Explanation

(b)
Given, $\overrightarrow{\mathbf{AD}} + \overrightarrow{\mathbf{EB}} + \overrightarrow{\mathbf{FC}} = \lambda\overrightarrow{\mathbf{ED}}$
[image:10_image11.emf]
$\Longrightarrow \left( \overrightarrow{\mathbf{AE}} + \overrightarrow{\mathbf{ED}} \right) + \left( \overrightarrow{\mathbf{ED}} + \overrightarrow{\mathbf{DB}} \right) + 2\overrightarrow{\mathbf{ED}} = \lambda\overrightarrow{\mathbf{ED}}$
$\Longrightarrow 4\overrightarrow{\mathbf{ED}} + \left( \overrightarrow{\mathbf{AE}} + \overrightarrow{\mathbf{DB}} \right) = \lambda\overrightarrow{\mathbf{ED}}$
$\Longrightarrow \ \ \ 4\overrightarrow{\mathbf{ED}} = \lambda\overrightarrow{\mathbf{ED}}\ \ \ \ \ \ \ \ \ \ \ \left( \because\overrightarrow{\mathbf{AE}} = - \overrightarrow{\mathbf{DB}} \right)$
Alternate
Now, $\overrightarrow{\mathbf{AD}} + \overrightarrow{\mathbf{EB}} + \overrightarrow{\mathbf{FC}} = 2\left( \overrightarrow{\mathbf{OD}} + \overrightarrow{\mathbf{EO}} + \overrightarrow{\mathbf{ED}} \right)$
$= 2\left( \overrightarrow{\mathbf{ED}} + \overrightarrow{\mathbf{ED}} \right) = 4\overrightarrow{\mathbf{ED}}\ \therefore\lambda = 4$