Question

Find non-zero scalars $\alpha ,\beta $ such that for all vectors a and b, $\alpha (2a - b) - \beta (a + 2b) = 4a - b$
$
  {\text{A}}{\text{. }}\alpha = \dfrac{7}{3},\beta = \dfrac{{ - 2}}{3} \\
  {\text{B}}{\text{. }}\alpha = \dfrac{{ - 7}}{3},\beta = \dfrac{2}{3} \\
  {\text{C}}{\text{. }}\alpha = \dfrac{9}{5},\beta = \dfrac{{ - 2}}{5} \\
  {\text{D}}{\text{. }}\alpha = \dfrac{9}{5},\beta = \dfrac{2}{3} \\
 $

Answer 1

Hint- A quantity having magnitude as well as direction is known as a vector necessary for finding the position of one point in space with respect to others. Vectors are represented by a symbol \[ \to \]that determines the direction and magnitude of a quantity.
A quantity only having magnitude but not direction is known as a scalar quantity.

Non-zero vectors are the vectors whose value is not zero. When a non-zero scalar is multiplied by a zero vector the result is zero.
In the given question coefficients of the equation are in the form of $\alpha ,\beta $and now we have to resolve the equation in the coefficient of a and b by equating the coefficient where equating of the coefficient is the way of solving a function equation of two expressions.

Complete step by step solution:
For the equation $\alpha (2a - b) - \beta (a + 2b) = 4a - b$, simplifying the equation and comparing the coefficients will result in the value of the non-zero scalars $\alpha ,\beta $.
$
  \alpha (2a - b) - \beta (a + 2b) = 4a - b \\
  2a\alpha - b\alpha - a\beta - 2b\beta = 4a - b \\
  a(2\alpha - \beta ) - b(\alpha + 2\beta ) = 4a - b \\
 $

Now comparing the coefficients of a and b, we get:
$2\alpha - \beta = 4{\text{ and, }}\alpha + 2\beta = 1$

Substitute $\beta = 2\alpha - 4$ in the equation $\alpha + 2\beta = 1$ to determine the values of $\alpha $ we get:
$
  \alpha + 2\beta = 1 \\
  \alpha + 2(2\alpha - 4) = 1 \\
  \alpha + 4\alpha - 8 = 1 \\
  5\alpha = 9 \\
  \alpha = \dfrac{9}{5} \\
 $

Again, substitute $\alpha = \dfrac{9}{5}$ in the equation $\beta = 2\alpha - 4$to determine the value of $\beta $ we get:
$
  \beta = 2\alpha - 4 \\
   = 2\left( {\dfrac{9}{5}} \right) - 4 \\
   = \dfrac{{18}}{5} - 4 \\
   = \dfrac{{18 - 20}}{5} \\
   = \dfrac{{ - 2}}{5} \\
 $

Hence, the value of $\alpha = \dfrac{9}{5}{\text{ and, }}\beta = \dfrac{{ - 2}}{5}$ such that for all vectors a and b, $\alpha (2a - b) - \beta (a + 2b) = 4a - b$.
Option C is correct.

Note: As the question is asking for the value of $\alpha ,\beta $, the given equation needs to be resolved in the form of a and b by equating the coefficient in the expression.