Vectors and the Dot Product

Study Sheet

Vectors and the Dot Product

Components, magnitude & direction, the dot product, projection, and applications

Vectors: Geometric and Component Form

Tip

Rounding for this topic: Give exact values (fractions or radicals) whenever they are clean. Otherwise round magnitudes and components to two decimal places and angles to the nearest 0.1∘0.1^\circ. Keep full precision in your calculator until the final step.

A vector has magnitude and direction; in components it is an arrow from the origin to (a,b)(a, b). Add vectors component-wise (tip to tail), scale them by multiplying each component, and use the dot product u⋅v=∣u∣∣v∣cos⁡θ\mathbf u\cdot\mathbf v = |\mathbf u||\mathbf v|\cos\theta to find angles and projections. A zero dot product means perpendicular.

Concept
What a vector is
xyP(1, 2)Q(5, 5)43
PQ⃗=⟨4,3⟩\vec{PQ}=\langle 4,3\rangle: over 44, up 33. The faint copy from the origin is the same vector.

A vector has both magnitude (length) and direction. Draw it as an arrow. Written in component form as v⃗=⟨a,b⟩\vec v=\langle a,b\rangle, where aa is the horizontal change and bb the vertical change.

From P(p1,p2) to Q(q1,q2):PQ⃗=⟨q1−p1, q2−p2⟩.\text{From }P(p_1,p_2)\text{ to }Q(q_1,q_2):\quad \vec{PQ}=\langle q_1-p_1,\ q_2-p_2\rangle .

Two vectors are equal when they have the same components — same magnitude and same direction — no matter where they are drawn.

Concept
Magnitude and direction angle
∥v⃗∥=∥⟨a,b⟩∥=a2+b2,tan⁡θ=ba.\|\vec v\|=\|\langle a,b\rangle\|=\sqrt{a^2+b^2},\qquad \tan\theta=\frac{b}{a}.

The direction angle θ\theta is measured counterclockwise from the positive xx-axis. Use the signs of aa and bb to place θ\theta in the correct quadrant.

Example
Full magnitude and direction
xy36.9°‖v‖ = 543
The components are the legs, the magnitude is the hypotenuse.

Find the magnitude and direction angle of v⃗=⟨4,3⟩\vec v=\langle 4,3\rangle.

∥v⃗∥=42+32=16+9=25=5.\|\vec v\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5.

Since a=4>0a=4>0 and b=3>0b=3>0, v⃗\vec v is in Quadrant I:

θ=tan⁡−1 ⁣(34)=36.9∘.\theta=\tan^{-1}\!\left(\tfrac{3}{4}\right)=36.9^\circ .

Answer: magnitude 55, direction ≈36.9∘\approx 36.9^\circ.

Side note
Where it goes wrong

Adding magnitudes to get the magnitude of a sum. ∣u+v∣|\mathbf u + \mathbf v| depends on the angle between them; only parallel vectors add their lengths.

Tip

Quadrant tip: tan⁡−1\tan^{-1} on a calculator only returns −90∘-90^\circ to 90∘90^\circ. If a<0a<0 (Quadrant II or III), add 180∘180^\circ to the calculator value. If a>0, b<0a>0,\ b<0 (Quadrant IV), add 360∘360^\circ to get a positive angle.

Try it
Quick check

Find the component form, magnitude, and direction angle of the vector from P(−2,5)P(-2,5) to Q(4,−3)Q(4,-3).

Show the answer

Answer: ⟨6,−8⟩\langle 6,-8\rangle, magnitude 1010, direction ≈306.9∘\approx 306.9^\circ. Subtract tail from tip: ⟨4−(−2), −3−5⟩=⟨6,−8⟩\langle 4-(-2),\,-3-5\rangle=\langle 6,-8\rangle, and 36+64=10\sqrt{36+64}=10. The calculator gives tan⁡−1 ⁣(−86)≈−53.1∘\tan^{-1}\!\left(\tfrac{-8}{6}\right)\approx-53.1^\circ; the vector points into Quadrant IV, so add 360∘360^\circ: 306.9∘306.9^\circ.

