Q.If $|\vec{a}|=4$ and $-3\le\lambda\le2$, then the range of $|\lambda\vec{a}|$ is
(A) $[0, 8]$
(B) $[-12, 8]$
(C) $[0, 12]$
(D) $[8, 12]$

The key idea is that scaling a vector by $\lambda$ multiplies its magnitude by $|\lambda|$, so $|\lambda \vec{a}| = |\lambda| \, |\vec{a}|$.

Given $|\vec{a}| = 4$, we have $|\lambda \vec{a}| = 4|\lambda|$. …

The key idea here is that magnitude is always non-negative, and scaling a vector by a scalar $\lambda$ multiplies its length by $|\lambda|$, not by $\lambda$ itself. Many students mistakenly plug the endpoints $-3$ and $2$ directly into $\lambda \cdot 4$ and get $[-12, 8]$, forgetting that a magnitude can never be negative. Let’s build the correct reasoning from the ground up.

This is a fundamental property: the absolute value of the scalar comes in because length is a non-negative quantity. So the problem reduces to finding the range of $|\lambda| \cdot 4$ as $\lambda$ varies over $[-3, 2]$.

1. Identify the possible values of $|\lambda|$.

   Since $\lambda$ can be any real number between $-3$ and $2$, the absolute value $|\lambda|$ takes values from $0$ (when $\lambda = 0$) up to the maximum distance from zero in that interval. The farthest point from $0$ in $[-3, 2]$ is $-3$, giving $|\lambda| = 3$. So $|\lambda| \in [0, 3]$.

2. Multiply by $|\vec{a}| = 4$.

   The expression $|\lambda \vec{a}| = |\lambda| \cdot 4$ is a linear function of $|\lambda|$. As $|\lambda|$ goes from $0$ to $3$, the product goes from $0$ to $12$. Because $|\lambda|$ can take every value in between (it’s a continuous interval), the range of $|\lambda \vec{a}|$ is exactly $[0, 12]$. …

Method: Finding the range of a scaled vector's magnitude

Use this whenever a scalar varies over an interval and you need the range of $|\lambda\vec{a}|$.

Steps

Step 1: Apply the scaling property of magnitude

The absolute value $|\lambda|$ appears because a length can never be negative — a negative $\lambda$ flips direction but the length still grows by $|\lambda|$.

Step 2: Find the range of $|\lambda|$ over the given interval

For $\lambda\in[p,q]$, the smallest $|\lambda|$ is $0$ if the interval contains $0$ (otherwise the endpoint nearest $0$), and the largest $|\lambda|$ is the endpoint farthest from $0$ in absolute value. …

Common Mistakes

Mistake 1: Plugging the endpoints straight into $\lambda|\vec a|$ and getting a negative range

Why it's wrong: this yields $[-12,8]$, but a magnitude can never be negative. Correct approach: use $|\lambda\vec a|=|\lambda|\,|\vec a|$ with the absolute value, giving $[0,12]$.

Mistake 2: Taking $\lambda=2$ as the maximum …