Worked examples — Resolution of vectors — into components (any axes)

The scenario matrix

Here is the full landscape of cases this topic can throw at you. Every worked example below is tagged with the cell it covers.

(In cell C7 the symbols $\hat{u}$ and $\hat{v}$ mean the two chosen axis directions — each an arrow of length $1$ pointing along one oblique axis. They are properly defined in Ex 7 before any calculation uses them.)

Ex 1 — Quadrant I, the base case (C1)

Figure 1 (below): a blue arrow of length $20$ in Quadrant I; its horizontal shadow (red, adjacent) and vertical shadow (yellow, opposite) form a right triangle, with the $37^{\circ }$ angle marked at the origin.

1. Draw the right triangle (Figure 1): hypotenuse $=v$, adjacent side along $x$, opposite side along $y$. Why this step? Components are the two perpendicular legs of this triangle — you can't read them off until you've drawn the triangle.
2. $v_x=v\cos \theta =20\cos 37^{\circ }=20(0.7986)=15.97\text{m/s}$. Why cos? The horizontal leg touches (is adjacent to) the angle, and cosine $=\frac{\text{adjacent}}{\text{hyp}}$ — it measures the shadow along the direction the angle is hugging.
3. $v_y=v\sin \theta =20\sin 37^{\circ }=20(0.6018)=12.04\text{m/s}$. Why sin? The vertical leg is across from the angle (opposite), and sine $=\frac{\text{opposite}}{\text{hyp}}$.

Ex 2 — Quadrant II, cosine goes negative (C2)

Figure 2 (below): a blue arrow at $120^{\circ }$ landing in Quadrant II; its horizontal shadow (red) now points backward onto the $-x$ side, while its vertical shadow (yellow) still points up. The $120^{\circ }$ sweep is drawn as a white arc from $+x$.

1. Apply the same formulas — the signs take care of themselves if $\theta$ is measured all the way round from $+x$. Why trust them past $90^{\circ }$? $\cos$ and $\sin$ are defined for any angle by where the arrow lands on the unit circle, not just inside a triangle. Past $90^{\circ }$ the tip is left of the $y$-axis, so its $x$-shadow points backward → cosine is negative.
2. $F_x=10\cos 120^{\circ }=10(-0.5)=-5.0\text{N}$. Why negative? The shadow lies on the $-x$ side — the minus sign is the direction, encoded automatically.
3. $F_y=10\sin 120^{\circ }=10(0.8660)=8.66\text{N}$ (positive, as forecast).

Ex 3 — Recovering the angle in Quadrant III (C3a)

Figure 3 (below): the true vector (red) points down-left into Quadrant III; a dashed yellow arrow shows where the naive ${\tan }^{-1}$ answer wrongly lands, up in Quadrant I. Both share the same slope — that's exactly why ${\tan }^{-1}$ is fooled.

1. Magnitude by Pythagoras: $A=\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5$. Why Pythagoras? The components are perpendicular legs; the vector is the hypotenuse of that same right triangle.
2. Naive angle: ${\tan }^{-1}\left(\frac{A_y}{A_x}\right)={\tan }^{-1}\left(\frac{-4}{-3}\right)={\tan }^{-1}(1.333)=53.13^{\circ }$. Why is this wrong? $\tan$ repeats every $180^{\circ }$: $\tan (53.13^{\circ })$ and $\tan (53.13^{\circ }+180^{\circ })$ are identical, so ${\tan }^{-1}$ can't tell QI from QIII. It always returns the QI/QIV branch.
3. Fix from the quadrant picture: we know the arrow is in QIII, so add $180^{\circ }$: $\theta =53.13^{\circ }+180^{\circ }=233.13^{\circ }.$ Why add $180^{\circ }$? Rotating half a turn flips both shadows' signs, sending $(+,+)$ to $(-,-)$ — exactly our quadrant, with the same tan value.

Ex 3b — Recovering the angle in Quadrant IV (C3b)

1. Magnitude by Pythagoras: $A=\sqrt{3^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5$. Why? Same right triangle; the signs of the legs don't change their squared lengths.
2. Naive angle: ${\tan }^{-1}\left(\frac{-4}{3}\right)={\tan }^{-1}(-1.333)=-53.13^{\circ }$. Why is this the QIV branch? ${\tan }^{-1}$ always returns an answer between $-90^{\circ }$ and $+90^{\circ }$. A negative result means "below the $+x$ axis" — which is Quadrant IV. So here the naive branch is already correct.
3. Express as a positive angle (optional): add $360^{\circ }$ to get $-53.13^{\circ }+360^{\circ }=306.87^{\circ }$. Why? $-53.13^{\circ }$ and $306.87^{\circ }$ are the same direction — one measured clockwise, the other the full counter-clockwise sweep from $+x$.