Vector Operations

Concept
Adding, subtracting, scaling, unit vectors
⟨a,b⟩+⟨c,d⟩=⟨a+c, b+d⟩(add tip-to-tail)⟨a,b⟩−⟨c,d⟩=⟨a−c, b−d⟩(u⃗−v⃗=u⃗+(−v⃗))k⟨a,b⟩=⟨ka, kb⟩(scalar multiple)\begin{aligned} \langle a,b\rangle+\langle c,d\rangle&=\langle a+c,\ b+d\rangle &\text{(add tip-to-tail)}\\ \langle a,b\rangle-\langle c,d\rangle&=\langle a-c,\ b-d\rangle &\text{($\vec u-\vec v=\vec u+(-\vec v)$)}\\ k\langle a,b\rangle&=\langle ka,\ kb\rangle &\text{(scalar multiple)} \end{aligned}

A unit vector has magnitude 11. To find the unit vector in the direction of v⃗\vec v:

u^=v⃗∥v⃗∥=⟨a∥v⃗∥, b∥v⃗∥⟩.\hat u=\frac{\vec v}{\|\vec v\|}=\left\langle \frac{a}{\|\vec v\|},\ \frac{b}{\|\vec v\|}\right\rangle .

Standard basis / i,j\mathbf{i},\mathbf{j} form: i=⟨1,0⟩\mathbf i=\langle 1,0\rangle, j=⟨0,1⟩\mathbf j=\langle 0,1\rangle, so ⟨a,b⟩=a i+b j.\langle a,b\rangle=a\,\mathbf i+b\,\mathbf j.

Adding vectors means doing one displacement and then the other. The horizontal changes simply add, and so do the vertical ones, which is why the sum is computed component by component and why the picture is tip-to-tail. Scaling by kk stretches the arrow (and flips it if k<0k<0) because every component is multiplied by the same factor. Dividing by the length gives a unit vector: same direction, length 11.

Example
Operations and a unit vector
xy2u = ⟨6, −4⟩−v⟨5, −8⟩
Tip to tail: go along 2u⃗2\vec u, then along −v⃗-\vec v; the dashed arrow is 2u⃗−v⃗2\vec u-\vec v.

Let u⃗=⟨3,−2⟩\vec u=\langle 3,-2\rangle and v⃗=⟨1,4⟩\vec v=\langle 1,4\rangle.

2u⃗−v⃗=⟨6,−4⟩−⟨1,4⟩=⟨5,−8⟩.2\vec u-\vec v=\langle 6,-4\rangle-\langle 1,4\rangle=\langle 5,-8\rangle .

Unit vector for u⃗\vec u: ∥u⃗∥=9+4=13\|\vec u\|=\sqrt{9+4}=\sqrt{13}, so

u^=⟨313, −213⟩≈⟨0.83, −0.55⟩.\hat u=\left\langle \tfrac{3}{\sqrt{13}},\ \tfrac{-2}{\sqrt{13}}\right\rangle\approx\langle 0.83,\,-0.55\rangle .

In i,j\mathbf i,\mathbf j form: u⃗=3 i−2 j\vec u=3\,\mathbf i-2\,\mathbf j.

Try it
Quick check

Let u⃗=⟨−1,4⟩\vec u=\langle -1,4\rangle and v⃗=⟨3,2⟩\vec v=\langle 3,2\rangle. Find 3u⃗+2v⃗3\vec u+2\vec v and the unit vector in the direction of v⃗\vec v.

Show the answer

Answer: 3u⃗+2v⃗=⟨3,16⟩3\vec u+2\vec v=\langle 3,16\rangle; v^=⟨313,213⟩≈⟨0.83,0.55⟩\hat v=\left\langle \tfrac{3}{\sqrt{13}},\tfrac{2}{\sqrt{13}}\right\rangle\approx\langle 0.83,0.55\rangle. 3u⃗=⟨−3,12⟩3\vec u=\langle -3,12\rangle and 2v⃗=⟨6,4⟩2\vec v=\langle 6,4\rangle add to ⟨3,16⟩\langle 3,16\rangle. ∥v⃗∥=9+4=13\|\vec v\|=\sqrt{9+4}=\sqrt{13}; divide each component by it.

Writing a Vector from Magnitude and Direction

Concept
Resolving into components

A vector with magnitude r=∥v⃗∥r=\|\vec v\| and direction angle θ\theta has components

v⃗=⟨rcos⁡θ, rsin⁡θ⟩.\vec v=\langle r\cos\theta,\ r\sin\theta\rangle .

This is the reverse of finding magnitude and direction — it resolves the vector into its horizontal and vertical parts.

Example
Magnitude and direction to components
xy30°1010 cos 30° = 5√310 sin 30° = 5
Resolving: the horizontal leg is rcos⁡θr\cos\theta, the vertical leg rsin⁡θr\sin\theta.

