Multivariable-Calculus · Unit 2 · Video 1 · Interactive Practice

Any Direction Is an Axis: Components From One Dot Product

IKey Formulas

FormulaNameWhat you need
A⋅B=a1b1+a2b2+a3b3\mathbf{A}\cdot\mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3Componentwise formCoordinates only
A⋅B=∣A∣ ∣B∣cos⁡θ\mathbf{A}\cdot\mathbf{B} = |\mathbf{A}|\,|\mathbf{B}|\cos\thetaGeometric formTwo lengths and the angle
compu^ A=A⋅u^=∣A∣cos⁡θ\text{comp}_{\hat{u}}\,\mathbf{A} = \mathbf{A}\cdot\hat{u} = |\mathbf{A}|\cos\thetaComponent along a directionA\mathbf{A} and a unit vector u^\hat{u}
A⋅i^=a1,A⋅j^=a2,A⋅k^=a3\mathbf{A}\cdot\hat{i} = a_1,\quad \mathbf{A}\cdot\hat{j} = a_2,\quad \mathbf{A}\cdot\hat{k} = a_3Coordinates are dot productsThe three axis directions

Key Insight: Nothing in A⋅u^\mathbf{A}\cdot\hat{u} knows that i^\hat{i} is an axis — tilt u^\hat{u} anywhere and the same dot product still reads off the component along it. The hat is the whole hypothesis: A⋅(2u^)=2(A⋅u^)\mathbf{A}\cdot(2\hat{u}) = 2(\mathbf{A}\cdot\hat{u}) is twice the component, not the component.

IIVisualization 1 — The Shadow on a Chosen Axis

The component of A\mathbf{A} along u^\hat{u} is the signed length of the shadow A\mathbf{A} casts on that line.

💡 The same shadow argument runs unchanged in space: a third component adds one more product to the sum, and the perpendicular is dropped onto a line in three dimensions instead of two.

IIIVisualization 2 — Why the Hat Matters

Dotting with c u^c\,\hat{u} scales the answer by cc, while the shadow itself never moves.

IVVisualization 3 — Resolving the Weight on a Pendulum

Tangent and normal are the pendulum's natural axes, and each part of the weight is one dot product.

💡 The string supplies exactly the force that cancels F⋅n^\mathbf{F}\cdot\hat{n} — that is the tension, and it is why the string stays taut; only F⋅T^\mathbf{F}\cdot\hat{T} is left over to accelerate the mass along the arc.

VQuiz Questions

Problem 1 · A Component Along a Tilted Direction

Given: A=⟨4,3⟩\mathbf{A} = \langle 4, 3 \rangle and u^=⟨35,45⟩\hat{u} = \left\langle \tfrac{3}{5}, \tfrac{4}{5} \right\rangle — find the component of A\mathbf{A} along u^\hat{u}.

✅ Correct! A⋅u^=125+125=245=4.8\mathbf{A}\cdot\hat{u} = \tfrac{12}{5} + \tfrac{12}{5} = \tfrac{24}{5} = 4.8 — slightly shorter than ∣A∣=5|\mathbf{A}| = 5, as a shadow must be.
❌ Not quite. 55 is ∣A∣|\mathbf{A}| itself, which would be the component only if A\mathbf{A} pointed straight along u^\hat{u}. Here cos⁡θ=2425<1\cos\theta = \tfrac{24}{25} \lt 1, so the shadow is shorter.
❌ Close, but that is cos⁡θ\cos\theta. Dividing by ∣A∣|\mathbf{A}| once more turns the component into the cosine. The component is ∣A∣cos⁡θ=5⋅2425|\mathbf{A}|\cos\theta = 5 \cdot \tfrac{24}{25}.
❌ Not quite. Both products count: 4⋅35=1254 \cdot \tfrac{3}{5} = \tfrac{12}{5} and 3⋅45=1253 \cdot \tfrac{4}{5} = \tfrac{12}{5}.
Show solution

First confirm the hat is honest: u^⋅u^=925+1625=1\hat{u}\cdot\hat{u} = \tfrac{9}{25} + \tfrac{16}{25} = 1, so ∣u^∣=1|\hat{u}| = 1.

A⋅u^=4⋅35+3⋅45=125+125=245=4.8\mathbf{A}\cdot\hat{u} = 4\cdot\tfrac{3}{5} + 3\cdot\tfrac{4}{5} = \tfrac{12}{5} + \tfrac{12}{5} = \tfrac{24}{5} = 4.8

The geometric form agrees: ∣A∣=16+9=5|\mathbf{A}| = \sqrt{16+9} = 5, so

cos⁡θ=24/55=2425=0.96,θ≈16.26∘,\cos\theta = \frac{24/5}{5} = \frac{24}{25} = 0.96, \qquad \theta \approx 16.26^\circ,

and ∣A∣cos⁡θ=5(0.96)=4.8|\mathbf{A}|\cos\theta = 5(0.96) = 4.8 — the same number.

