In triangle ABC if `bar(AB) =u/(|bar u|)-v/(|bar v|) and bar(AC)=(2bar u)/(|baru|)` where `|bar u| != |bar |v|` then

Options

A. 1+ cos 2A +cos 2B + cos 2C=0
B. sin A = cos C
C. projection of AC on BC is equal to BC
D. projection of AB on BC is equal to AB

Answer

To solve the problem step by step, we will analyze the given vectors and their relationships in triangle ABC.

### Step 1: Understand the Given Vectors
We have:
- \( \overline{AB} = \frac{\mathbf{u}}{|\mathbf{u}|} - \frac{\mathbf{v}}{|\mathbf{v}|} \)
- \( \overline{AC} = \frac{2\mathbf{u}}{|\mathbf{u}|} \)

### Step 2: Rewrite the Vectors in Unit Vector Form
The vectors \( \frac{\mathbf{u}}{|\mathbf{u}|} \) and \( \frac{\mathbf{v}}{|\mathbf{v}|} \) are unit vectors. Let's denote:
- \( \mathbf{u}_{\text{cap}} = \frac{\mathbf{u}}{|\mathbf{u}|} \)
- \( \mathbf{v}_{\text{cap}} = \frac{\mathbf{v}}{|\mathbf{v}|} \)

Thus, we can rewrite:
- \( \overline{AB} = \mathbf{u}_{\text{cap}} - \mathbf{v}_{\text{cap}} \)
- \( \overline{AC} = 2\mathbf{u}_{\text{cap}} \)

### Step 3: Assign Points in the Triangle
Assuming point A is at the origin (0, 0):
- Let \( A = \mathbf{0} \)
- \( B = \overline{AB} = \mathbf{u}_{\text{cap}} - \mathbf{v}_{\text{cap}} \)
- \( C = \overline{AC} = 2\mathbf{u}_{\text{cap}} \)

### Step 4: Calculate the Magnitudes of the Vectors
1. **Magnitude of \( \overline{AB} \)**:
\[ |\overline{AB}| = |\mathbf{u}_{\text{cap}} - \mathbf{v}_{\text{cap}}| \]
Since both \( \mathbf{u}_{\text{cap}} \) and \( \mathbf{v}_{\text{cap}} \) are unit vectors:
\[ |\overline{AB}| = \sqrt{|\mathbf{u}_{\text{cap}}|^2 + |\mathbf{v}_{\text{cap}}|^2 - 2|\mathbf{u}_{\text{cap}}||\mathbf{v}_{\text{cap}}|\cos(\theta)} \]
where \( \theta \) is the angle between \( \mathbf{u}_{\text{cap}} \) and \( \mathbf{v}_{\text{cap}} \).

2. **Magnitude of \( \overline{AC} \)**:
\[ |\overline{AC}| = |2\mathbf{u}_{\text{cap}}| = 2 \]

### Step 5: Calculate \( \overline{BC} \)
To find \( \overline{BC} \):
\[ \overline{BC} = \overline{AC} - \overline{AB} = 2\mathbf{u}_{\text{cap}} - (\mathbf{u}_{\text{cap}} - \mathbf{v}_{\text{cap}}) = \mathbf{u}_{\text{cap}} + \mathbf{v}_{\text{cap}} \]

### Step 6: Magnitude of \( \overline{BC} \)
\[ |\overline{BC}| = |\mathbf{u}_{\text{cap}} + \mathbf{v}_{\text{cap}}| = \sqrt{|\mathbf{u}_{\text{cap}}|^2 + |\mathbf{v}_{\text{cap}}|^2 + 2|\mathbf{u}_{\text{cap}}||\mathbf{v}_{\text{cap}}|\cos(\phi)} \]
where \( \phi \) is the angle between \( \mathbf{u}_{\text{cap}} \) and \( \mathbf{v}_{\text{cap}} \).

### Step 7: Apply Pythagorean Theorem
Since we have a triangle, we can apply the Pythagorean theorem:
\[ |\overline{AC}|^2 = |\overline{AB}|^2 + |\overline{BC}|^2 \]
Substituting the magnitudes we found:
\[ 2^2 = |\overline{AB}|^2 + |\overline{BC}|^2 \]

### Step 8: Determine Angles
From the triangle properties, we can find angles:
- If \( |\overline{AB}| = |\overline{BC}| \), then angles at A and C are equal.

### Step 9: Final Calculation
Using the angles, we can find \( 2A, 2B, 2C \) and calculate \( 1 + \cos(2A) + \cos(2B) + \cos(2C) \).