In a trapezium ABCD the vector `B vec C = lambda 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=lamda+1`
B. `lamda=mu+1`
C. `lamda+mu=1`
D. `mu=2+lamda`

Explanation

To solve the problem step by step, we will analyze the given information and use vector algebra to derive the required relationship between the scalars \( \lambda \) and \( \mu \).

### Step 1: Understand the Geometry of the Trapezium
We have a trapezium \( ABCD \) with the following properties:
- \( \vec{BC} = \lambda \vec{AD} \)
- \( \vec{p} = \vec{AC} + \vec{BD} \) is collinear with \( \vec{AD} \) such that \( \vec{p} = \mu \vec{AD} \)

### Step 2: Express Vectors in Terms of Points
Let’s denote the position vectors of points \( A, B, C, D \) as \( \vec{A}, \vec{B}, \vec{C}, \vec{D} \).
From the given information, we can express \( \vec{BC} \) as:
\[ \vec{BC} = \vec{C} - \vec{B} = \lambda (\vec{D} - \vec{A}) \]

### Step 3: Write the Expression for \( \vec{p} \)
Next, we express \( \vec{p} \):
\[ \vec{p} = \vec{AC} + \vec{BD} \]
This can be rewritten as:
\[ \vec{p} = (\vec{C} - \vec{A}) + (\vec{D} - \vec{B}) \]
Substituting \( \vec{A} \) as the origin (i.e., \( \vec{A} = \vec{0} \)):
\[ \vec{p} = \vec{C} + \vec{D} - \vec{B} \]

### Step 4: Substitute \( \vec{B} \) and Rearrange
Since \( \vec{B} = \mu \vec{AD} \), we can express \( \vec{B} \) as:
\[ \vec{B} = \mu (\vec{D} - \vec{A}) = \mu \vec{D} \]
Now substituting this into the expression for \( \vec{p} \):
\[ \vec{p} = \vec{C} + \vec{D} - \mu \vec{D} \]
This simplifies to:
\[ \vec{p} = \vec{C} + (1 - \mu) \vec{D} \]

### Step 5: Set Up the Collinearity Condition
Since \( \vec{p} \) is collinear with \( \vec{AD} \), we can write:
\[ \vec{p} = \mu \vec{AD} = \mu \vec{D} \]
Equating the two expressions for \( \vec{p} \):
\[ \vec{C} + (1 - \mu) \vec{D} = \mu \vec{D} \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ \vec{C} = \mu \vec{D} - (1 - \mu) \vec{D} \]
This simplifies to:
\[ \vec{C} = (2\mu - 1) \vec{D} \]

### Step 7: Substitute Back to Find the Relationship
Now we can use the earlier relationship \( \vec{BC} = \lambda \vec{AD} \):
\[ \vec{C} - \vec{B} = \lambda \vec{D} \]
Substituting \( \vec{B} = \mu \vec{D} \) into this gives:
\[ (2\mu - 1) \vec{D} - \mu \vec{D} = \lambda \vec{D} \]
This simplifies to:
\[ (2\mu - 1 - \mu) \vec{D} = \lambda \vec{D} \]
Thus, we have:
\[ (\mu - 1) \vec{D} = \lambda \vec{D} \]

### Step 8: Final Relationship
Since \( \vec{D} \neq \vec{0} \), we can equate the coefficients:
\[ \mu - 1 = \lambda \]
Therefore, we find:
\[ \mu = \lambda + 1 \]

### Conclusion
The required relationship is:
\[ \mu = \lambda + 1 \]