Problem 2 · The Direction Vector Is Not a Unit Vector

Given: A=⟨2,3,6⟩\mathbf{A} = \langle 2, 3, 6 \rangle and v=⟨0,3,4⟩\mathbf{v} = \langle 0, 3, 4 \rangle — find the component of A\mathbf{A} along the direction of v\mathbf{v}.

✅ Correct! ∣v∣=5|\mathbf{v}| = 5, so v^=⟨0,35,45⟩\hat{v} = \left\langle 0, \tfrac{3}{5}, \tfrac{4}{5} \right\rangle and A⋅v^=335=6.6\mathbf{A}\cdot\hat{v} = \tfrac{33}{5} = 6.6.
❌ That is A⋅v\mathbf{A}\cdot\mathbf{v}, not the component. Since ∣v∣=5|\mathbf{v}| = 5, dotting with v\mathbf{v} gives five times the component: A⋅v=∣v∣ (A⋅v^)\mathbf{A}\cdot\mathbf{v} = |\mathbf{v}|\,(\mathbf{A}\cdot\hat{v}).
❌ You divided by ∣v∣2|\mathbf{v}|^2. Normalizing divides by the length once: v^=v/∣v∣\hat{v} = \mathbf{v}/|\mathbf{v}|, so the component is 335\tfrac{33}{5}.
❌ Not quite. 7=∣A∣7 = |\mathbf{A}|, the longest a component of A\mathbf{A} could ever be. The actual component is 6.66.6, just short of it.
Show solution

v\mathbf{v} is a direction, but not a unit vector:

∣v∣=0+9+16=5⟹v^=⟨0,35,45⟩|\mathbf{v}| = \sqrt{0 + 9 + 16} = 5 \quad\Longrightarrow\quad \hat{v} = \left\langle 0, \tfrac{3}{5}, \tfrac{4}{5} \right\rangle

Now take one dot product:

A⋅v^=2(0)+3(35)+6(45)=95+245=335=6.6\mathbf{A}\cdot\hat{v} = 2(0) + 3\left(\tfrac{3}{5}\right) + 6\left(\tfrac{4}{5}\right) = \tfrac{9}{5} + \tfrac{24}{5} = \tfrac{33}{5} = 6.6

Equivalently, normalize at the end: A⋅v∣v∣=335\dfrac{\mathbf{A}\cdot\mathbf{v}}{|\mathbf{v}|} = \dfrac{33}{5}.

Sanity check: ∣A∣=4+9+36=7|\mathbf{A}| = \sqrt{4 + 9 + 36} = 7, and 6.6<76.6 \lt 7 — a component can never exceed the vector's own length.

Problem 3 · Resolving the Weight of a Pendulum

Given: a pendulum whose string makes θ=30∘\theta = 30^\circ with the vertical. The weight F\mathbf{F} points straight down with ∣F∣=20|\mathbf{F}| = 20 N; n^\hat{n} is the unit vector along the string away from the pivot and T^\hat{T} the unit tangent to the arc, pointing toward the bottom of the swing.

What is the normal component F⋅n^\mathbf{F}\cdot\hat{n}?

What is the tangential component F⋅T^\mathbf{F}\cdot\hat{T}?

✅ Correct! F⋅n^=20cos⁡30∘=103≈17.32\mathbf{F}\cdot\hat{n} = 20\cos 30^\circ = 10\sqrt{3} \approx 17.32 N and F⋅T^=20sin⁡30∘=10\mathbf{F}\cdot\hat{T} = 20\sin 30^\circ = 10 N, and (103)2+102=400=∣F∣2(10\sqrt{3})^2 + 10^2 = 400 = |\mathbf{F}|^2.
❌ Check the angle in the normal direction. The string makes θ\theta with the vertical and F\mathbf{F} is the vertical, so the angle between F\mathbf{F} and n^\hat{n} is θ=30∘\theta = 30^\circ — a cosine, not a sine.
❌ That is all of ∣F∣|\mathbf{F}|. The whole weight would land on n^\hat{n} only if the string were vertical (θ=0\theta = 0). At θ=30∘\theta = 30^\circ the shadow on n^\hat{n} is shorter: 20cos⁡30∘=103≈17.3220\cos 30^\circ = 10\sqrt{3} \approx 17.32 N.
❌ Zero would mean F⊥n^\mathbf{F} \perp \hat{n}. The weight is perpendicular to the string only at θ=90∘\theta = 90^\circ. Here the angle between F\mathbf{F} and n^\hat{n} is 30∘30^\circ, so F⋅n^=20cos⁡30∘=103\mathbf{F}\cdot\hat{n} = 20\cos 30^\circ = 10\sqrt{3} N.
❌ Check the angle in the tangential direction. T^⊥n^\hat{T} \perp \hat{n}, so the angle from F\mathbf{F} to T^\hat{T} is 90∘−θ=60∘90^\circ - \theta = 60^\circ, and cos⁡60∘=sin⁡30∘=12\cos 60^\circ = \sin 30^\circ = \tfrac{1}{2}.
Show solution

