Q.Identify the unit vector in the following.

For parallel vectors, corresponding components must be in the same ratio; use the $\hat i$ or $\hat j$ components to find the scale factor, then apply it to $\hat k$.

Given $\vec a=\hat i+3\hat j-\hat k$, $\vec b=2\hat i+6\hat j+\lambda\hat k$, with $\vec a\parallel\vec b$, so $\vec b=k\vec a$ for some scalar $k$.

From the $\hat i$-component: $2=k(1)\Rightarrow k=2$. Check $\hat j$: $6=k(3)=6$. ✓ Consistent.

Then $\hat k$-component: $\lambda=k(-1)=2(-1)=-2$.

Setting $|m\vec a|=1$ and using $|m\vec a|=|m||\vec a|$ shows $m\vec a$ is a unit vector exactly when $|\vec a|=1/|m|$.

1. Let $a=|\vec a|$ denote the magnitude (length) of the non-zero vector $\vec a$.
2. For a scalar $m$, the magnitude of $m\vec a$ is $|m\vec a|=|m|\cdot|\vec a|=|m|\,a$.
3. $m\vec a$ is a unit vector precisely when its magnitude equals $1$: $|m|\,a=1$.
4. Solve for $a$: $a=\dfrac{1}{|m|}$ (valid since $m\neq0$).
5. This is the required condition, matching option (a) exactly (with $a$ meaning $|\vec a|$).

Since a unit vector has magnitude $1$, $|m\vec a|=1$ directly forces $|\vec a|=1/|m|$.

1. $m\vec a$ is a unit vector precisely when its magnitude equals $1$: $|m\vec a|=1$.
2. Magnitude of a scalar multiple: $|m\vec a|=|m|\,|\vec a|$.
3. So the condition becomes $|m|\,|\vec a|=1$.
4. Solving for $|\vec a|$ (denoted $a$ in the options, since $m\ne0$): $|\vec a|=\dfrac{1}{|m|}$, i.e. $a=\dfrac1{|m|}$.
5. Distractors: (a) $m=\pm1$ would only work if $|\vec a|=1$ already, not the general condition; (b) $a=|m|$ inverts the required relation; (d) $a=1$ ignores $m$ entirely.