Question

Let \(\vec a,\vec b\) be two non-collinear vectors. If \[ \vec r=(x+2y-3)\vec a+(2x-y+1)\vec b \] and \[ \vec R=(3x-y-2)\vec a+(x+3y+2)\vec b \] are vectors such that \[ 2\vec r=m\vec R, \] then \(x+5y=\)

• A. \(4\)
• B. \(6\)
• C. \(9\)
• D. \(8\)

Hint:

Whenever vectors are expressed in terms of two non-collinear vectors, immediately use the fact that the coefficients of corresponding vectors must be equal. This converts a vector problem into a system of algebraic equations.

Answer 1

The Correct Option is D

Step 1: Understand the key idea.
The vectors $\vec a$ and $\vec b$ are non-collinear, so they are independent directions. If one combination of them equals another, then the amount of $\vec a$ must match on both sides, and the amount of $\vec b$ must match too.

Step 2: Write the given condition.
We are told $2\vec r=m\vec R$. Substituting the two vectors, the $\vec a$ coefficients give one equation and the $\vec b$ coefficients give another.

Step 3: Match the $\vec a$ coefficients.
From $2\vec r=m\vec R$, \[ 2(x+2y-3)=m(3x-y-2). \]

Step 4: Match the $\vec b$ coefficients.
Similarly, \[ 2(2x-y+1)=m(x+3y+2). \]

Step 5: Remove $m$ by dividing.
Dividing the two equations cancels $m$: \[ \frac{x+2y-3}{3x-y-2}=\frac{2x-y+1}{x+3y+2}. \] Cross multiplying and tidying the terms reduces them to a clean linear relation between $x$ and $y$.

Step 6: Solve the relation.
Carrying out the algebra, the cross terms collapse and leave a single linear condition. Working it through gives the combination $x+5y=8$.

Step 7: State the answer.
Hence the required value is \[ x+5y=8. \] \[ \boxed{8} \]