Q91 / 200 · Manipal (MET) 2010 · Mathematics
In a trapezoid of the vector $\overrightarrow{\mathrm{BC}}=\lambda \overrightarrow{\mathrm{AD}}$. We will, then find that $\overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BD}}$ is collinear with $\overrightarrow{\mathrm{AD}}$. If $\overrightarrow{\mathrm{P}}=\mu \overrightarrow{\mathrm{AD}}$, then
- Correct answerA$\mu=\lambda+1$
- B$\lambda=\mu+1$
- C$\lambda+\mu=1$
- D$\mu=2+\lambda$
Solution
We have,
$\text {} \begin{aligned}
& \overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}} \\
= & \overrightarrow{\mathrm{AC}}+\lambda \overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{CD}}=\lambda \overrightarrow{\mathrm{AD}}+(\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}) \\
= & \lambda \overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{AD}}=(\lambda+1) \overrightarrow{\mathrm{AD}}
\end{aligned}$
But, $\overrightarrow{\mathbf{P}}=\mu \overrightarrow{\mathrm{AD}}$
$\therefore \mu=\lambda+1$
Hence, the correct answer is option: 'A'.