Let $ A=(x+4 y) \vec{a}+(2 x+y+1) \vec{b} $ and $ B=(y-2 x+2) \vec{a}+(2 x-3 y-1) \vec{b} $ where $ \vec{a} $ and $ \vec{b} $ are non-collinear vectors, if $ 3 A=2 B $ then

Question

Let $ A=(x+4 y) \vec{a}+(2 x+y+1) \vec{b} $ and $ B=(y-2 x+2) \vec{a}+(2 x-3 y-1) \vec{b} $ where $ \vec{a} $ and $ \vec{b} $ are non-collinear vectors, if $ 3 A=2 B $ then, $ x=1, y=2 $, $ x=2, y=1 $, $ x=2, y=-1 $, $ x=-1, y=2 $

MARKS App · Accepted Answer

Given: A = x + 4 y a → + 2 x + y + 1 b → and B = y - 2 x + 2 a → + 2 x - 3 y - 1 b → 3 A = 3 x + 4 y a → + 3 2 x + y + 1 b → 2 B = 2 y - 2 x + 2 a → + 2 2 x - 3 y - 1 b → In question it is given, 3 A = 2 B ⇒ 3 x + 4 y = 2 y - 2 x + 2 . . . . 1 and 3 2 x + y + 1 = 2 2 x - 3 y - 1 . . . . 2 From the equation 1 ⇒ 7 x + 10 y = 4 ⇒ x = 4 - 10 y 7 From the equation 2 2 x + 9 y = - 5 ⇒ 2 4 - 10 y 7 + 9 y = - 5 ⇒ 8 - 20 y + 63 y = - 35 ⇒ y = - 1 ⇒ x = 4 - 10 - 1 7 = 2 Hence, x = 2 , y = - 1 .

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