Question

Let a,b be two noncoffinear vectors. If $\overline{OA}=(x+4y)\overline{a}+(2x+y+1)\overline{b},\overline{OB}=(y-2x+2)\overline{a}+(2x-3y-1)\overline{b}$ and $3\overline{OA}=2\overline{OB},$ then $(x,y)=$

• A. $(1,2)$
• B. $(1,-2)$
• C. $(2,-1)$
• D. $(-2,-1)$

Answer 1

The correct option is C $(2,-1)$

Given,

$OA=(x+4y)a+(2x+y+1)b$

$OB=(y-2x+2)a+(2x-3y-1)b$

$3OA=2OB$

Equating coefficients of a and b separately, we get,

$3(x+4y)=2(y-2x+2)$

$7x+10y=4$......(1)

$3(2x+y+1)=2(2x-3y-1)$

$2x+9y=-5$........(2)

$2\times \left(1\right)-7\times \left(2\right)$ gives,

$-43y=43$

$\Rightarrow y=-1$

From (1), we get,

$7x=4-10(-1)$

$\Rightarrow x=2$

$\therefore (x,y)=(2,-1)$

Hence, this is the answer.