Write the vector of magnitude 1010 pointing at 30∘30^\circ.

v⃗=⟨10cos⁡30∘, 10sin⁡30∘⟩=⟨10⋅32, 10⋅12⟩=⟨53, 5⟩≈⟨8.66, 5⟩.\vec v=\langle 10\cos 30^\circ,\ 10\sin 30^\circ\rangle=\left\langle 10\cdot\tfrac{\sqrt3}{2},\ 10\cdot\tfrac12\right\rangle=\langle 5\sqrt3,\ 5\rangle\approx\langle 8.66,\ 5\rangle .
Tip

Check: resolving and then re-computing the magnitude must return rr: (53)2+52=75+25=100=10.\sqrt{(5\sqrt3)^2+5^2}=\sqrt{75+25}=\sqrt{100}=10. ✓

Try it
Quick check

Write the vector of magnitude 88 with direction angle 135∘135^\circ in component form.

Show the answer

Answer: ⟨−42, 42⟩≈⟨−5.66, 5.66⟩\langle -4\sqrt2,\ 4\sqrt2\rangle\approx\langle -5.66,\ 5.66\rangle. ⟨8cos⁡135∘, 8sin⁡135∘⟩=⟨8⋅(−22), 8⋅22⟩\langle 8\cos135^\circ,\ 8\sin135^\circ\rangle=\left\langle 8\cdot\left(-\tfrac{\sqrt2}{2}\right),\ 8\cdot\tfrac{\sqrt2}{2}\right\rangle. The signs come out right automatically: 135∘135^\circ is in Quadrant II, left and up.

The Dot Product, Angle Between Vectors, Orthogonality

Concept
Dot product and the angle between vectors
u⃗⋅v⃗=⟨a,b⟩⋅⟨c,d⟩=ac+bd(a scalar, not a vector).\vec u\cdot\vec v=\langle a,b\rangle\cdot\langle c,d\rangle=ac+bd \qquad(\text{a scalar, not a vector}).
u⃗⋅v⃗=∥u⃗∥ ∥v⃗∥cos⁡θ⟹cos⁡θ=u⃗⋅v⃗∥u⃗∥ ∥v⃗∥,0∘≤θ≤180∘.\vec u\cdot\vec v=\|\vec u\|\,\|\vec v\|\cos\theta \qquad\Longrightarrow\qquad \cos\theta=\frac{\vec u\cdot\vec v}{\|\vec u\|\,\|\vec v\|},\quad 0^\circ\le\theta\le 180^\circ .

Orthogonal (perpendicular): u⃗⊥v⃗\vec u\perp\vec v exactly when u⃗⋅v⃗=0\vec u\cdot\vec v=0.

The two formulas meet in the Law of Cosines. The third side of the triangle formed by u⃗\vec u and v⃗\vec v is u⃗−v⃗\vec u-\vec v, so ∥u⃗−v⃗∥2=∥u⃗∥2+∥v⃗∥2−2∥u⃗∥∥v⃗∥cos⁡θ\|\vec u-\vec v\|^2=\|\vec u\|^2+\|\vec v\|^2-2\|\vec u\|\|\vec v\|\cos\theta. Expanding the left side in components gives ∥u⃗∥2+∥v⃗∥2−2(ac+bd)\|\vec u\|^2+\|\vec v\|^2-2(ac+bd). Matching the two, ac+bd=∥u⃗∥∥v⃗∥cos⁡θac+bd=\|\vec u\|\|\vec v\|\cos\theta. When θ=90∘\theta=90^\circ the cosine is 00, which is the orthogonality test.

Example
Angle between two vectors
xyu = ⟨3, 1⟩v = ⟨2, 4⟩45°
The dot product 1010 and the lengths 10,20\sqrt{10},\sqrt{20} give cos⁡θ=22\cos\theta=\tfrac{\sqrt2}{2}.

Find the angle between u⃗=⟨3,1⟩\vec u=\langle 3,1\rangle and v⃗=⟨2,4⟩\vec v=\langle 2,4\rangle.

u⃗⋅v⃗=3(2)+1(4)=10,∥u⃗∥=10,∥v⃗∥=20.\vec u\cdot\vec v=3(2)+1(4)=10,\qquad \|\vec u\|=\sqrt{10},\quad \|\vec v\|=\sqrt{20}.
cos⁡θ=101020=10200=1014.142=0.7071 ⟹ θ=45∘.\cos\theta=\frac{10}{\sqrt{10}\sqrt{20}}=\frac{10}{\sqrt{200}}=\frac{10}{14.142}=0.7071 \ \Longrightarrow\ \theta=45^\circ .
Tip

Orthogonal check: ⟨2,3⟩⋅⟨3,−2⟩=6−6=0\langle 2,3\rangle\cdot\langle 3,-2\rangle=6-6=0, so these vectors are perpendicular. A negative dot product means the angle is obtuse; a positive one means acute.

