The value of non-zero scalars α and β for all vectors \(\vec a\) and \(\vec b\), such that \(\rm \alpha\left(2\vec a-\vec b \right)+\beta\left(\vec a + 2\vec b\right)=8\vec b - \vec a \) is:

Concept:

If two vectors \(\rm \vec a = {a_1}\hat i + {a_2}\hat j+{a_3}\hat k\) and \(\rm \vec b = {b_1}\hat i + {b_2}\hat j+{b_3}\hat k\) are equal, then a1 = b1, a2 = b2 and c1 = c2.

Calculation:

It is given that \(\rm \alpha\left(2\vec a-\vec b \right)+\beta\left(\vec a + 2\vec b\right)=8\vec b - \vec a\).

⇒ \(\rm (2\alpha+\beta)\vec a+(-\alpha+2\beta)\vec b=8\vec b - \vec a\)

Comparing the scalar coefficients on both sides, we get:

2α + β = -1             ... (1)

-α + 2β = 8             ... (2)

Adding twice the equation (2) with the equation (1), we get:

5β = 15

⇒ β = 3

And using either equation (1) or (2), we get:

α = -2

Hence, α = -2, β = 3.