A straight line L cuts the sides AB, AC, AD of a parallelogram ABCD at `B_(1), C_(1), d_(1)` respectively. If `vec(AB_(1))=lambda_(1)vec(AB), vec(AD_(1))=lambda_(2)vec(AD) and vec(AC_(1))=lambda_(3)vec(AC),` then `(1)/(lambda_(3))` equal to

Options

A. `(1)/(lambda_(1))+(1)/(lambda_(2))`
B. `(1)/(lambda_(1))-(1)/(lambda_(2))`
C. `-lambda_(1)+lambda_(2)`
D. `lambda_(1)+lambda_(2)`

Answer

To solve the problem, we need to analyze the given information about the parallelogram ABCD and the points where the line L intersects the sides AB, AC, and AD. We are given the following relations:

1. \(\vec{AB_1} = \lambda_1 \vec{AB}\)
2. \(\vec{AD_1} = \lambda_2 \vec{AD}\)
3. \(\vec{AC_1} = \lambda_3 \vec{AC}\)

We need to find the value of \(\frac{1}{\lambda_3}\).

### Step 1: Define the vectors
Let:
- \(\vec{A} = \vec{0}\) (the origin)
- \(\vec{B} = \vec{b}\)
- \(\vec{C} = \vec{b} + \vec{d}\)
- \(\vec{D} = \vec{d}\)

From this, we can express the vectors:
- \(\vec{AB} = \vec{B} - \vec{A} = \vec{b}\)
- \(\vec{AD} = \vec{D} - \vec{A} = \vec{d}\)
- \(\vec{AC} = \vec{C} - \vec{A} = \vec{b} + \vec{d}\)

### Step 2: Express the intersection points
Using the given relations:
- \(\vec{AB_1} = \lambda_1 \vec{b}\)
- \(\vec{AD_1} = \lambda_2 \vec{d}\)
- \(\vec{AC_1} = \lambda_3 (\vec{b} + \vec{d})\)

### Step 3: Find the vector relations
The points \(B_1\), \(C_1\), and \(D_1\) can be expressed as:
- \(\vec{B_1} = \lambda_1 \vec{b}\)
- \(\vec{C_1} = \lambda_3 (\vec{b} + \vec{d})\)
- \(\vec{D_1} = \lambda_2 \vec{d}\)

### Step 4: Establish collinearity
Since the points \(B_1\), \(C_1\), and \(D_1\) are collinear, we can set up the relationship:
\[ \vec{C_1} - \vec{D_1} = k(\vec{B_1} - \vec{D_1}) \]
for some scalar \(k\).

Substituting the expressions for \(\vec{C_1}\), \(\vec{D_1}\), and \(\vec{B_1}\):
\[ \lambda_3 (\vec{b} + \vec{d}) - \lambda_2 \vec{d} = k(\lambda_1 \vec{b} - \lambda_2 \vec{d}) \]

### Step 5: Rearranging the equation
Expanding and rearranging gives:
\[ \lambda_3 \vec{b} + (\lambda_3 - \lambda_2) \vec{d} = k(\lambda_1 \vec{b} - \lambda_2 \vec{d}) \]

### Step 6: Equate coefficients
From the above equation, we can equate the coefficients of \(\vec{b}\) and \(\vec{d}\):
1. For \(\vec{b}\): \(\lambda_3 = k \lambda_1\)
2. For \(\vec{d}\): \(\lambda_3 - \lambda_2 = -k \lambda_2\)

### Step 7: Solve for \(k\)
From the first equation, we have:
\[ k = \frac{\lambda_3}{\lambda_1} \]

Substituting \(k\) into the second equation:
\[ \lambda_3 - \lambda_2 = -\frac{\lambda_3}{\lambda_1} \lambda_2 \]
Rearranging gives:
\[ \lambda_3 (1 + \frac{\lambda_2}{\lambda_1}) = \lambda_2 \]

### Step 8: Solve for \(\lambda_3\)
Thus, we can express \(\lambda_3\) as:
\[ \lambda_3 = \frac{\lambda_2}{1 + \frac{\lambda_2}{\lambda_1}} = \frac{\lambda_2 \lambda_1}{\lambda_1 + \lambda_2} \]

### Step 9: Find \(\frac{1}{\lambda_3}\)
Finally, we find:
\[ \frac{1}{\lambda_3} = \frac{\lambda_1 + \lambda_2}{\lambda_2} \]

### Conclusion
Thus, the required value of \(\frac{1}{\lambda_3}\) is:
\[ \frac{1}{\lambda_3} = \frac{\lambda_1 + \lambda_2}{\lambda_2} \]