Try it
Quick check

(a) Find the angle between ⟨2,−1⟩\langle 2,-1\rangle and ⟨1,3⟩\langle 1,3\rangle. (b) Are ⟨4,6⟩\langle 4,6\rangle and ⟨−3,2⟩\langle -3,2\rangle orthogonal?

Show the answer

(a) ≈98.1∘\approx 98.1^\circ. The dot product is 2−3=−12-3=-1; the lengths are 5\sqrt5 and 10\sqrt{10}. So cos⁡θ=−150≈−0.1414\cos\theta=\dfrac{-1}{\sqrt{50}}\approx-0.1414 and θ≈98.1∘\theta\approx 98.1^\circ, just past a right angle, as the negative dot product warned. (b) Yes. 4(−3)+6(2)=−12+12=04(-3)+6(2)=-12+12=0.

Vector Projection and Work

Concept
Projection and parallel/orthogonal decomposition

The (vector) projection of u⃗\vec u onto v⃗\vec v:

proj⁡v⃗u⃗=(u⃗⋅v⃗∥v⃗∥2)v⃗.\operatorname{proj}_{\vec v}\vec u=\left(\frac{\vec u\cdot\vec v}{\|\vec v\|^2}\right)\vec v .

The scalar component of u⃗\vec u along v⃗\vec v is u⃗⋅v⃗∥v⃗∥\dfrac{\vec u\cdot\vec v}{\|\vec v\|}. Every vector splits into a part parallel to v⃗\vec v and a part orthogonal to it:

u⃗=proj⁡v⃗u⃗⏟w⃗1 (∥v⃗)+(u⃗−proj⁡v⃗u⃗)⏟w⃗2 (⊥v⃗).\vec u=\underbrace{\operatorname{proj}_{\vec v}\vec u}_{\vec w_1\ (\parallel\vec v)}+\underbrace{\big(\vec u-\operatorname{proj}_{\vec v}\vec u\big)}_{\vec w_2\ (\perp\vec v)} .

Think of the projection as the shadow of u⃗\vec u on the line of v⃗\vec v when light shines perpendicular to that line. Its signed length is ∥u⃗∥cos⁡θ\|\vec u\|\cos\theta, which by the dot-product formula is u⃗⋅v⃗∥v⃗∥\dfrac{\vec u\cdot\vec v}{\|\vec v\|}. Multiplying that length by the unit vector v⃗∥v⃗∥\dfrac{\vec v}{\|\vec v\|} turns it into a vector, and the two factors of ∥v⃗∥\|\vec v\| combine into ∥v⃗∥2\|\vec v\|^2.

Example
Full projection and decomposition
xyuvproj = ⟨3.12, 4.16⟩w₂
The projection runs along v⃗\vec v; the leftover w⃗2\vec w_2 is perpendicular to it.

Project u⃗=⟨6,2⟩\vec u=\langle 6,2\rangle onto v⃗=⟨3,4⟩\vec v=\langle 3,4\rangle.

u⃗⋅v⃗=6(3)+2(4)=26,∥v⃗∥2=32+42=25.\vec u\cdot\vec v=6(3)+2(4)=26,\qquad \|\vec v\|^2=3^2+4^2=25.
proj⁡v⃗u⃗=2625⟨3,4⟩=⟨7825, 10425⟩=⟨3.12, 4.16⟩.\operatorname{proj}_{\vec v}\vec u=\frac{26}{25}\langle 3,4\rangle=\left\langle \tfrac{78}{25},\ \tfrac{104}{25}\right\rangle=\langle 3.12,\ 4.16\rangle .

Orthogonal part: w⃗2=u⃗−w⃗1=⟨6−3.12, 2−4.16⟩=⟨2.88, −2.16⟩\vec w_2=\vec u-\vec w_1=\langle 6-3.12,\ 2-4.16\rangle=\langle 2.88,\ -2.16\rangle. Check: w⃗1⋅w⃗2=3.12(2.88)+4.16(−2.16)=8.9856−8.9856=0\vec w_1\cdot\vec w_2=3.12(2.88)+4.16(-2.16)=8.9856-8.9856=0. ✓

Concept
Work done by a force

When a constant force F⃗\vec F moves an object through displacement d⃗\vec d,

W=F⃗⋅d⃗=∥F⃗∥ ∥d⃗∥cos⁡θ,W=\vec F\cdot\vec d=\|\vec F\|\,\|\vec d\|\cos\theta,

where θ\theta is the angle between the force and the direction of motion.

Example
Work along a ramp
50 lb30°displacement 20 ft
Only the horizontal part of the pull, 50cos⁡30∘50\cos 30^\circ, does work along the floor.

