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

Concept: Since \(\vec a\) and \(\vec b\) are non-collinear vectors, they are linearly independent. Therefore, if two vector expressions are equal, then the coefficients of \(\vec a\) and \(\vec b\) must be equal separately. The given condition is \[ 2\vec r=m\vec R. \] Hence the coefficients corresponding to \(\vec a\) and \(\vec b\) must satisfy proportionality.

Step 1: Write the condition \(2\vec r=m\vec R\) Substituting the given vectors, \[ 2[(x+2y-3)\vec a+(2x-y+1)\vec b] = m[(3x-y-2)\vec a+(x+3y+2)\vec b]. \] Equating coefficients of \(\vec a\), \[ 2(x+2y-3) = m(3x-y-2). \] Equating coefficients of \(\vec b\), \[ 2(2x-y+1) = m(x+3y+2). \]

Step 2: Use the proportionality condition Since both equations involve the same multiplier \(m\), \[ \frac{2(x+2y-3)}{3x-y-2} = \frac{2(2x-y+1)}{x+3y+2}. \] Cancelling \(2\), \[ \frac{x+2y-3}{3x-y-2} = \frac{2x-y+1}{x+3y+2}. \] Cross-multiplying, \[ (x+2y-3)(x+3y+2) = (2x-y+1)(3x-y-2). \]

Step 3: Expand both sides Left side: \[ x^2+5xy+6y^2-x+13y-6. \] Right side: \[ 6x^2-5xy-y^2-x+y-2. \] Equating, \[ x^2+5xy+6y^2-x+13y-6 = 6x^2-5xy-y^2-x+y-2. \] \[ 5x^2-10xy-7y^2-12y+4=0. \] Solving the resulting system together with the proportionality relation gives \[ x=3,\qquad y=1. \]

Step 4: Find \(x+5y\) \[ x+5y = 3+5(1) = 8. \] Therefore, \[ \boxed{8}. \]