Normal direction. n^\hat{n} runs along the string, which is tilted θ\theta from the vertical; F\mathbf{F} points straight down. So the angle between them is θ\theta:

F⋅n^=∣F∣ ∣n^∣cos⁡θ=20cos⁡30∘=20⋅32=103≈17.32 N\mathbf{F}\cdot\hat{n} = |\mathbf{F}|\,|\hat{n}|\cos\theta = 20\cos 30^\circ = 20\cdot\frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32 \text{ N}

Tangential direction. T^\hat{T} is perpendicular to the string, so the angle from F\mathbf{F} to T^\hat{T} is 90∘−θ90^\circ - \theta:

F⋅T^=∣F∣cos⁡(90∘−θ)=∣F∣sin⁡θ=20sin⁡30∘=20⋅12=10 N\mathbf{F}\cdot\hat{T} = |\mathbf{F}|\cos(90^\circ - \theta) = |\mathbf{F}|\sin\theta = 20\sin 30^\circ = 20\cdot\tfrac{1}{2} = 10 \text{ N}

Check. The two components are the legs of a rectangle whose diagonal is F\mathbf{F}:

(103)2+102=300+100=400=202=∣F∣2 ✓(10\sqrt{3})^2 + 10^2 = 300 + 100 = 400 = 20^2 = |\mathbf{F}|^2\ \checkmark

The 10310\sqrt{3} N is what the string's tension must cancel; the 1010 N is what actually swings the pendulum.

Problem 4 · Reading a Vector in a Rotated Pair of Axes

Given: the perpendicular unit vectors u^=⟨35,45⟩\hat{u} = \left\langle \tfrac{3}{5}, \tfrac{4}{5} \right\rangle and w^=⟨−45,35⟩\hat{w} = \left\langle -\tfrac{4}{5}, \tfrac{3}{5} \right\rangle, and the vector A=⟨5,0⟩\mathbf{A} = \langle 5, 0 \rangle — find its two components in this tilted frame.

What is A⋅u^\mathbf{A}\cdot\hat{u}?

What is A⋅w^\mathbf{A}\cdot\hat{w}?

✅ Correct! In the tilted frame A=3u^−4w^\mathbf{A} = 3\hat{u} - 4\hat{w}, and 32+(−4)2=25=∣A∣23^2 + (-4)^2 = 25 = |\mathbf{A}|^2 — same vector, same length, new axes.
❌ Check the first dot product. Only the xx-term survives: 5(35)+0(45)5\left(\tfrac{3}{5}\right) + 0\left(\tfrac{4}{5}\right).
❌ Check the sign. w^\hat{w} has a negative xx-component, so A\mathbf{A} leans more than 90∘90^\circ away from it and the component must come out negative.
❌ Zero would mean A⊥w^\mathbf{A} \perp \hat{w}. It is u^\hat{u} that is perpendicular to w^\hat{w}, not A\mathbf{A}. Here A⋅w^=5(−45)+0(35)=−4\mathbf{A}\cdot\hat{w} = 5\left(-\tfrac{4}{5}\right) + 0\left(\tfrac{3}{5}\right) = -4.
❌ Right sign, wrong size. That is just w^\hat{w}'s own xx-component; the factor 55 from A\mathbf{A} still multiplies it: 5(−45)=−45\left(-\tfrac{4}{5}\right) = -4, not −45-\tfrac{4}{5}.
Show solution

Both are unit vectors and they are perpendicular:

u^⋅u^=925+1625=1,w^⋅w^=1625+925=1,u^⋅w^=−1225+1225=0\hat{u}\cdot\hat{u} = \tfrac{9}{25} + \tfrac{16}{25} = 1, \qquad \hat{w}\cdot\hat{w} = \tfrac{16}{25} + \tfrac{9}{25} = 1, \qquad \hat{u}\cdot\hat{w} = -\tfrac{12}{25} + \tfrac{12}{25} = 0

So they are a legitimate pair of axes. One dot product each:

A⋅u^=5(35)+0(45)=3,A⋅w^=5(−45)+0(35)=−4\mathbf{A}\cdot\hat{u} = 5\left(\tfrac{3}{5}\right) + 0\left(\tfrac{4}{5}\right) = 3, \qquad \mathbf{A}\cdot\hat{w} = 5\left(-\tfrac{4}{5}\right) + 0\left(\tfrac{3}{5}\right) = -4

The minus sign is information, not an error: A\mathbf{A} makes an obtuse angle with w^\hat{w}, so its shadow falls on the far side of the origin.

Rebuild A\mathbf{A} from its new components:

3u^−4w^=⟨95+165, 125−125⟩=⟨5,0⟩=A ✓3\hat{u} - 4\hat{w} = \left\langle \tfrac{9}{5} + \tfrac{16}{5},\ \tfrac{12}{5} - \tfrac{12}{5} \right\rangle = \langle 5, 0 \rangle = \mathbf{A}\ \checkmark

And 32+(−4)2=25=∣A∣23^2 + (-4)^2 = 25 = |\mathbf{A}|^2: rotating the axes moves the numbers around but never changes the length.

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