If in triangle ABC, ` vec(AB) = vecu/|vecu|-vecv/|vecv| and vec(AC) = (2vecu)/|vecu| , " where " |vecu| ne |vecv|` , then `(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

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

Explanation

To solve the problem, we start by analyzing the vectors given in the triangle ABC.

### Step 1: Define the vectors
Given:
- \( \vec{AB} = \frac{\vec{u}}{|\vec{u}|} - \frac{\vec{v}}{|\vec{v}|} \)
- \( \vec{AC} = \frac{2\vec{u}}{|\vec{u}|} \)

### Step 2: Find the vector \( \vec{BC} \)
Using the property of triangles that \( \vec{AB} + \vec{BC} + \vec{CA} = 0 \), we can express \( \vec{BC} \) as:
\[ \vec{BC} = -\vec{AB} - \vec{AC} \]
Substituting the expressions for \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{BC} = -\left(\frac{\vec{u}}{|\vec{u}|} - \frac{\vec{v}}{|\vec{v}|}\right) - \frac{2\vec{u}}{|\vec{u}|} \]
Simplifying this, we get:
\[ \vec{BC} = -\frac{\vec{u}}{|\vec{u}|} + \frac{\vec{v}}{|\vec{v}|} - \frac{2\vec{u}}{|\vec{u}|} = \frac{\vec{v}}{|\vec{v}|} - \frac{3\vec{u}}{|\vec{u}|} \]

### Step 3: Calculate the magnitudes of the vectors
Now we will find the magnitudes of the vectors:
- \( |\vec{AB}| = \left| \frac{\vec{u}}{|\vec{u}|} - \frac{\vec{v}}{|\vec{v}|} \right| \)
- \( |\vec{AC}| = \left| \frac{2\vec{u}}{|\vec{u}|} \right| = 2 \)
- \( |\vec{BC}| = \left| \frac{\vec{v}}{|\vec{v}|} - \frac{3\vec{u}}{|\vec{u}|} \right| \)

### Step 4: Establish relationships between the angles
Using the cosine rule in triangle ABC:
\[ |\vec{AC}|^2 = |\vec{AB}|^2 + |\vec{BC}|^2 - 2|\vec{AB}||\vec{BC}|\cos C \]
Substituting the known values and simplifying will help us establish relationships between angles A, B, and C.

### Step 5: Check the options
1. **Option (a)**: \( 1 + \cos 2A + \cos 2B + \cos 2C = 0 \)
- Since angle B is 90 degrees, we can substitute \( \cos B = 0 \) and find that the equation holds true.

2. **Option (b)**: \( \sin A = \cos C \)
- Since \( A + C = 90^\circ \), this relationship holds true.

3. **Option (c)**: Projection of \( AC \) on \( BC \) is equal to \( BC \)
- Since \( AC \) is the hypotenuse, its projection on the side \( BC \) will always equal \( BC \).

4. **Option (d)**: Projection of \( AB \) on \( BC \) is equal to \( AB \)
- Since \( AB \) and \( BC \) are perpendicular, the projection will be 0, not equal to \( AB \).

### Conclusion
The correct options are (a), (b), and (c). The incorrect option is (d).