Exercises — Unit vectors — î, ĵ, k̂; constructing unit vector

A unit vector has magnitude exactly $1$. Check each length with $\sqrt{\text{sum of squares}}$.

• (a) $|\vec{P}|=\sqrt{1^2+1^2}=\sqrt{2}\approx 1.414$ → not unit.
• (b) $|\vec{Q}|=\sqrt{0.6^2+(-0.8{)}^2}=\sqrt{0.36+0.64}=\sqrt{1}=1$ → unit ✓ (the minus sign only sets direction, it squares away).
• (c) $|\vec{R}|=\sqrt{0^2+0^2+1^2}=1$ → unit ✓ ($\hat{k}$ is a standard basis vector, length $1$ by definition).
• (d) $|\vec{S}|=\sqrt{2^2}=2$ → not unit.

Answer: (b) and (c).

False. Having one component says nothing about length. The length is $|\vec{A}|=\sqrt{5^2}=5$, not $1$. Its unit vector would be $\hat{A}=\frac{5\hat{i}}{5}=\hat{i}$. The number of components and the magnitude are two separate facts.

Step 1 (WHAT): find the length. $|\vec{A}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$. WHY: we can only "divide out" a length once we know it. Step 2: divide each component by $10$: $\hat{A}=0.6\hat{i}+0.8\hat{j}$. Check: $\sqrt{0.6^2+0.8^2}=\sqrt{1}=1$ ✓.

Step 1: $|\vec{B}|=\sqrt{2^2+(-3{)}^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$. WHY the squares kill the minus: squaring makes every term positive; direction is stored in the sign of the component, not in the length. Step 2: $\hat{B}=\frac{2}{7}\hat{i}-\frac{3}{7}\hat{j}+\frac{6}{7}\hat{k}$. The minus on $\hat{j}$ stays — it says "point toward $-y$".

Step 1: strip $\vec{d}$ to a pure direction. $|\vec{d}|=\sqrt{(-4{)}^2+3^2}=\sqrt{25}=5$, so $\hat{d}=-0.8\hat{i}+0.6\hat{j}$. WHY: we want only $\vec{d}$'s aim, not its size. Step 2: $\vec{v}=(20\text{m/s})\hat{d}=-16\hat{i}+12\hat{j}\text{m/s}$. Check: $|\vec{v}|=\sqrt{16^2+12^2}=\sqrt{400}=20\text{m/s}$ ✓.

Set up the length equation: $|\vec{A}|=\sqrt{3^2+c^2}=5$. Square both sides (to undo the root): $9+c^2=25\Rightarrow c^2=16$. Take the root — keep BOTH signs: $c=+4$ or $c=-4$. WHY two answers: magnitude only fixes $c^2$. Geometrically, an arrow of length $5$ with horizontal reach $3$ can tilt up ($c=4$) or down ($c=-4$) — two mirror-image vectors, both length $5$. Don't discard the negative one.

Idea: two 2D vectors are perpendicular when their dot product is $0$. A quick way to rotate $(A_x,A_y)$ by $90^{\circ }$ is to swap the components and flip one sign: $(A_x,A_y)\to (-A_y,A_x)$. Step 1: a perpendicular vector is $\vec{n}=-4\hat{i}+3\hat{j}$. Check perpendicularity: $\vec{A}\cdot \vec{n}=(3)(-4)+(4)(3)=-12+12=0$ ✓. Step 2: normalize. $|\vec{n}|=\sqrt{(-4{)}^2+3^2}=5$, so $\hat{n}=-0.8\hat{i}+0.6\hat{j}$. (The opposite direction $\hat{n}=0.8\hat{i}-0.6\hat{j}$ is equally valid — there are two perpendicular unit vectors.)

The "fraction along an axis" is exactly the matching component of the unit vector — because $\hat{A}$ is $\vec{A}$ scaled to length $1$, each of its components is that axis's share of one unit. Step 1: $|\vec{A}|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$. Step 2: $\hat{A}=\frac{2}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}$. Answer: the $z$-share is $\frac{1}{3}\approx 0.333$, i.e. about $33.3\%$ of the length points along $z$.