A rope pulls a crate 2020 ft horizontally with a 5050-lb force directed 30∘30^\circ above the horizontal.

W=∥F⃗∥ ∥d⃗∥cos⁡θ=50(20)cos⁡30∘=1000(0.8660)=866.03 ft-lb.W=\|\vec F\|\,\|\vec d\|\cos\theta=50(20)\cos 30^\circ=1000(0.8660)=866.03\ \text{ft-lb}.
Try it
Quick check

(a) Find proj⁡v⃗u⃗\operatorname{proj}_{\vec v}\vec u for u⃗=⟨5,5⟩\vec u=\langle 5,5\rangle, v⃗=⟨1,2⟩\vec v=\langle 1,2\rangle, and the perpendicular part. (b) A 3030 N force at 40∘40^\circ above horizontal drags a box 1212 m along level ground. Find the work.

Show the answer

(a) u⃗⋅v⃗=15\vec u\cdot\vec v=15 and ∥v⃗∥2=5\|\vec v\|^2=5, so the projection is 3⟨1,2⟩=⟨3,6⟩3\langle1,2\rangle=\langle 3,6\rangle. The perpendicular part is ⟨5,5⟩−⟨3,6⟩=⟨2,−1⟩\langle 5,5\rangle-\langle 3,6\rangle=\langle 2,-1\rangle; check: ⟨2,−1⟩⋅⟨1,2⟩=0\langle 2,-1\rangle\cdot\langle 1,2\rangle=0. (b) W=30⋅12cos⁡40∘≈275.78W=30\cdot 12\cos40^\circ\approx 275.78 J.

Applications: Resultant Forces, Velocity, Navigation, Inclined Planes

Concept
Combining vectors in the real world

Resultant force / velocity: resolve each vector into components, add the components, then re-compute magnitude and direction.

R⃗=F⃗1+F⃗2,∥R⃗∥=Rx2+Ry2,θ=tan⁡−1 ⁣RyRx (adjust quadrant).\vec R=\vec F_1+\vec F_2,\qquad \|\vec R\|=\sqrt{R_x^2+R_y^2},\qquad \theta=\tan^{-1}\!\frac{R_y}{R_x}\ (\text{adjust quadrant}).

Inclined plane: for weight WW on a ramp of angle α\alpha,

force along ramp=Wsin⁡α,force into ramp=Wcos⁡α.\text{force along ramp}=W\sin\alpha,\qquad \text{force into ramp}=W\cos\alpha.
Example
Resultant of two forces
F₁ = 40 NF₂ = 60 NR ≈ 72.11 N
The resultant is the diagonal of the parallelogram on F⃗1\vec F_1 and F⃗2\vec F_2.

F⃗1\vec F_1: 4040 N at 30∘30^\circ. F⃗2\vec F_2: 6060 N at 120∘120^\circ.

F⃗1=⟨40cos⁡30∘,40sin⁡30∘⟩=⟨34.64, 20⟩,F⃗2=⟨60cos⁡120∘,60sin⁡120∘⟩=⟨−30, 51.96⟩,R⃗=⟨4.64, 71.96⟩.\begin{aligned} \vec F_1&=\langle 40\cos 30^\circ,40\sin 30^\circ\rangle=\langle 34.64,\ 20\rangle,\\ \vec F_2&=\langle 60\cos 120^\circ,60\sin 120^\circ\rangle=\langle -30,\ 51.96\rangle,\\ \vec R&=\langle 4.64,\ 71.96\rangle. \end{aligned}
∥R⃗∥=4.642+71.962=5199.7=72.11 N,θ=tan⁡−1 ⁣71.964.64=86.3∘.\|\vec R\|=\sqrt{4.64^2+71.96^2}=\sqrt{5199.7}=72.11\ \text{N},\quad \theta=\tan^{-1}\!\frac{71.96}{4.64}=86.3^\circ .
Tip

Navigation reminder: a compass bearing is measured clockwise from North, but a direction angle is measured counterclockwise from the positive xx-axis (East). Convert carefully, or set East =x=x, North =y=y and resolve each velocity into components before adding.

Try it
Quick check

A boat points due north at 1212 km/h while the current carries it due east at 55 km/h. Find its actual speed and its bearing.

Show the answer

Answer: 1313 km/h on a bearing of about N22.6∘22.6^\circE. With East =x=x, North =y=y, the ground velocity is ⟨5,12⟩\langle 5,12\rangle, of length 25+144=13\sqrt{25+144}=13. It leans east of north by tan⁡−1512≈22.6∘\tan^{-1}\tfrac{5}{12}\approx 22.6^\circ.

