If ka¯=3b¯+2c¯, a¯=−2b¯+5c¯ and if b¯ and c¯ have opposite direction then

k(−2b¯+5c¯)=3b¯+2c¯⇒2k+3b¯=5k−2c¯⇒c¯=2k+35k−2b¯b¯ and c¯ have opposite signs ⇒2k+35k−2<0i.e.2k+35k−2<0⇒k∈−32,25

If ka¯=3b¯+2c¯, a¯=−2b¯+5c¯ and if b¯ and c¯ have opposite direction then

k(−2b¯+5c¯)=3b¯+2c¯⇒2k+3b¯=5k−2c¯⇒c¯=2k+35k−2b¯b¯ and c¯ have opposite signs ⇒2k+35k−2<0i.e.2k+35k−2<0⇒k∈−32,25

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If $k \bar{a} = 3 \bar{b} + 2 \bar{c}, \bar{a} = - 2 \bar{b} + 5 \bar{c}$ and if $\bar{b}$ and $\bar{c}$ have opposite direction then

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$k \in (\frac{- 2}{3}, \frac{1}{4})$

$k \in (\frac{- 3}{2}, \frac{2}{5})$

$k \in (\frac{- 3}{2}, \frac{1}{5})$

$k \in (\frac{- 1}{2}, \frac{2}{3})$

answer is B.

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Detailed Solution

$k (- 2 \bar{b} + 5 \bar{c}) = 3 \bar{b} + 2 \bar{c}$
$\Rightarrow (2 k + 3) \bar{b} = (5 k - 2) \bar{c} \Rightarrow \bar{c} = (\frac{2 k + 3}{5 k - 2}) \bar{b}$
$\bar{b}$ and $\bar{c}$ have opposite signs $\Rightarrow \frac{2 k + 3}{5 k - 2} < 0$
i.e. $(2 k + 3) (5 k - 2) < 0 \Rightarrow k \in (- \frac{3}{2}, \frac{2}{5})$