Step 1 — displacement direction (displacement = end minus start): $\overrightarrow{PQ}={\vec{r}}_Q-{\vec{r}}_P=(4-1)\hat{i}+(6-2)\hat{j}=3\hat{i}+4\hat{j}$. Step 2 — pure direction: $|\overrightarrow{PQ}|=\sqrt{3^2+4^2}=5$, so $\hat{u}=0.6\hat{i}+0.8\hat{j}$. WHY: "toward $Q$" means we only borrow $\overrightarrow{PQ}$'s aim, not its full length of $5$. Step 3 — the $10$-unit step: $\Delta \vec{r}=10\hat{u}=6\hat{i}+8\hat{j}$. Step 4 — add to the start (tip-to-tail): ${\vec{r}}_{\text{final}}={\vec{r}}_P+\Delta \vec{r}=(1+6)\hat{i}+(2+8)\hat{j}=7\hat{i}+10\hat{j}$. Note $Q$ itself is only $5$ units from $P$, so moving $10$ overshoots past $Q$ — as our answer confirms.

Step 1 — add components: $\vec{F}={\vec{F}}_1+{\vec{F}}_2=(3-1)\hat{i}+(4+2)\hat{j}=2\hat{i}+6\hat{j}\text{N}$. Step 2 — magnitude: $|\vec{F}|=\sqrt{2^2+6^2}=\sqrt{40}=2\sqrt{10}\approx 6.32\text{N}$. Step 3 — direction: $\hat{F}=\frac{2\hat{i}+6\hat{j}}{2\sqrt{10}}=\frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j}\approx 0.316\hat{i}+0.949\hat{j}$. WHY normalize the sum, not the parts: the net direction depends on how the two forces combine, so we must add first, then strip the length.

Let $\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$ with $|\vec{A}|=m=\sqrt{A_x^2+A_y^2+A_z^2}$, and $m\ne 0$. Then $\hat{A}=\frac{A_x}{m}\hat{i}+\frac{A_y}{m}\hat{j}+\frac{A_z}{m}\hat{k}$. Compute its magnitude: $|\hat{A}|=\sqrt{{\left(\frac{A_x}{m}\right)}^2+{\left(\frac{A_y}{m}\right)}^2+{\left(\frac{A_z}{m}\right)}^2}=\sqrt{\frac{A_x^2+A_y^2+A_z^2}{m^2}}=\sqrt{\frac{m^2}{m^2}}=\sqrt{1}=1.$ The dimension never mattered: the numerator inside the root is precisely $m^2$ by definition. $■$

Why $m\ne 0$ is required: dividing by $m=0$ is undefined — see Problem 5.2.

$|\vec{0}|=\sqrt{0+0+0}=0$. The recipe would demand $\hat{0}=\frac{\vec{0}}{0}$, which is undefined — you cannot divide by zero. Geometric reason: the zero vector is a point, not an arrow — it has no direction to point in. A unit vector's only job is to encode a direction, so there is nothing for it to represent. Conclusion: the zero vector has no unit vector. This is the one input where the construction fails, and it fails because the very concept (a direction) is absent.

Equal angles with $x$, $y$, $z$ and all-positive means equal positive components: take $\vec{A}=\hat{i}+\hat{j}+\hat{k}$ (any positive multiple gives the same direction). Step 1: $|\vec{A}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$. Step 2: $\hat{A}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\approx 0.577(\hat{i}+\hat{j}+\hat{k})$. Sanity check: $3\times {\left(\frac{1}{\sqrt{3}}\right)}^2=3\times \frac{1}{3}=1$ ✓. This is the famous "body diagonal" direction of a unit cube.

Length first ($\sqrt{\text{sum of squares}}$), then divide to point ($\hat{A}=\vec{A}/|\vec{A}|$), keep every sign, expect $\pm$ from magnitude conditions, add full vectors before normalizing — and never normalize $\vec{0}$.