Let a¯,b¯,and c¯ are 3 non zero vectors such that no 2 of these are collinear. If a¯+2b¯ is collinear with c¯ and b¯+3c¯ is collinear with a¯ and a¯+2b¯+6c¯

a¯+2b¯∥lel to c¯⇒a¯+2b¯=λc¯→(1) b¯+3c¯∥lel to a¯⇒b¯+3c¯=μa¯→(2) (1)−2×(2)⇒a¯−6c¯=λc¯−2μa¯ ⇒λ=−6a a¯+2b¯+6c¯=0¯

Let a¯,b¯,and c¯ are 3 non zero vectors such that no 2 of these are collinear. If a¯+2b¯ is collinear with c¯ and b¯+3c¯ is collinear with a¯ and a¯+2b¯+6c¯

a¯+2b¯∥lel to c¯⇒a¯+2b¯=λc¯→(1) b¯+3c¯∥lel to a¯⇒b¯+3c¯=μa¯→(2) (1)−2×(2)⇒a¯−6c¯=λc¯−2μa¯ ⇒λ=−6a a¯+2b¯+6c¯=0¯

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Let $\bar{a}, \bar{b}, a n d \bar{c}$ are 3 non zero vectors such that no 2 of these are collinear. If $\bar{a} + 2 \bar{b}$ is collinear with $\bar{c}$ and $\bar{b} + 3 \bar{c}$ is collinear with $\bar{a}$ and $\bar{a} + 2 \bar{b} + 6 \bar{c}$

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$\bar{0}$

$λ \bar{b}$

$λ \bar{a}$

$λ \bar{c}$

answer is C.

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Detailed Solution

$\bar{a} + 2 \bar{b} ∥^{l e l} t o \bar{c}$
$\Rightarrow \bar{a} + 2 \bar{b} = λ \bar{c} \to (1) \bar{b} + 3 \bar{c} ∥^{l e l} t o \bar{a}$
$\Rightarrow \bar{b} + 3 \bar{c} = μ \bar{a} \to (2) (1) - 2 \times (2) \Rightarrow \bar{a} - 6 \bar{c} = λ \bar{c} - 2 μ \bar{a} \Rightarrow λ = - 6 a \bar{a} + 2 \bar{b} + 6 \bar{c} = \bar{0}$