In a trapezium ABCD the vector `B vec C = alpha vec(AD).` If `vec p = A vec C + vec(BD)` is coillinear with `vec(AD)` such that `vec p = mu vec (AD),` then

Options

A. `mu= alpha +2`
B. `mu+ alpha =1`
C. `alpha = mu +1`
D. `mu =alpha +1`

Explanation

To solve the problem step by step, we will analyze the given information and derive the necessary relationships.

### Step 1: Understand the given vectors and relationships
We have a trapezium ABCD with the vector relationship given as:
\[ \vec{BC} = \alpha \vec{AD} \]
This means that the vector from B to C is a scalar multiple of the vector from A to D.

### Step 2: Express the vectors in terms of position vectors
Let’s denote the position vectors of points A, B, C, and D as follows:
- \(\vec{A} = \vec{0}\) (taking A as the origin)
- \(\vec{B} = \vec{b}\)
- \(\vec{C} = \vec{c}\)
- \(\vec{D} = \vec{d}\)

From the relationship \(\vec{BC} = \alpha \vec{AD}\), we can express this as:
\[ \vec{C} - \vec{B} = \alpha (\vec{D} - \vec{A}) \implies \vec{c} - \vec{b} = \alpha \vec{d} \]

### Step 3: Express vector \(\vec{p}\)
The vector \(\vec{p}\) is given as:
\[ \vec{p} = \vec{AC} + \vec{BD} \]
Expressing this in terms of position vectors:
\[ \vec{p} = (\vec{C} - \vec{A}) + (\vec{D} - \vec{B}) = \vec{c} + \vec{d} - \vec{b} \]

### Step 4: Use the collinearity condition
We know that \(\vec{p}\) is collinear with \(\vec{AD}\), which means:
\[ \vec{p} = \mu \vec{AD} \implies \vec{p} = \mu (\vec{D} - \vec{A}) = \mu \vec{d} \]
Thus, we have:
\[ \vec{c} + \vec{d} - \vec{b} = \mu \vec{d} \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ \vec{c} - \vec{b} = \mu \vec{d} - \vec{d} \]
This simplifies to:
\[ \vec{c} - \vec{b} = (\mu - 1) \vec{d} \]

### Step 6: Substitute \(\vec{c} - \vec{b}\) from Step 2
From Step 2, we know that:
\[ \vec{c} - \vec{b} = \alpha \vec{d} \]
Substituting this into the equation from Step 5 gives:
\[ \alpha \vec{d} = (\mu - 1) \vec{d} \]

### Step 7: Equating coefficients
Since \(\vec{d} \neq \vec{0}\), we can equate the coefficients:
\[ \alpha = \mu - 1 \]
Rearranging this gives:
\[ \mu = \alpha + 1 \]

### Conclusion
Thus, the relationship between \(\mu\) and \(\alpha\) is:
\[ \mu = \alpha + 1 \]