If `vec(A)` and `vec(B)` are two non`-`zero vectors such that `|vec(A)+vec(B)|=(|vec(A)-vec(B)|)/(2)` and `|vec(A)|=2|vec(B)|` then the angle between `vec(A)` and `vec(B)` is `:`

Options

A. `37^(@)`
B. `53^(@)`
C. `cos^(-1)(-3//4)`
D. `cos^(-1)(-4//3)`

Explanation

To solve the problem, we need to find the angle between two vectors \(\vec{A}\) and \(\vec{B}\) given the conditions:

1. \(|\vec{A} + \vec{B}| = \frac{|\vec{A} - \vec{B}|}{2}\)
2. \(|\vec{A}| = 2 |\vec{B}|\)

Let's denote:
- \(|\vec{A}| = A\)
- \(|\vec{B}| = B\)

From the second condition, we have:
\[ A = 2B \]

Now, substituting this into the first condition:

### Step 1: Express the magnitudes in terms of \(A\) and \(B\)

Using the formula for the magnitude of the sum and difference of two vectors, we have:

\[ |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
\[ |\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta} \]

### Step 2: Substitute the expressions into the first condition

According to the first condition:
\[ \sqrt{A^2 + B^2 + 2AB \cos \theta} = \frac{1}{2} \sqrt{A^2 + B^2 - 2AB \cos \theta} \]

### Step 3: Square both sides to eliminate the square roots

Squaring both sides gives:
\[ A^2 + B^2 + 2AB \cos \theta = \frac{1}{4} (A^2 + B^2 - 2AB \cos \theta) \]

### Step 4: Clear the fraction by multiplying through by 4

\[ 4(A^2 + B^2 + 2AB \cos \theta) = A^2 + B^2 - 2AB \cos \theta \]

### Step 5: Rearranging the equation

This simplifies to:
\[ 4A^2 + 4B^2 + 8AB \cos \theta = A^2 + B^2 - 2AB \cos \theta \]
\[ 4A^2 + 4B^2 - A^2 - B^2 + 10AB \cos \theta = 0 \]
\[ (4A^2 + 3B^2) + 10AB \cos \theta = 0 \]

### Step 6: Substitute \(A = 2B\)

Substituting \(A = 2B\):
\[ 4(2B)^2 + 3B^2 + 10(2B)B \cos \theta = 0 \]
\[ 16B^2 + 3B^2 + 20B^2 \cos \theta = 0 \]
\[ 19B^2 + 20B^2 \cos \theta = 0 \]

### Step 7: Solve for \(\cos \theta\)

Factoring out \(B^2\) (since \(B \neq 0\)):
\[ 19 + 20 \cos \theta = 0 \]
\[ 20 \cos \theta = -19 \]
\[ \cos \theta = -\frac{19}{20} \]

### Step 8: Find the angle \(\theta\)

Thus, the angle \(\theta\) is:
\[ \theta = \cos^{-1}\left(-\frac{19}{20}\right) \]

### Final Answer

The angle between \(\vec{A}\) and \(\vec{B}\) is \(\cos^{-1}\left(-\frac{19}{20}\right)\).

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