MCQ
If $\vec{a},\vec{b},\vec{c}$ are non-coplanar vectors and $(\vec{a} - \lambda \vec{b}) . (\vec{b} - 2\vec{c})  \times (\vec{c} + 2 \vec{a}) =0$ ,then $\lambda$ is equal to
  • A
    $1$
  • B
    $1/4$
  • C
    $0$
  • $-1/4$

Answer

Correct option: D.
$-1/4$
d
 Given, $\quad (\vec a - \lambda ,\vec b) \cdot (\vec b - 2\vec c) \times (\vec c + 2\vec a) = 0$

$ \Rightarrow (\vec a - \lambda \vec b) - \{ \vec b \times \vec c + \vec b \times 2\vec a - 4(\vec c \times \vec a)\}  = 0$

$ \Rightarrow \vec a \cdot (\vec b \times \vec c) + \vec a \cdot (\vec b \times 2\vec a) - \vec a \cdot 4(\vec c \times \vec a)$

$ - \lambda \vec b \cdot (\vec b \times \vec c) - \lambda \vec b \cdot (\vec b \times 2\vec a) + 4\lambda \vec b - (\vec c \times \vec a) = 0$

$ \Rightarrow \vec a \cdot (\vec b \times \vec c) + 4\lambda \vec b \cdot (\vec c \times \vec a) = 0$

$ \Rightarrow \{ \vec a \cdot (\vec b \times \vec c)\} (1 + 4\lambda ) = 0$

$ \Rightarrow \quad \lambda  =  - \frac{1}{4}$       [$\because $  $\vec a \cdot (\vec b \times \vec c) \ne 0,$    given]

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