Q.If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors, then find $x$ such that $\vec{\alpha} = (x - 2)\vec{a} + \vec{b}$ and $\vec{\beta} = (3 + 2x)\vec{a} - 2\vec{b}$ are collinear.

The key idea here is the property of Collinear Vectors. Two non-zero vectors $\vec{u}$ and $\vec{v}$ are collinear if and only if one is a scalar multiple of the other, i.e., $\vec{u} = k\vec{v}$ for some scalar $k$. Additionally, if $\vec{a}$ and $\vec{b}$ are non-collinear vectors, then $p\vec{a} + q\vec{b} = \vec{0}$ implies $p=0$ and $q=0$.

Since $\vec{\alpha}$ and $\vec{\beta}$ are collinear, we can write $\vec{\alpha} = k\vec{\beta}$ for some scalar $k$.

Substituting the given expressions:

$(x - 2)\vec{a} + \vec{b} = k((3 + 2x)\vec{a} - 2\vec{b})$

$(x - 2)\vec{a} + \vec{b} = k(3 + 2x)\vec{a} - 2k\vec{b}$

Rearranging the terms to form a linear combination of $\vec{a}$ and $\vec{b}$ equal to $\vec{0}$:

$((x - 2) - k(3 + 2x))\vec{a} + (1 + 2k)\vec{b} = \vec{0}$

Since $\vec{a}$ and $\vec{b}$ are non-collinear, their coefficients must both be zero: …

When we say two vectors are collinear, it means they lie on the same line or on parallel lines. Geometrically, they point in the same direction or exactly opposite directions. Mathematically, this implies a very specific relationship: one vector must be a scalar multiple of the other.

Consider two vectors, $\vec{u}$ and $\vec{v}$. If they are collinear, then there exists some scalar $k$ such that $\vec{v} = k\vec{u}$. This scalar $k$ can be positive (same direction), negative (opposite direction), or zero (if one of the vectors is the zero vector).

The problem provides $\vec{\alpha}$ and $\vec{\beta}$ in terms of two other vectors, $\vec{a}$ and $\vec{b}$. A crucial piece of information is that $\vec{a}$ and $\vec{b}$ are non-collinear. This means they form a basis for a 2D plane; you cannot express one as a scalar multiple of the other. This property is key: if we have an equation $P\vec{a} + Q\vec{b} = \vec{0}$ where $\vec{a}$ and $\vec{b}$ are non-collinear, then the only way for this equation to hold is if both coefficients $P$ and $Q$ are zero. This is analogous to saying that if $Px + Qy = 0$ for all $x, y$ and $x, y$ are not proportional, then $P=0$ and $Q=0$.

Let's apply this understanding to solve the problem.

1. Set up the collinearity condition for $\vec{\alpha}$ and $\vec{\beta}$.

   Since $\vec{\alpha}$ and $\vec{\beta}$ are collinear, one must be a scalar multiple of the other. Let's assume $\vec{\beta} = k\vec{\alpha}$ for some scalar $k$.

   If $\vec{u}$ and $\vec{v}$ are collinear, then $\vec{v} = k\vec{u}$ for some scalar $k$.

2. Substitute the given expressions for $\vec{\alpha}$ and $\vec{\beta}$.

   We are given $\vec{\alpha} = (x - 2)\vec{a} + \vec{b}$ and $\vec{\beta} = (3 + 2x)\vec{a} - 2\vec{b}$.

   Substituting these into $\vec{\beta} = k\vec{\alpha}$:

3. Distribute the scalar $k$ and rearrange the equation. Expand the right side and then move all terms to one side to set the equation to $\vec{0}$:

Now, group the terms involving $\vec{a}$ and $\vec{b}$:

4. Apply the non-collinearity condition of $\vec{a}$ and $\vec{b}$. We are given that $\vec{a}$ and $\vec{b}$ are non-collinear vectors. For a linear combination of non-collinear vectors to be the zero vector, the coefficients of each vector must individually be zero. This is a fundamental property of linearly independent vectors. …