Let a¯,b¯ be two non–collinear vectors. If OA¯=(x+4y)a¯+(2x+y+1)b¯ and OB¯=(y−2x+2)a¯+(2x−3y−1)b¯ , 9OA¯−6OB¯=0 then (x,y)=

9[(x+4y)a¯+(2x+y+1)b¯]=6[(y−2x+2)a¯+(2x+3y−1)b¯] On comparing the coefficients of a¯&b¯ ⇒7x+10y=4 &2x+9y=−5 ⇒(x,y)=(2,−1)

Let a¯,b¯ be two non–collinear vectors. If OA¯=(x+4y)a¯+(2x+y+1)b¯ and OB¯=(y−2x+2)a¯+(2x−3y−1)b¯ , 9OA¯−6OB¯=0 then (x,y)=

9[(x+4y)a¯+(2x+y+1)b¯]=6[(y−2x+2)a¯+(2x+3y−1)b¯] On comparing the coefficients of a¯&b¯ ⇒7x+10y=4 &2x+9y=−5 ⇒(x,y)=(2,−1)

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Let $\bar{a}, \bar{b}$ be two non–collinear vectors. If $\bar{O A} = (x + 4 y) \bar{a} + (2 x + y + 1) \bar{b}$ and $\bar{O B} = (y - 2 x + 2) \bar{a} + (2 x - 3 y - 1) \bar{b}$ , $9 \bar{O A} - 6 \bar{O B} = 0$ then $(x, y) =$

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$(- 2, - 1)$

$(1, - 2)$

$(1, 2)$

$(2, - 1)$

answer is C.

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Detailed Solution

$9 [(x + 4 y) \bar{a} + (2 x + y + 1) \bar{b}] = 6 [(y - 2 x + 2) \bar{a} + (2 x + 3 y - 1) \bar{b}] O n c o m p a r i n g t h e c o e f f i c i e n t s o f \bar{a} & \bar{b} \Rightarrow 7 x + 10 y = 4 & 2 x + 9 y = - 5 \Rightarrow (x, y) = (2, - 1)$