Going Deeper: Advanced Vector Ideas

Tip

Same rounding rules apply: keep exact radicals when clean, otherwise round magnitudes/components to two decimals and angles to 0.1∘0.1^\circ. In proofs, prefer exact arithmetic so a dot product that should be zero comes out exactly zero.

Concept
Magnitude of a sum or difference from the angle

You do not need components to find ∥u⃗+v⃗∥\|\vec u+\vec v\| — only the two magnitudes and the angle θ\theta between the vectors. Expanding ∥u⃗±v⃗∥2=(u⃗±v⃗)⋅(u⃗±v⃗)\|\vec u\pm\vec v\|^2=(\vec u\pm\vec v)\cdot(\vec u\pm\vec v) gives the Law of Cosines in vector form:

∥u⃗±v⃗∥2=∥u⃗∥2+∥v⃗∥2±2 ∥u⃗∥ ∥v⃗∥cos⁡θ.\|\vec u\pm\vec v\|^2=\|\vec u\|^2+\|\vec v\|^2\pm 2\,\|\vec u\|\,\|\vec v\|\cos\theta .

This is exactly the parallelogram law: u⃗+v⃗\vec u+\vec v is the diagonal and u⃗−v⃗\vec u-\vec v is the other diagonal of the parallelogram built on u⃗\vec u and v⃗\vec v.

Reminder — Law of Cosines:  c2=a2+b2−2abcos⁡Cc^{2}=a^{2}+b^{2}-2ab\cos C
Example
Both diagonals from magnitudes and the angle
60°537√19
The long diagonal is u⃗+v⃗\vec u+\vec v, the short one u⃗−v⃗\vec u-\vec v.

Vectors u⃗\vec u and v⃗\vec v have ∥u⃗∥=5\|\vec u\|=5, ∥v⃗∥=3\|\vec v\|=3, and the angle between them is θ=60∘\theta=60^\circ.

∥u⃗+v⃗∥2=52+32+2(5)(3)cos⁡60∘=25+9+30(0.5)=49 ⟹ ∥u⃗+v⃗∥=7.\|\vec u+\vec v\|^2=5^2+3^2+2(5)(3)\cos 60^\circ=25+9+30(0.5)=49\ \Longrightarrow\ \|\vec u+\vec v\|=7.
∥u⃗−v⃗∥2=25+9−30(0.5)=19 ⟹ ∥u⃗−v⃗∥=19≈4.36.\|\vec u-\vec v\|^2=25+9-30(0.5)=19\ \Longrightarrow\ \|\vec u-\vec v\|=\sqrt{19}\approx 4.36.

Answer: the diagonals measure 77 and ≈4.36\approx 4.36.

Concept
Proving geometry with vectors

Turn a geometric claim about points A,B,C,…A,B,C,\dots into an algebra statement about the vectors between them:

  • Perpendicular / right angle at BB: show BA⃗⋅BC⃗=0\vec{BA}\cdot\vec{BC}=0.
  • Equal lengths (isosceles): show ∥AB⃗∥=∥AC⃗∥\|\vec{AB}\|=\|\vec{AC}\|, i.e. equal v⃗⋅v⃗\vec v\cdot\vec v.
  • Parallel / collinear: show one vector is a scalar multiple of the other, u⃗=kv⃗\vec u=k\vec v.
  • Midpoint MM of ABAB: OM⃗=12(OA⃗+OB⃗)\vec{OM}=\tfrac12(\vec{OA}+\vec{OB}).

Because the dot product is exact, these tests are cleaner than measuring slopes or distances.

Example
Proving a triangle is right and isosceles
xyA(1, 2)B(5, 4)C(3, 8)
BA⃗⋅BC⃗=0\vec{BA}\cdot\vec{BC}=0 and ∥BA⃗∥=∥BC⃗∥\|\vec{BA}\|=\|\vec{BC}\|: a right angle and equal legs.

Triangle with A(1,2)A(1,2), B(5,4)B(5,4), C(3,8)C(3,8). Form the edge vectors at BB:

BA⃗=A−B=⟨−4,−2⟩,BC⃗=C−B=⟨−2,4⟩.\vec{BA}=A-B=\langle -4,-2\rangle,\qquad \vec{BC}=C-B=\langle -2,4\rangle .
BA⃗⋅BC⃗=(−4)(−2)+(−2)(4)=8−8=0 ⟹ right angle at B.\vec{BA}\cdot\vec{BC}=(-4)(-2)+(-2)(4)=8-8=0\ \Longrightarrow\ \text{right angle at }B.

Now compare the two legs:

∥BA⃗∥=(−4)2+(−2)2=20,∥BC⃗∥=(−2)2+42=20.\|\vec{BA}\|=\sqrt{(-4)^2+(-2)^2}=\sqrt{20},\qquad \|\vec{BC}\|=\sqrt{(-2)^2+4^2}=\sqrt{20}.

