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

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a

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

b

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

c

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

d

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

answer is A.

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Detailed Solution

$a = \frac{3 b + 2 c}{k} a = - 2 b + 5 c 3 b + 2 c = - 2 k b + 5 k c (3 + 2 k) b = (5 k - 2) c \Rightarrow b = \frac{5 k - 2}{3 + 2 k} = c \frac{5 k - 2}{2 k + 3} < 0 \Rightarrow (5 k - 2) (2 k + 3) < 0 \Rightarrow k \in (\frac{- 3}{2}, \frac{2}{5})$

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