Ex 4 — Degenerate & limiting cases (C4)

1. (a) $A_x=7\cos 0^{\circ }=7(1)=7$, $A_y=7\sin 0^{\circ }=7(0)=0$. Why? The arrow lies flat along $x$; its vertical shadow has length zero. Cosine of $0$ is $1$ (full shadow), sine of $0$ is $0$ (no rise).
2. (b) $B_x=7\cos 90^{\circ }=7(0)=0$, $B_y=7\sin 90^{\circ }=7(1)=7$. Why? Now the arrow stands straight up; nothing spills onto $x$. This is the limiting mirror of (a).
3. (c) $|\vec{0}|=0$, and ${\tan }^{-1}(0/0)$ is undefined — division by zero. Why? A zero-length arrow has no tip to point anywhere, so "its angle" is a meaningless question. In code you must guard against $A_x=A_y=0$ before calling ${\tan }^{-1}$.

Ex 5 — Tilted perpendicular axes: inclined plane (C5)

Figure 4 (below): a wedge inclined at $30^{\circ }$ with a block on it. The blue arrow is the straight-down weight $W$; the yellow arrow is its down-slope part $W\sin 30^{\circ }$; the red arrow is its into-surface part $W\cos 30^{\circ }$. The $30^{\circ }$ base angle is marked.

1. Tilt the axes to align with the slope (Figure 4): one axis down the slope, one perpendicular (into the surface). Why tilt? Motion happens along the slope. Tilted axes make one component the driving force and the other a balanced force — much simpler than plain $x$–$y$. See Inclined plane dynamics.
2. Geometry (Figure 4): the angle between the vertical weight and the perpendicular-to-slope axis equals $\theta$. So the along-slope leg is opposite to $\theta$: $W_{\parallel }=W\sin \theta =50\sin 30^{\circ }=50(0.5)=25\text{N}.$ Why sin? Opposite side → sine.
3. The perpendicular leg is adjacent to $\theta$: $W_{\perp }=W\cos \theta =50\cos 30^{\circ }=50(0.8660)=43.30\text{N}.$ Why cos? Adjacent side → cosine. This is what the normal force $N$ balances.

Ex 6 — Projection onto an arbitrary direction (C6)

Figure 5 (below): the blue arrow $\vec{A}=(6,8)$, a dashed yellow ray for the $\hat{n}$ direction at $30^{\circ }$, and a green arrow showing the shadow of $\vec{A}$ dropped perpendicularly onto that ray — its length is the projection $9.196$.

1. Build the unit vector: $\hat{n}=(\cos 30^{\circ },\sin 30^{\circ })=(0.8660,0.5)$. Why a unit vector? A component is a length; the dot product $\vec{A}\cdot \hat{n}$ only reads a pure length if $\hat{n}$ has length $1$.
2. Dot product = projection: $A_{\hat{n}}=\vec{A}\cdot \hat{n}=6(0.8660)+8(0.5)=5.196+4.0=\mathrm{9.196.}$ Why the dot product? As proven in the parent note, $\vec{A}\cdot \hat{n}=A\cos \varphi$ is exactly the shadow of $\vec{A}$ along $\hat{n}$, where $\varphi$ is the vector-to-vector angle just defined. See Dot product & scalar projection.
3. Cross-check via that angle $\varphi$: here $\varphi =53.13^{\circ }-30^{\circ }=23.13^{\circ }$ (the gap between $\vec{A}$'s $53.13^{\circ }$ and $\hat{n}$'s $30^{\circ }$), so $A\cos \varphi =10\cos 23.13^{\circ }=10(0.9196)=9.196$. ✓ Why both routes? They must agree — that's the whole content of "dot product IS projection."

Ex 7 — Oblique axes: projection ≠ component (C7)

Figure 6 (below): the blue diagonal $\vec{A}=(4,3)$; the yellow side $a\hat{u}$ laid along the horizontal axis; the green side $b\hat{v}$ laid along the $60^{\circ }$ axis, drawn from the tip of $a\hat{u}$ up to $\vec{A}$'s tip so the two sides close the parallelogram back onto $\vec{A}$.

