If `vecOA=veca,vecOB=vecb,vecOC=2veca+3vecb,vecOD=veca-2vecb`, the length of `vecOA` is three times the length of `vecOB` and `vecOA` is perpendicular to `vecDB`, then `(vecBDxxvecAC).(vecODxxvecOC)` is

Options

A. `7vecaxxvecb|^(2)`
B. `42|vecaxxvecb|^(2)`
C. 0
D. 1

Answer

To solve the problem step by step, we will start by defining the vectors based on the information given in the question.

### Step 1: Define the vectors
Given:
- \( \vec{OA} = \vec{a} \)
- \( \vec{OB} = \vec{b} \)
- \( \vec{OC} = 2\vec{a} + 3\vec{b} \)
- \( \vec{OD} = \vec{a} - 2\vec{b} \)

### Step 2: Use the length condition
We know that the length of \( \vec{OA} \) is three times the length of \( \vec{OB} \):
\[ |\vec{a}| = 3|\vec{b}| \]
Let \( |\vec{b}| = k \), then \( |\vec{a}| = 3k \).

### Step 3: Find \( \vec{BD} \) and \( \vec{AC} \)
- The vector \( \vec{BD} \) can be calculated as:
\[ \vec{BD} = \vec{OD} - \vec{OB} = (\vec{a} - 2\vec{b}) - \vec{b} = \vec{a} - 3\vec{b} \]

- The vector \( \vec{AC} \) can be calculated as:
\[ \vec{AC} = \vec{OC} - \vec{OA} = (2\vec{a} + 3\vec{b}) - \vec{a} = \vec{a} + 3\vec{b} \]

### Step 4: Compute \( \vec{BD} \times \vec{AC} \)
Now we compute the cross product:
\[ \vec{BD} \times \vec{AC} = (\vec{a} - 3\vec{b}) \times (\vec{a} + 3\vec{b}) \]
Using the distributive property of the cross product:
\[ = \vec{a} \times \vec{a} + 3\vec{a} \times \vec{b} - 3\vec{b} \times \vec{a} - 9\vec{b} \times \vec{b} \]
Since \( \vec{u} \times \vec{u} = \vec{0} \):
\[ = 0 + 3\vec{a} \times \vec{b} - 3\vec{b} \times \vec{a} + 0 \]
Using the property \( \vec{b} \times \vec{a} = -\vec{a} \times \vec{b} \):
\[ = 3\vec{a} \times \vec{b} + 3\vec{a} \times \vec{b} = 6\vec{a} \times \vec{b} \]

### Step 5: Compute \( \vec{OD} \times \vec{OC} \)
Now we compute the cross product:
\[ \vec{OD} \times \vec{OC} = (\vec{a} - 2\vec{b}) \times (2\vec{a} + 3\vec{b}) \]
Using the distributive property:
\[ = \vec{a} \times 2\vec{a} + 3\vec{a} \times \vec{b} - 4\vec{b} \times \vec{a} - 6\vec{b} \times \vec{b} \]
Again, using \( \vec{u} \times \vec{u} = \vec{0} \):
\[ = 0 + 3\vec{a} \times \vec{b} - 4\vec{b} \times \vec{a} + 0 \]
Using \( \vec{b} \times \vec{a} = -\vec{a} \times \vec{b} \):
\[ = 3\vec{a} \times \vec{b} + 4\vec{a} \times \vec{b} = 7\vec{a} \times \vec{b} \]

### Step 6: Compute the dot product
Now we need to find:
\[ (\vec{BD} \times \vec{AC}) \cdot (\vec{OD} \times \vec{OC}) = (6\vec{a} \times \vec{b}) \cdot (7\vec{a} \times \vec{b}) \]
Using the property of dot product:
\[ = 6 \cdot 7 \cdot |\vec{a} \times \vec{b}|^2 = 42 |\vec{a} \times \vec{b}|^2 \]

### Final Answer
Thus, the final answer is:
\[ 42 |\vec{a} \times \vec{b}|^2 \]