Question

In a trapezium the vector $\overline{BC}=\alpha \overline{AD}$. We will then find that $\overline{p}=\overline{AC}+\overline{BD}$ is collinear with $\overline{AD}$. if $\overline{p}=\mu \overline{AD}$ then

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

Answer 1

Answer: (D)

$\mu =\alpha +1$

Let $A$ be the origin.
So,

$\alpha \overrightarrow{d}=\overrightarrow{c}-\overrightarrow{b}$
$p=\overrightarrow{c}+\overrightarrow{d}-\overrightarrow{b}=\mu \overrightarrow{d}$

Combining two equations, we get $(1+\alpha )\overrightarrow{d}=\mu \overrightarrow{d}$
$\Rightarrow \mu =\alpha +1$