Question

Let ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and $ (\vec{AB}-\vec{BC})+(\vec{AD}-\vec{DC})=k \vec{FE} $, then k is equal to

• A. -4
• B. -2
• C. 2
• D. 4

Answer 1

The Correct Option is A

The question involves finding the value of \( k \) in the given vector equation. Let's solve it step-by-step:

Given a quadrilateral \( ABCD \), with \( E \) and \( F \) as the midpoints of diagonals \( AC \) and \( BD \) respectively, we are given the equation:

(\vec{AB} - \vec{BC}) + (\vec{AD} - \vec{DC}) = k \vec{FE}

Let's analyze and simplify the expression. According to the properties of vector algebra and midpoints:

1. We know \vec{FE} = \vec{F} - \vec{E}, where \( \vec{E} \) is the midpoint of \( AC \) and \( \vec{F} \) is the midpoint of \( BD \).
2. From midpoint properties, we have:
   • \vec{E} = \frac{\vec{A} + \vec{C}}{2}
   • \vec{F} = \frac{\vec{B} + \vec{D}}{2}
3. Thus, \vec{FE} = \left(\frac{\vec{B} + \vec{D}}{2}\right) - \left(\frac{\vec{A} + \vec{C}}{2}\right) = \frac{\vec{B} + \vec{D} - \vec{A} - \vec{C}}{2}
4. Substitute \(\vec{FE}\) into the equation: (\vec{AB} - \vec{BC}) + (\vec{AD} - \vec{DC}) = k \cdot \frac{\vec{B} + \vec{D} - \vec{A} - \vec{C}}{2}
5. Simplify the left-hand side:
   • \vec{AB} - \vec{BC} = \vec{A} - \vec{C} (since \vec{AB} = \vec{B} - \vec{A} and \vec{BC} = \vec{C} - \vec{B})
   • \vec{AD} - \vec{DC} = \vec{A} - \vec{C} (since \vec{AD} = \vec{D} - \vec{A} and \vec{DC} = \vec{C} - \vec{D})
   • Therefore, (\vec{A} - \vec{C}) + (\vec{A} - \vec{C}) = 2(\vec{A} - \vec{C})
6. Equating and solving for \( k \): 2(\vec{A} - \vec{C}) = k \cdot \frac{\vec{B} + \vec{D} - \vec{A} - \vec{C}}{2}
7. Therefore, 4(\vec{A} - \vec{C}) = k (\vec{B} + \vec{D} - \vec{A} - \vec{C})
8. Comparing the vectors on both sides, 4\vec{A} - 4\vec{C} = k(\vec{B} + \vec{D} - \vec{A} - \vec{C})
9. To satisfy this equality, we need k = -4 with cancellation of similar terms in the direction of \vec{A} - \vec{C}.

Thus, the correct answer is -4.