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