Equal legs, so the triangle is an isosceles right triangle. ✓

Concept
Direction of an angle bisector

The bisector of the angle between u⃗\vec u and v⃗\vec v points along the sum of the two unit vectors:

b⃗=u^+v^=u⃗∥u⃗∥+v⃗∥v⃗∥.\vec b=\hat u+\hat v=\frac{\vec u}{\|\vec u\|}+\frac{\vec v}{\|\vec v\|}.

Using unit vectors is essential — adding u⃗+v⃗\vec u+\vec v directly leans toward the longer vector. Any positive scalar multiple of b⃗\vec b gives the same bisector direction.

Example
Bisector direction between two vectors
xyuvû + v̂u + v
u^+v^=⟨1,1⟩\hat u+\hat v=\langle1,1\rangle bisects the right angle; u⃗+v⃗\vec u+\vec v (dashed) leans toward the longer v⃗\vec v.

Bisect the angle between u⃗=⟨3,0⟩\vec u=\langle 3,0\rangle and v⃗=⟨0,4⟩\vec v=\langle 0,4\rangle.

u^=⟨1,0⟩,v^=⟨0,1⟩,b⃗=u^+v^=⟨1,1⟩.\hat u=\langle 1,0\rangle,\qquad \hat v=\langle 0,1\rangle,\qquad \vec b=\hat u+\hat v=\langle 1,1\rangle .

Its direction angle is tan⁡−1(1/1)=45∘\tan^{-1}(1/1)=45^\circ, exactly halfway between 0∘0^\circ and 90∘90^\circ. ✓ Note that u⃗+v⃗=⟨3,4⟩\vec u+\vec v=\langle 3,4\rangle (angle 53.1∘53.1^\circ) would have been wrong — it tilts toward the longer v⃗\vec v.

Tip

Work along a path with a constant force: for a constant F⃗\vec F, the work over a path made of straight legs d⃗1,d⃗2,…\vec d_1,\vec d_2,\dots is

W=F⃗⋅d⃗1+F⃗⋅d⃗2+⋯=F⃗⋅(d⃗1+d⃗2+⋯ )=F⃗⋅d⃗total.W=\vec F\cdot\vec d_1+\vec F\cdot\vec d_2+\cdots=\vec F\cdot(\vec d_1+\vec d_2+\cdots)=\vec F\cdot\vec d_{\text{total}} .

So a constant force does path-independent work: only the straight-line displacement from start to finish matters.

Concept
Equilibrium: resultant of several forces is zero

An object is in equilibrium when the forces on it sum to the zero vector:

∑F⃗=0⃗⟺∑Fx=0  and  ∑Fy=0.\sum \vec F=\vec 0\quad\Longleftrightarrow\quad \sum F_x=0\ \text{ and }\ \sum F_y=0 .

For a weight hung by two ropes, resolve each tension into components and force both totals to zero. On an inclined plane of angle α\alpha, split gravity into the part Wsin⁡αW\sin\alpha down the ramp (balanced by friction/tension) and Wcos⁡αW\cos\alpha into the ramp (balanced by the normal force).

Example
Two rope tensions holding a weight
100 lbT₁T₂30°60°
Rope 1 at 120∘120^\circ (that is, 60∘60^\circ above the leftward horizontal) and rope 2 at 30∘30^\circ.

A 100100-lb weight hangs in equilibrium from two ropes: rope 1 pulls up-left at 120∘120^\circ, rope 2 up-right at 30∘30^\circ (angles from the positive xx-axis). Find the tensions T1,T2T_1,T_2. Gravity is ⟨0,−100⟩\langle 0,-100\rangle. Equilibrium in each axis:

x:T1cos⁡120∘+T2cos⁡30∘=0⇒ −0.5 T1+0.8660 T2=0,y:T1sin⁡120∘+T2sin⁡30∘−100=0⇒ 0.8660 T1+0.5 T2=100.\begin{aligned} x:\quad & T_1\cos 120^\circ+T_2\cos 30^\circ=0 &&\Rightarrow\ -0.5\,T_1+0.8660\,T_2=0,\\ y:\quad & T_1\sin 120^\circ+T_2\sin 30^\circ-100=0 &&\Rightarrow\ 0.8660\,T_1+0.5\,T_2=100. \end{aligned}

From the xx-equation, T1=1.7320 T2T_1=1.7320\,T_2. Substitute into the yy-equation:

0.8660(1.7320 T2)+0.5 T2=100 ⟹ 2.0 T2=100 ⟹ T2=50 lb,  T1≈86.60 lb.0.8660(1.7320\,T_2)+0.5\,T_2=100\ \Longrightarrow\ 2.0\,T_2=100\ \Longrightarrow\ T_2=50\ \text{lb},\ \ T_1\approx 86.60\ \text{lb}.
Example
Navigation: correcting a heading for wind
air 250 mphwind 40groundφ
Heading φ≈9.2∘\varphi\approx 9.2^\circ west of north cancels the eastward wind.

