Question

In a trapezoid of the vector $\overrightarrow{ B C }=\lambda \overrightarrow{ A D }$. We will, then find that $\vec{ P }=\overrightarrow{ A C }+\overrightarrow{ B D }$ is collinear with $\overrightarrow{ A D }$. If $\vec{ P }=\mu\, \overrightarrow{ A D }$, then

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

Answer 1

Answer: (A)

We have, $\vec{ P }=\overrightarrow{ A C }+\overrightarrow{ B D }=\overrightarrow{ A C }+\overrightarrow{ B C }+\overrightarrow{ C D }$
$=\overrightarrow{ A C }+\lambda \overrightarrow{ A D }+\overrightarrow{ C D }=\lambda \overrightarrow{ A D }+(\overrightarrow{ A C }+\overrightarrow{ C D })$
$=\lambda \overrightarrow{AD } +\overrightarrow{ A D }=(\lambda+1) \overrightarrow{ A D }$
But, $\vec{ P }=\mu\, \overrightarrow{ A D } $
$\therefore \mu=\lambda+1$