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 will follow these steps:

### Step 1: Understand the given information
We have two unit vectors \(\vec{a}\) and \(\vec{b}\) such that \(|\vec{a}| = 1\), \(|\vec{b}| = 1\), and \(|\vec{a} + \vec{b}| = \frac{3}{2}\).

### Step 2: Square the equation
We square the equation \(|\vec{a} + \vec{b}| = \frac{3}{2}\):
\[ |\vec{a} + \vec{b}|^2 = \left(\frac{3}{2}\right)^2 \]
This gives us:
\[ |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = \frac{9}{4} \]
Since \(|\vec{a}|^2 = 1\) and \(|\vec{b}|^2 = 1\), we have:
\[ 1 + 1 + 2\vec{a} \cdot \vec{b} = \frac{9}{4} \]
Thus,
\[ 2 + 2\vec{a} \cdot \vec{b} = \frac{9}{4} \]
Subtracting 2 from both sides:
\[ 2\vec{a} \cdot \vec{b} = \frac{9}{4} - 2 = \frac{9}{4} - \frac{8}{4} = \frac{1}{4} \]
So,
\[ \vec{a} \cdot \vec{b} = \frac{1}{8} \]

### Step 3: Calculate the dot product
Now we need to find the value of \((2\vec{a} + 7\vec{b}) \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b})\).

Expanding this dot product:
\[ (2\vec{a} + 7\vec{b}) \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b}) = 2\vec{a} \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b}) + 7\vec{b} \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b}) \]

### Step 4: Calculate each term
Calculating the first term:
\[ 2\vec{a} \cdot (4\vec{a}) + 2\vec{a} \cdot (3\vec{b}) + 2\vec{a} \cdot (2020\vec{a} \times \vec{b}) \]
\[ = 8 + 6\vec{a} \cdot \vec{b} + 0 \quad (\text{since } \vec{a} \cdot (\vec{a} \times \vec{b}) = 0) \]
Substituting \(\vec{a} \cdot \vec{b} = \frac{1}{8}\):
\[ = 8 + 6 \cdot \frac{1}{8} = 8 + \frac{6}{8} = 8 + \frac{3}{4} = 8.75 \]

Now for the second term:
\[ 7\vec{b} \cdot (4\vec{a}) + 7\vec{b} \cdot (3\vec{b}) + 7\vec{b} \cdot (2020\vec{a} \times \vec{b}) \]
\[ = 28\vec{b} \cdot \vec{a} + 21 + 0 \quad (\text{since } \vec{b} \cdot (\vec{a} \times \vec{b}) = 0) \]
Substituting \(\vec{b} \cdot \vec{a} = \frac{1}{8}\):
\[ = 28 \cdot \frac{1}{8} + 21 = \frac{28}{8} + 21 = 3.5 + 21 = 24.5 \]

### Step 5: Combine the results
Now, we combine both parts:
\[ 8.75 + 24.5 = 33.25 \]

### Final Result
Thus, the value of \((2\vec{a} + 7\vec{b}) \cdot (4\vec{a} + 3\vec{b} + 2020\vec{a} \times \vec{b})\) is:
\[ \boxed{33.25} \]