Q.If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect: (A) $\vec{b} = \lambda\vec{a}$, for some scalar $\lambda$ (B) $\vec{a} = \pm\vec{b}$ (C) the respective components of $\vec{a}$ and $\vec{b}$ are not proportional (D) both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes.

Concept: Collinear Vectors Properties

Two vectors are collinear if they lie along the same or parallel lines — one is a scalar multiple of the other.

Step 1: By definition, if $\vec{a}$ and $\vec{b}$ are collinear, there exists a scalar $\lambda$ such that $\vec{b} = \lambda \vec{a}$. So (A) is correct.

Step 2: (B) says $\vec{a} = \pm \vec{b}$. This is a special case where $|\lambda| = 1$, but collinearity allows any $\lambda$ (e.g., $\vec{b} = 2\vec{a}$). So (B) is not always true — it is incorrect as a general statement. …

Collinear vectors lie along the same line — they are parallel or anti-parallel. This means one vector can be written as a scalar multiple of the other. The scalar can be positive (same direction), negative (opposite direction), or zero (if one vector is the zero vector). The key is that the direction is either exactly the same or exactly opposite; the magnitudes can differ.

Let’s examine each option carefully.

1. Option (A): $\vec{b} = \lambda\vec{a}$, for some scalar $\lambda$

   This is the definition of collinear vectors. If two vectors are collinear, one is always a scalar multiple of the other. This statement is correct.

2. Option (B): $\vec{a} = \pm\vec{b}$

   This says the vectors are either equal or exact negatives. But collinearity only requires one to be a scalar multiple — the scalar can be any real number, not just $1$ or $-1$. For example, $\vec{a} = 2\vec{b}$ is perfectly collinear but doesn’t satisfy $\vec{a} = \pm\vec{b}$. So this statement is incorrect.

3. Option (C): the respective components of $\vec{a}$ and $\vec{b}$ are not proportional

   If $\vec{a} = (a_1, a_2, a_3)$ and $\vec{b} = (b_1, b_2, b_3)$ are collinear, then $\vec{b} = \lambda\vec{a}$ means $b_1 = \lambda a_1$, $b_2 = \lambda a_2$, $b_3 = \lambda a_3$. So the components are proportional (with the same $\lambda$). Saying they are “not proportional” is false. This statement is incorrect. …

Method: Judging Statements About Collinear Vectors

This is a definition-checking task. The technique is to hold each statement against the precise definition of collinearity and keep or reject it.

Steps

Step 1: State the exact definition.

$\vec{a}$ and $\vec{b}$ are collinear $\iff \vec{b}=\lambda\vec{a}$ for some scalar $\lambda$ — any real number, positive, negative, or (for the zero vector) zero. Equivalently, their components are proportional.

Step 2: Test each statement against "any $\lambda$".

A statement is correct only if it holds for every collinear pair, not just a special case. Watch for statements that secretly restrict $\lambda$ (to $\pm1$, or to positive values only).

Step 3: Flag the over-restrictive or contradictory ones. …

Common Mistakes

Mistake 1: Believing collinear means "same direction only".

Why it's wrong: a negative $\lambda$ (opposite direction) is still collinear, so "same direction, different magnitudes" is too restrictive. Correct approach: allow both same-sense and opposite-sense — any real $\lambda$.

Mistake 2: Thinking $\vec{a}=\pm\vec{b}$ captures collinearity.

Why it's wrong: that only covers $\lambda=\pm1$; $\vec{b}=2\vec{a}$ is collinear but not $\pm\vec{a}$. Correct approach: the scalar can be any value, not just $1$ or $-1$. …