If $\vec{a}$ is a nonzero vector of magnitude '$a$' and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is a unit vector if
(A) $\lambda = 1$
(B) $\lambda = -1$
(C) $a = |\lambda|$
(D) $a = \dfrac{1}{|\lambda|}$

Question

Accepted Answer

> [!TLDR]
> A unit vector has magnitude 1. Scaling a vector $\vec{a}$ of magnitude $a$ by $\lambda$ gives magnitude $|\lambda| a$. For this to be 1, we need $a = 1/|\lambda|$. The correct option is **(D)**.

The core idea here is simple: a **unit vector** is any vector whose length (magnitude) is exactly 1. When you scale a vector by a scalar, you scale its magnitude by the absolute value of that scalar. So the question reduces to: what condition on $a$ and $\lambda$ makes $|\lambda| a = 1$?

Let’s unpack this carefully.

1. **Recall the definition.**  
   A vector $\vec{v}$ is a unit vector if its magnitude $|\vec{v}| = 1$. The problem tells us $\vec{a}$ has magnitude $a$ (so $|\vec{a}| = a$), and $\lambda$ is a nonzero scalar.

2. **What happens when you multiply a vector by a scalar?**  
   If you take $\vec{a}$ and multiply it by $\lambda$, you get $\lambda \vec{a}$. The magnitude of this new vector is:
   $$|\lambda \vec{a}| = |\lambda| \, |\vec{a}| = |\lambda| \, a$$
   This is a fundamental property: scaling stretches (or shrinks) the length by the absolute value of the scalar. The sign of $\lambda$ only flips direction, not magnitude.

3. **Set the condition for a unit vector.**  
   We want $\lambda \vec{a}$ to be a unit vector, so:
   $$|\lambda \vec{a}| = 1 \quad \Rightarrow \quad |\lambda| \, a = 1$$

4. **Solve for the relationship between $a$ and $\lambda$.**  
   Since $a > 0$ (it’s a magnitude) and $|\lambda| > 0$ (nonzero scalar), we can divide:
   $$a = \frac{1}{|\lambda|}$$

That’s it. The magnitude of the original vector must be the reciprocal of the absolute value of the scalar. …

Q.If $\vec{a}$ is a nonzero vector of magnitude ' $a$ ' and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is a unit vector if
(A) $\lambda = 1$
(B) $\lambda = -1$
(C) $a = |\lambda|$
(D) $a = \dfrac{1}{|\lambda|}$

Concept understanding — Unit Vector Scaling

Unit Vector Scaling: From Intuition to Precision

The Intuition First

The Precise Statement

Example in 2D

Example in 3D

Why This Matters …

Method: When Is a Scaled Vector a Unit Vector?

Steps

Common Mistakes

Q.If a⃗\vec{a}a is a nonzero vector of magnitude 'aaa' and λ\lambdaλ a nonzero scalar, then λa⃗\lambda\vec{a}λa is a unit vector if (A) λ=1\lambda = 1λ=1 (B) λ=−1\lambda = -1λ=−1 (C) a=∣λ∣a = |\lambda|a=∣λ∣ (D) a=1∣λ∣a = \dfrac{1}{|\lambda|}a=∣λ∣1​

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Q.If $\vec{a}$ is a nonzero vector of magnitude ' $a$ ' and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is a unit vector if
(A) $\lambda = 1$
(B) $\lambda = -1$
(C) $a = |\lambda|$
(D) $a = \dfrac{1}{|\lambda|}$

Unlock everything free for 14 days