Let `veca and vecb` are unit vectors such that `|veca+vecb|=(3)/(2)`, then the value of `(2veca+7vecb).(4veca+3vecb+2020vecaxxvecb)` is equal to

Options

A. `(133)/(4)`
B. 133
C. 30
D. 120

Answer

To solve the problem, we need to find the value of \((2\vec{a} + 7\vec{b}) \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b})\) given that \(|\vec{a} + \vec{b}| = \frac{3}{2}\) and \(\vec{a}\) and \(\vec{b}\) are unit vectors.

### Step-by-step Solution:

1. **Given Information:**
- \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\) (since they are unit vectors).
- \(|\vec{a} + \vec{b}| = \frac{3}{2}\).

2. **Square the Magnitude:**
\[ |\vec{a} + \vec{b}|^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]
Expanding the left side using the formula for the magnitude of a vector:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} \]
Substituting the values:
\[ 1 + 1 + 2\vec{a} \cdot \vec{b} = \frac{9}{4} \]
This simplifies to:
\[ 2 + 2\vec{a} \cdot \vec{b} = \frac{9}{4} \]

3. **Solve for \(\vec{a} \cdot \vec{b}\):**
\[ 2\vec{a} \cdot \vec{b} = \frac{9}{4} - 2 = \frac{9}{4} - \frac{8}{4} = \frac{1}{4} \]
Therefore:
\[ \vec{a} \cdot \vec{b} = \frac{1}{8} \]

4. **Calculate the Dot Product:**
We need to evaluate:
\[ (2\vec{a} + 7\vec{b}) \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b}) \]
Expanding this:
\[ = 2\vec{a} \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b}) + 7\vec{b} \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b}) \]

5. **Calculate Each Dot Product:**
- For \(2\vec{a} \cdot (4\vec{a})\):
\[ 2 \cdot 4(\vec{a} \cdot \vec{a}) = 8 \cdot 1 = 8 \]
- For \(2\vec{a} \cdot (3\vec{b})\):
\[ 2 \cdot 3(\vec{a} \cdot \vec{b}) = 6 \cdot \frac{1}{8} = \frac{3}{4} \]
- For \(2\vec{a} \cdot (2020\vec{a} \times \vec{b})\):
\[ \text{This is zero since } \vec{a} \cdot (\vec{a} \times \vec{b}) = 0. \]

Now for \(7\vec{b} \cdot (4\vec{a})\):
- For \(7\vec{b} \cdot (4\vec{a})\):
\[ 7 \cdot 4(\vec{b} \cdot \vec{a}) = 28 \cdot \frac{1}{8} = \frac{7}{2} \]
- For \(7\vec{b} \cdot (3\vec{b})\):
\[ 7 \cdot 3(\vec{b} \cdot \vec{b}) = 21 \cdot 1 = 21 \]
- For \(7\vec{b} \cdot (2020\vec{a} \times \vec{b})\):
\[ \text{This is zero since } \vec{b} \cdot (\vec{a} \times \vec{b}) = 0. \]

6. **Combine All Results:**
\[ 8 + \frac{3}{4} + \frac{7}{2} + 21 \]
Converting everything to a common denominator (4):
\[ = 8 + \frac{3}{4} + \frac{14}{4} + \frac{84}{4} = \frac{32}{4} + \frac{3}{4} + \frac{14}{4} + \frac{84}{4} = \frac{133}{4} \]

### Final Answer:
\[ \frac{133}{4} \]