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.

Concept
What a vector is

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

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.

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.

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.

Example
Operations and a unit vector

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.

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

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. ✓

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.

Example
Angle between two vectors

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.

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)} .
Example
Full projection and decomposition

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

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}.

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

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.

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

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:

  • 2pt
  • 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

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

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

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

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.

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.