Q.The values of $k$ for which $|k\vec{a}|<|\vec{a}|$ and $k\vec{a}+\dfrac{1}{2}\vec{a}$ is parallel to $\vec{a}$ holds true are ________.

Concept: Vector Magnitude Properties — scaling a vector changes its magnitude by $|k|$, and two vectors are parallel if one is a scalar multiple of the other.

Step 1: For $k\vec{a} + \frac{1}{2}\vec{a}$ to be parallel to $\vec{a}$, it must be a scalar multiple of $\vec{a}$:

This is always parallel to $\vec{a}$ for any real $k$, so no restriction from this condition. …

The key here is to separate two different ideas: magnitude scaling and direction. When you multiply a vector by a scalar, the magnitude gets scaled by the absolute value of that scalar, but the direction either stays the same (if the scalar is positive) or reverses (if negative). A vector is parallel to another if one is a scalar multiple of the other — direction can be same or opposite.

Let’s unpack the problem piece by piece.

1. First condition: $|k\vec{a}| < |\vec{a}|$ The magnitude of $k\vec{a}$ is $|k|\,|\vec{a}|$. So the inequality becomes

Assuming $\vec{a}$ is not the zero vector (otherwise the inequality would be $0 < 0$, which is false), we can divide both sides by $|\vec{a}| > 0$ to get

This means $k$ must lie strictly between $-1$ and $1$:

2. Second condition: $k\vec{a} + \frac12\vec{a}$ is parallel to $\vec{a}$ Combine the two terms:

This is simply a scalar multiple of $\vec{a}$. Any scalar multiple of a vector is always parallel to that vector (including the case where the scalar is zero, which gives the zero vector — the zero vector is considered parallel to every vector by convention in most Indian exam contexts).

So this condition holds for every real $k$. It imposes no restriction. …

Method: Combining a Magnitude Inequality with a Parallelism Condition

Use this pattern when constraints mix a magnitude inequality like $|k\vec{a}|<|\vec{a}|$ with a "parallel to $\vec{a}$" requirement on a scalar combination.

Steps

Step 1: Convert the magnitude condition using $|k\vec{a}|=|k|\,|\vec{a}|$

Divide through by $|\vec{a}|>0$ to reduce $|k\vec{a}|<|\vec{a}|$ to the scalar inequality $|k|<1$, i.e. $-1<k<1$. The absolute value is essential — a negative $k$ still shortens the vector.

Step 2: Test the parallelism condition …

Common Mistakes

Mistake 1: Solving $|k|<1$ as simply $k<1$

Why it's wrong: $|k\vec{a}|=|k|\,|\vec{a}|$, so the inequality is $|k|<1$, which means $-1<k<1$ — not $k<1$. Correct approach: keep the absolute value and write the open interval $(-1,1)$.

Mistake 2: Thinking the parallelism condition restricts $k$

Why it's wrong: $k\vec{a}+\tfrac12\vec{a}=(k+\tfrac12)\vec{a}$ is a scalar multiple of $\vec{a}$ for every real $k$, so it is always parallel. Correct approach: recognise this adds no constraint. …