A plane cruises at airspeed 250250 mph and must travel due north. A wind blows toward the east at 4040 mph, so w⃗=⟨40,0⟩\vec w=\langle 40,0\rangle (East =x=x, North =y=y). The pilot heads slightly west of north by angle φ\varphi, so the air velocity is a⃗=⟨−250sin⁡φ, 250cos⁡φ⟩\vec a=\langle -250\sin\varphi,\ 250\cos\varphi\rangle. The ground velocity g⃗=a⃗+w⃗\vec g=\vec a+\vec w must have zero east-west component:

−250sin⁡φ+40=0 ⟹ sin⁡φ=40250=0.16 ⟹ φ=9.2∘ west of north.-250\sin\varphi+40=0\ \Longrightarrow\ \sin\varphi=\frac{40}{250}=0.16\ \Longrightarrow\ \varphi=9.2^\circ\ \text{west of north}.

Groundspeed is the surviving north component:

∥g⃗∥=250cos⁡9.2∘=250(0.9871)=246.78 mph.\|\vec g\|=250\cos 9.2^\circ=250(0.9871)=246.78\ \text{mph}.

Answer: steer 9.2∘9.2^\circ west of north; the plane makes good ≈246.78\approx 246.78 mph northward.

Try it
Quick check

∥u⃗∥=4\|\vec u\|=4, ∥v⃗∥=6\|\vec v\|=6, and the angle between them is 120∘120^\circ. Find ∥u⃗+v⃗∥\|\vec u+\vec v\| and ∥u⃗−v⃗∥\|\vec u-\vec v\|.

Show the answer

Answer: ∥u⃗+v⃗∥=27≈5.29\|\vec u+\vec v\|=2\sqrt7\approx 5.29 and ∥u⃗−v⃗∥=219≈8.72\|\vec u-\vec v\|=2\sqrt{19}\approx 8.72. ∥u⃗±v⃗∥2=16+36±2(4)(6)cos⁡120∘=52∓24\|\vec u\pm\vec v\|^2=16+36\pm 2(4)(6)\cos120^\circ=52\mp 24. So the sum has square 2828 and the difference 7676. With an obtuse angle the difference is the longer diagonal.

Why It Works

The two dot-product formulas agree because of the law of cosines. For u\mathbf u and v\mathbf v with angle θ\theta between them, ∣u−v∣2=∣u∣2+∣v∣2−2∣u∣∣v∣cos⁡θ|\mathbf u - \mathbf v|^2 = |\mathbf u|^2 + |\mathbf v|^2 - 2|\mathbf u||\mathbf v|\cos\theta; expanding the left side in components gives ∣u∣2+∣v∣2−2(u1v1+u2v2)|\mathbf u|^2 + |\mathbf v|^2 - 2(u_1v_1 + u_2v_2). Comparing, u1v1+u2v2=∣u∣∣v∣cos⁡θu_1v_1 + u_2v_2 = |\mathbf u||\mathbf v|\cos\theta. The component formula is easy to compute; the cosine formula is what it means.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
The dot product
u⋅v=u1v1+u2v2=∣u∣∣v∣cos⁡θ\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2=|\mathbf{u}||\mathbf{v}|\cos\theta

What it means. Multiply matching components and add; the result also measures how aligned the vectors are.

Example. (1,2)⋅(3,4)=1⋅3+2⋅4=11(1,2)\cdot(3,4)=1\cdot3+2\cdot4=11.

Why it works. Apply the Law of Cosines to the triangle formed by u\mathbf u, v\mathbf v and v−u\mathbf v-\mathbf u. Expanding ∣v−u∣2|\mathbf v-\mathbf u|^2 in components and comparing with ∣u∣2+∣v∣2−2∣u∣∣v∣cos⁡θ|\mathbf u|^2+|\mathbf v|^2-2|\mathbf u||\mathbf v|\cos\theta leaves exactly u1v1+u2v2=∣u∣∣v∣cos⁡θu_1v_1+u_2v_2=|\mathbf u||\mathbf v|\cos\theta.

Tip. A dot product of 00 means the vectors are perpendicular — the fastest perpendicularity test there is.