If OACB is a parallelogramwith ` vec(OC) = vec a and vec (AB) = vec b, " then " vec(OA)=`

Options

A. `vec a + vec b `
B. `vec q - vec b `
C. `(1)/(2) (vec b - veca)`
D. `(1)/(2) (vec a - vec b )`

Answer

To find the vector \( \vec{OA} \) in the parallelogram \( OACB \) where \( \vec{OC} = \vec{a} \) and \( \vec{AB} = \vec{b} \), we can use the properties of vectors in a parallelogram.

### Step-by-Step Solution:

1. **Understand the Parallelogram Properties**:
In a parallelogram, opposite sides are equal in length and direction. Therefore, we have:
\[ \vec{OC} = \vec{OA} + \vec{AC} \]
and
\[ \vec{AB} = \vec{AC} - \vec{BC} \]

2. **Express \( \vec{AC} \)**:
From the properties of the parallelogram, we know that:
\[ \vec{AC} = \vec{AB} + \vec{BC} \]
Since \( \vec{BC} = -\vec{AB} \) (because \( \vec{AB} \) and \( \vec{BC} \) are in opposite directions), we can simplify this to:
\[ \vec{AC} = \vec{b} + (-\vec{b}) = 0 \]

3. **Set Up the Equation**:
We can rewrite the equation for \( \vec{OC} \):
\[ \vec{OC} = \vec{OA} + \vec{AC} \]
Plugging in \( \vec{AC} = \vec{b} \), we have:
\[ \vec{a} = \vec{OA} + \vec{b} \]

4. **Rearrange to Solve for \( \vec{OA} \)**:
Now, we can isolate \( \vec{OA} \):
\[ \vec{OA} = \vec{a} - \vec{b} \]

### Final Answer:
Thus, the vector \( \vec{OA} \) is given by:
\[ \vec{OA} = \vec{a} - \vec{b} \]