If $\vec{a}$ is a non-zero vector of magnitude '$a$' and '$\lambda$' is a non-zero scalar, then $\lambda\vec{a}$ is a unit vector if

**OR**

The value of $\hat{i}\cdot(\hat{j} imes\hat{k}) + \hat{j}\cdot(\hat{i} imes\hat{k}) + \hat{k}\cdot(\hat{i} imes\hat{j})$ is (a) 0 (b) -

Question

If $\vec{a}$ is a non-zero vector of magnitude '$a$' and '$\lambda$' is a non-zero scalar, then $\lambda\vec{a}$ is a unit vector if

**OR**

The value of $\hat{i}\cdot(\hat{j}	imes\hat{k}) + \hat{j}\cdot(\hat{i}	imes\hat{k}) + \hat{k}\cdot(\hat{i}	imes\hat{j})$ is (a) 0 (b) -1 (c) 1 (d) 3

(a) $\lambda=1$
(b) $\lambda=-1$
(c) $a=|\lambda|$
(d) $a=\dfrac{1}{|\lambda|}$

Accepted Answer

>

> [!TLDR]
> A vector is a unit vector when its magnitude equals $1$; use $|\lambda\vec a|=|\lambda||\vec a|$.
>
> Given $|\vec a|=a\ (
e0)$ and scalar $\lambda\ (
e0)$, the magnitude of $\lambda\vec a$ is
> $$|\lambda\vec a|=|\lambda|\,|\vec a|=|\lambda|\,a.$$
> For $\lambda\vec a$ to be a unit vector we need $|\lambda\vec a|=1$:
> $$|\lambda|\,a=1\ \Rightarrow\ a=\frac{1}{|\lambda|}.$$
>
> **Checking the options:** (a) $\lambda=1$ and (b) $\lambda=-1$ force a specific $\lambda$ but ignore $a$; (c) $a=|\lambda|$ is the reciprocal of the correct relation. Only (d) $a=\frac{1}{|\lambda|}$ works.
> …

Q.If $\vec{a}$ is a non-zero vector of magnitude ' $a$ ' and ' $\lambda$ ' is a non-zero scalar, then $\lambda\vec{a}$ is a unit vector if OR The value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$ is

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 …

🎓Unlock everything free for 14 days

Unlock everything free for 14 days