Let α = (λ - 2)a + b and β = (4λ - 2)a + 3b be two given vectors where vectors a and b are non-colli

Q: Why can't we just set $\vec{\alpha} = \vec{\beta}$?

That would imply the vectors are identical in both magnitude and direction. Collinearity is a broader condition; it only requires the vectors to be parallel, meaning one is a scalar multiple of the other. We must introduce a proportionality constant $k$ to account for this.

Q: What happens if $\vec{a}$ and $\vec{b}$ are collinear?

If $\vec{a}$ and $\vec{b}$ were collinear, they would not span a 2D plane. They would essentially be on the same line, and the concept of 'coefficients' would not be unique. The problem explicitly states they are non-collinear to ensure we have a well-defined basis.

Q: Why is the ratio of coefficients the same?

Because $\vec{a}$ and $\vec{b}$ are linearly independent. If $c_1\vec{a} + d_1\vec{b} = k(c_2\vec{a} + d_2\vec{b})$, then $(c_1 - kc_2)\vec{a} + (d_1 - kd_2)\vec{b} = 0$. For this to be true with non-zero $\vec{a}, \vec{b}$, the coefficients $(c_1 - kc_2)$ and $(d_1 - kd_2)$ must both be zero, which leads directly to the proportionality of the coefficients.

Question

Let $\vec{\alpha} = (\lambda - 2)\vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors where vectors $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear, is :

Options:

(A) -3

(B) 4

(C) 3

(D) -4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plane. You have two arrows, $\vec{a}$ and $\vec{b}$, pointing in different directions. Because they are non-collinear, they define the very fabric of this plane and serve as your basis vectors.

We are given two new vectors defined by the scalar $\lambda$:
$$\vec{\alpha} = (\lambda - 2)\vec{a} + \vec{b}$$
$$\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$$

Our goal is to find the specific value of $\lambda$ that makes $\vec{\alpha}$ and $\vec{\beta}$ collinear.

### The Golden Rule of Collinearity

Two vectors are collinear if they lie on the same line. Mathematically, this is equivalent to saying that one vector is a scalar multiple of the other:
$$\vec{\alpha} = k\vec{\beta}$$

Substituting our expressions into this condition, we obtain:
$$(\lambda - 2)\vec{a} + 1\vec{b} = k((4\lambda - 2)\vec{a} + 3\vec{b})$$

Because $\vec{a}$ and $\vec{b}$ are non-collinear, they are linearly independent. This implies that the coefficients of $\vec{a}$ and $\vec{b}$ must match on both sides of the equation.

This yields the following system of equations:
$$\lambda - 2 = k(4\lambda - 2)$$
$$1 = 3k$$

### Solving the Algebraic Puzzle

From the second equation, we immediately find the proportionality constant:
$$k = \frac{1}{3}$$

Now, we substitute this value back into the first equation:
$$\lambda - 2 = \frac{1}{3}(4\lambda - 2)$$

To solve for $\lambda$, we clear the fraction by multiplying both sides by $3$:
$$3(\lambda - 2) = 4\lambda - 2$$

Expanding the left side gives:
$$3\lambda - 6 = 4\lambda - 2$$

Rearranging the terms by subtracting $3\lambda$ from both sides results in:
$$-6 = \lambda - 2$$

Adding $2$ to both sides, we arrive at the final result:
$$\mathbf{\lambda = -4}$$

### The Elegance of the Result

When $\lambda = -4$, the vectors $\vec{\alpha}$ and $\vec{\beta}$ align perfectly.

The key takeaway for your JEE journey is this: whenever you see vectors expressed in terms of a non-collinear basis, think of their coefficients as coordinates. Collinearity is simply the condition that these coordinates are proportional.

Animated Solution for Mathematics - Vector Algebra: Let $α = (λ - 2) a + b$ and $β = (4 λ - 2) a + 3 b$ be two given vectors where vectors $a$ and $b$ are non-collinear. The value of $λ$ for which vectors $α$ and $β$ are collinear, is :

Select Answer:

Visualized Solution

Visualizing the Basis Vectors $a$ and $b$

Defining Vectors $α$ and $β$

The Condition for Collinearity

Setting up the Proportion

Cross-Multiplication

Expanding the Terms

Grouping Like Terms

Final Calculation for $λ$

Conclusion and Key Takeaway

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Golden Rule of Collinearity

Solving the Algebraic Puzzle

The Elegance of the Result

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