Question

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

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

Answer 1

Answer: (A)

We have, $\vec{P}=\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AC}+\overrightarrow{BC}+\overrightarrow{CD}$
$=\overrightarrow{AC}+\lambda \overrightarrow{AD}+\overrightarrow{CD}=\lambda \overrightarrow{AD}+(\overrightarrow{AC}+\overrightarrow{CD})$
$=\lambda \overrightarrow{AD}+\overrightarrow{AD}=(\lambda +1)\overrightarrow{AD}$
But, $\vec{P}=\mu \overrightarrow{AD}$
$\therefore \mu =\lambda +1$