Key Concepts and Formulas

• Collinearity of Vectors: Two non-zero vectors $\overrightarrow u$ and $\overrightarrow v$ are collinear if and only if $\overrightarrow u = k\overrightarrow v$ for some scalar $k$.
• Representation in Non-Collinear Basis: If $\overrightarrow a$ and $\overrightarrow b$ are non-collinear vectors, then any vector $\overrightarrow v$ can be uniquely expressed as $\overrightarrow v = x\overrightarrow a + y\overrightarrow b$.
• Collinearity Condition with Non-Collinear Basis: If $\overrightarrow \alpha = x_1 \overrightarrow a + y_1 \overrightarrow b$ and $\overrightarrow \beta = x_2 \overrightarrow a + y_2 \overrightarrow b$, where $\overrightarrow a$ and $\overrightarrow b$ are non-collinear, then $\overrightarrow \alpha$ and $\overrightarrow \beta$ are collinear if and only if the ratio of their corresponding coefficients is equal, i.e., $\frac{x_1}{x_2} = \frac{y_1}{y_2}$ (provided $x_2 \neq 0$ and $y_2 \neq 0$). Alternatively, this can be stated as $x_1 y_2 - x_2 y_1 = 0$.

Step-by-Step Solution

Step 1: Identify the given vectors and the condition. We are given two vectors: $\overrightarrow \alpha = \left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b$ $\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b$ We are also given that $\overrightarrow a$ and $\overrightarrow b$ are non-collinear. The problem states that $\overrightarrow \alpha$ and $\overrightarrow \beta$ are collinear.

Step 2: Apply the collinearity condition using the non-collinear basis. Since $\overrightarrow a$ and $\overrightarrow b$ are non-collinear, we can use the property that if $\overrightarrow \alpha$ and $\overrightarrow \beta$ are collinear, their coefficients with respect to the basis vectors $\overrightarrow a$ and $\overrightarrow b$ must be proportional.

The coefficient of $\overrightarrow a$ in $\overrightarrow \alpha$ is $(\lambda - 2)$. The coefficient of $\overrightarrow b$ in $\overrightarrow \alpha$ is $1$. The coefficient of $\overrightarrow a$ in $\overrightarrow \beta$ is $(4\lambda - 2)$. The coefficient of $\overrightarrow b$ in $\overrightarrow \beta$ is $3$.

Therefore, for $\overrightarrow \alpha$ and $\overrightarrow \beta$ to be collinear, we must have: $\frac{\text{Coefficient of } \overrightarrow a \text{ in } \overrightarrow \alpha}{\text{Coefficient of } \overrightarrow a \text{ in } \overrightarrow \beta} = \frac{\text{Coefficient of } \overrightarrow b \text{ in } \overrightarrow \alpha}{\text{Coefficient of } \overrightarrow b \text{ in } \overrightarrow \beta}$ $\frac{\lambda - 2}{4\lambda - 2} = \frac{1}{3}$ This equation is valid as long as the denominators are non-zero. The denominator for the coefficient of $\overrightarrow b$ is $3$, which is non-zero. We assume $4\lambda - 2 \neq 0$ for the proportionality to hold in this form.

Step 3: Solve the equation for $\lambda$. To solve the proportion, we cross-multiply: $3(\lambda - 2) = 1(4\lambda - 2)$ Distribute the constants on both sides: $3\lambda - 6 = 4\lambda - 2$ Now, we rearrange the terms to solve for $\lambda$. Subtract $3\lambda$ from both sides: $-6 = 4\lambda - 3\lambda - 2$ $-6 = \lambda - 2$ Add $2$ to both sides: $-6 + 2 = \lambda$ $-4 = \lambda$ So, the value of $\lambda$ is $-4$. Let's check if $4\lambda - 2$ is zero for $\lambda = -4$. $4(-4) - 2 = -16 - 2 = -18 \neq 0$. Thus, our assumption was valid.

Common Mistakes & Tips

• Non-collinear Basis: Always ensure that the basis vectors ($\overrightarrow a$ and $\overrightarrow b$ in this case) are indeed non-collinear. If they were collinear, the uniqueness of coefficient representation would be lost, and this method would not apply directly.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with negative signs and fractions. A small error can lead to an incorrect value of $\lambda$.
• Division by Zero: When using the ratio of coefficients, be mindful of potential division by zero. If a denominator term like $(4\lambda - 2)$ were zero for the calculated $\lambda$, you would need to re-examine the problem. In this specific problem, the denominator is non-zero for the obtained value of $\lambda$.

Summary

The problem asks for the value of $\lambda$ that makes two vectors, $\overrightarrow \alpha$ and $\overrightarrow \beta$, collinear. Both vectors are expressed as linear combinations of two non-collinear vectors $\overrightarrow a$ and $\overrightarrow b$. The key principle used is that if two vectors are collinear and expressed in terms of a non-collinear basis, the ratios of their corresponding coefficients must be equal. By setting up this proportion and solving the resulting linear equation for $\lambda$, we found that $\lambda = -4$.

The final answer is $\boxed{\text{-4}}$.