1. Why not just project? For $60^{\circ }$ axes, the $\hat{v}$ piece has its own shadow on $\hat{u}$ — a "leak." Projection would double-count that leak. We must instead demand the pieces add back to $\vec{A}$. This is the parallelogram idea (see the definition above): $\vec{A}$ is the diagonal; $a\hat{u}$ and $b\hat{v}$ are the two sides.
2. Write one equation per coordinate ($\vec{A}=a\hat{u}+b\hat{v}$): $x:4=a(1)+b(0.5),y:3=a(0)+b(0.8660).$ Why two equations, and why does this "solve the system"? The single vector statement $\vec{A}=a\hat{u}+b\hat{v}$ is really two statements — the $x$-parts must match AND the $y$-parts must match. Two matching conditions pin down the two unknowns $a,b$ exactly. Solving them forces the reassembled pieces to equal $\vec{A}$ in both directions at once, which is precisely the definition of a component.
3. From the $y$-equation: $b=\frac{3}{0.8660}=3.4641$. Substitute into the $x$-equation: $4=a+0.5(3.4641)=a+1.7320$, so $a=4-1.7320=2.2680$.
4. Compare with the naive projection: $\vec{A}\cdot \hat{u}=4$. But the true component is $a=2.268$ — much smaller. Why smaller? The $\hat{v}$-piece already supplied $1.732$ of horizontal reach; the parallelogram must not recount it. Projection $=4$; component $=2.268$; they differ precisely because the axes are not orthogonal.

Ex 8 — Word problem + equilibrium (C8)

1. Translate to components. Rope 1: $(+T_1\cos 50^{\circ },+T_1\sin 50^{\circ })$. Rope 2: $(-T_2\cos 40^{\circ },+T_2\sin 40^{\circ })$. Weight: $(0,-100)$. Why resolve? Equilibrium means the vector sum is zero; but the parent note's rule says components along a fixed axis just add as numbers. So "sum = 0" becomes two scalar equations. See Equilibrium of concurrent forces.
2. $x$-balance: $T_1\cos 50^{\circ }-T_2\cos 40^{\circ }=0\Rightarrow T_1(0.6428)=T_2(0.7660)$, so $T_1=1.1918T_2$. Why? No horizontal motion → horizontal shadows cancel.
3. $y$-balance: $T_1\sin 50^{\circ }+T_2\sin 40^{\circ }=100\Rightarrow T_1(0.7660)+T_2(0.6428)=100$. Why? Vertical shadows must together lift the $100\text{N}$ weight.
4. Substitute $T_1=1.1918T_2$: $1.1918T_2(0.7660)+0.6428T_2=100\Rightarrow (0.9129+0.6428)T_2=100\Rightarrow T_2=\frac{100}{1.5557}=64.28\text{N}.$ Then $T_1=1.1918(64.28)=76.60\text{N}$.

Ex 9 — Exam twist: angle measured from the y-axis (C9)

Figure 7 (below): a blue launch-velocity arrow leaning only $20^{\circ }$ away from straight-up. The vertical shadow (yellow) is the long, adjacent leg; the horizontal shadow (red) is the short, opposite leg. The $20^{\circ }$ angle is drawn hugging the vertical axis.

1. Redraw the triangle with the angle at the correct corner (Figure 7). The $20^{\circ }$ sits between the velocity and the vertical axis. Why redraw? The rule "adjacent → cos" is about which leg touches the angle, not about which is horizontal. Here the vertical leg is adjacent to $20^{\circ }$.
2. Vertical component (adjacent to the angle): $v_y=v\cos 20^{\circ }=25\cos 20^{\circ }=25(0.9397)=23.49\text{m/s}$. Why cos? The vertical leg hugs the $20^{\circ }$ angle → it is the adjacent side → cosine.
3. Horizontal component (opposite the angle): $v_x=v\sin 20^{\circ }=25\sin 20^{\circ }=25(0.3420)=8.55\text{m/s}$. Why sin? The horizontal leg is across from the angle → opposite side → sine. This value feeds straight into Projectile motion as the constant horizontal speed.

Self-test

Connections

• Resolution of vectors — into components (any axes) (parent)
• Vectors — addition (parallelogram & triangle law)
• Dot product & scalar projection
• Unit vectors and Cartesian coordinates
• Inclined plane dynamics
• Projectile motion
• Equilibrium of concurrent forces