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Question: Find non-zero scalars \(\alpha ,\beta \) such that for all vectors a and b, \(\alpha (2a - b) - \bet...

Find non-zero scalars α,β\alpha ,\beta such that for all vectors a and b, α(2a−b)−β(a+2b)=4a−b\alpha (2a - b) - \beta (a + 2b) = 4a - b
A. α=73,β=−23 B. α=−73,β=23 C. α=95,β=−25 D. α=95,β=23  {\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} \\\

Explanation

Solution

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 →\tothat 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 α,β\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 α(2a−b)−β(a+2b)=4a−b\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 α,β\alpha ,\beta .
α(2a−b)−β(a+2b)=4a−b 2aα−bα−aβ−2bβ=4a−b a(2α−β)−b(α+2β)=4a−b  \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:
2α−β=4 and, α+2β=12\alpha - \beta = 4{\text{ and, }}\alpha + 2\beta = 1

Substitute β=2α−4\beta = 2\alpha - 4 in the equation α+2β=1\alpha + 2\beta = 1 to determine the values of α\alpha we get:
α+2β=1 α+2(2α−4)=1 α+4α−8=1 5α=9 α=95  \alpha + 2\beta = 1 \\\ \alpha + 2(2\alpha - 4) = 1 \\\ \alpha + 4\alpha - 8 = 1 \\\ 5\alpha = 9 \\\ \alpha = \dfrac{9}{5} \\\

Again, substitute α=95\alpha = \dfrac{9}{5} in the equation β=2α−4\beta = 2\alpha - 4to determine the value of β\beta we get:
β=2α−4 =2(95)−4 =185−4 =18−205 =−25  \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 α=95 and, β=−25\alpha = \dfrac{9}{5}{\text{ and, }}\beta = \dfrac{{ - 2}}{5} such that for all vectors a and b, α(2a−b)−β(a+2b)=4a−b\alpha (2a - b) - \beta (a + 2b) = 4a - b.
Option C is correct.

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