450 = 8 - 12 + 12 + 30 - 72 + 90 - 30 + 90 - 450 \]
\[ = 8 + 30 + 12 + 90 + 90 - 12 - 72 - 30 - 450 = 8 + 30 + 12 + 90 + 90 - 12 - 72 - 30 - 450 = 8 - 72 - 450 + 180 \]
\[ = 8 - 72 - 450 + 180 = -334 \]

Thus, the scalar product of the vectors is \(-334\).