Question

In a trapezium, the vector $\overline{BC}=\lambda \overline{AD}$ and $\overline{P}=\overline{AC}+\overline{BD}=\mu \overline{AD}$, then

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

Answer 1

Answer: (A)

$\mu =\lambda +1$

$\overrightarrow{AC}+\overrightarrow{BD}$ $=(\overrightarrow{AD}+\overrightarrow{DC})+(\overrightarrow{AD}-\overrightarrow{AB})$
$=2\overrightarrow{AD}+\overrightarrow{DC}-\overrightarrow{AB}$
Now, $\overrightarrow{AB}=\overrightarrow{D{B}^{\prime }}$
Hence, $2\overrightarrow{AD}+\overrightarrow{DC}-\overrightarrow{AB}$ $=2\overrightarrow{AD}+\overrightarrow{DC}-\overrightarrow{D{B}^{\prime }}$

Consider triangle DCB'
$=2\overrightarrow{AD}+\overrightarrow{DC}-(\overrightarrow{DC}+\overrightarrow{C{B}^{\prime }})$
$=2\overrightarrow{AD}-\overrightarrow{C{B}^{\prime }}$
$=2\overrightarrow{AD}-(\overrightarrow{B{B}^{\prime }}-\overrightarrow{BC})$
$=2\overrightarrow{AD}-(\overrightarrow{AD}-\overrightarrow{BC})$
$=\overrightarrow{AD}+\overrightarrow{BC}$
$=\overrightarrow{AD}+\lambda \overrightarrow{AD}$
$=(1+\lambda )\overrightarrow{AD}$

Hence,
$\mu \overrightarrow{AD}=(1+\lambda )\overrightarrow{AD}$
$\mu =1+\lambda$