Question

In a trapezoid of the vector \( \vec{BC} = \lambda \vec{AD} \). We will, then find that \( \vec{P} = \vec{AC} + \vec{BD} \) is collinear with \( \vec{AD} \). If \( \vec{P} = \mu \vec{AD} \), then

• A. $\mu=\lambda+1$
• B. $\lambda=\mu+1$
• C. $\lambda+\mu=1$
• D. $\mu=2+\lambda$

Hint:

In a trapezoid of the vector $\vecBC=λ\vecAD$. We will, then find that $\vecP=\vecAC+\vecBD$ is collinear with $\vecAD$. If $\vecP=μ\vecAD$, then

Answer 1

The Correct Option is A

Step 1: Concept
Use vector addition and the given relationship $\vec{BC} = \lambda\vec{AD}$.
Step 2: Analysis
We have $\vec{P} = \vec{AC} + \vec{BD}$. Expressing $\vec{BD}$ as $\vec{BC} + \vec{CD}$, we get $\vec{P} = \vec{AC} + \vec{BC} + \vec{CD}$.
Step 3: Evaluation
Rearranging: $\vec{P} = \vec{BC} + (\vec{AC} + \vec{CD}) = \vec{BC} + \vec{AD}$. Substituting $\vec{BC} = \lambda\vec{AD}$, we get $\vec{P} = \lambda\vec{AD} + \vec{AD} = (\lambda + 1)\vec{AD}$.
Step 4: Conclusion
Since $\vec{P} = \mu\vec{AD}$ , comparing the coefficients gives $\mu = \lambda + 1$.
Final Answer: (a)