Mathematics · Book 2 · Grades 10–12

High School Mathematics

High School Mathematics · Grades 10–12

16The Scalar Product in the Plane

The scalar product attaches a number to a pair of vectors, and that number sees geometry: it vanishes exactly for perpendicular directions, and it measures angles and lengths. It is the engine behind the law of cosines, equations of circles, and perpendicularity computations. The same tool, in space, is the subject of Chapter 31.

16.1 Definition and first computations

Definition 16.1 (Scalar product)

The scalar product of two vectors u⃗\vec u and v⃗\vec v is the number

u⃗⋅v⃗=12(∥u⃗+v⃗∥2−∥u⃗∥2−∥v⃗∥2),\vec u \cdot \vec v = \tfrac12\left(\norm{\vec u + \vec v}^2 - \norm{\vec u}^2 - \norm{\vec v}^2\right),

where ∥u⃗∥\norm{\vec u} denotes the length (norm) of u⃗\vec u.

Theorem 16.2 (The four faces of the scalar product)

Let u⃗ (x,y)\vec u\,(x, y) and v⃗ (x′,y′)\vec v\,(x', y') be nonzero vectors, and θ\theta the angle between them. Then

  1. (coordinates) u⃗⋅v⃗=xx′+yy′\vec u \cdot \vec v = xx' + yy';
  2. (cosine) u⃗⋅v⃗=∥u⃗∥ ∥v⃗∥cos⁡θ\vec u \cdot \vec v = \norm{\vec u}\,\norm{\vec v}\cos\theta;
  3. (projection) u⃗⋅v⃗=u⃗⋅v⃗′\vec u \cdot \vec v = \vec u \cdot \vec v', where v⃗′\vec v' is the orthogonal projection of v⃗\vec v on the direction of u⃗\vec u; if u⃗=OA→\vec u = \vect{OA} and v⃗=OB→\vec v = \vect{OB}, then u⃗⋅v⃗=OA‾×OH‾\vec u \cdot \vec v = \overline{OA} \times \overline{OH}, a product of signed lengths, where HH is the foot of the perpendicular from BB to the line (OA)(OA);
  4. (norms) the defining formula of Definition 16.1.

Proof of 1 and 2. 1. In coordinates, ∥u⃗∥2=x2+y2\norm{\vec u}^2 = x^2 + y^2 (Pythagoras) and u⃗+v⃗\vec u + \vec v has coordinates (x+x′,y+y′)(x + x', y + y'), so

2 u⃗⋅v⃗=(x+x′)2+(y+y′)2−(x2+y2)−(x′2+y′2)=2xx′+2yy′,\begin{align*} 2\,\vec u \cdot \vec v &= (x + x')^2 + (y + y')^2 - (x^2 + y^2) - (x'^2 + y'^2)\\ &= 2xx' + 2yy', \end{align*}

after expanding the squares and cancelling.

2. Choose the coordinate system so that u⃗\vec u points along the positive xx-axis: u⃗ (∥u⃗∥,0)\vec u\,(\norm{\vec u}, 0) and v⃗ (∥v⃗∥cos⁡θ, ∥v⃗∥sin⁡θ)\vec v\,(\norm{\vec v}\cos\theta,\ \norm{\vec v}\sin\theta) (Chapter 14). Formula 1 then gives u⃗⋅v⃗=∥u⃗∥ ∥v⃗∥cos⁡θ+0\vec u \cdot \vec v = \norm{\vec u}\,\norm{\vec v}\cos\theta + 0. Formula 3 reads off the same computation: the projection of v⃗\vec v on the xx-axis has signed length ∥v⃗∥cos⁡θ=OH‾\norm{\vec v}\cos\theta = \overline{OH}. ∎

The projection view: u v = OA × OH, where H is the foot of the perpendicular from the tip of v. The orange segment OH has signed length v.
The projection view: u⃗⋅v⃗=OA‾×OH‾\vec u \cdot \vec v = \overline{OA} \times \overline{OH}, where HH is the foot of the perpendicular from the tip of v⃗\vec v. The orange segment OHOH has signed length ∥v⃗∥cos⁡θ\norm{\vec v}\cos\theta.

Example 16.3

u⃗ (3,1)\vec u\,(3, 1) and v⃗ (2,−4)\vec v\,(2, -4): u⃗⋅v⃗=6−4=2\vec u \cdot \vec v = 6 - 4 = 2. The norms are ∥u⃗∥=10\norm{\vec u} = \sqrt{10} and ∥v⃗∥=20\norm{\vec v} = \sqrt{20}, so cos⁡θ=21020=2102=210\cos\theta = \frac{2}{\sqrt{10}\sqrt{20}} = \frac{2}{10\sqrt2} = \frac{\sqrt2}{10}: the angle is a bit less than 82∘82^\circ.

Proposition 16.4 (Algebraic rules)

For all vectors and all λ∈R\lambda \in \R:

u⃗⋅v⃗=v⃗⋅u⃗,u⃗⋅(v⃗+w⃗)=u⃗⋅v⃗+u⃗⋅w⃗,(λu⃗)⋅v⃗=λ (u⃗⋅v⃗),\vec u \cdot \vec v = \vec v \cdot \vec u, \qquad \vec u \cdot (\vec v + \vec w) = \vec u \cdot \vec v + \vec u \cdot \vec w, \qquad (\lambda \vec u)\cdot \vec v = \lambda\,(\vec u \cdot \vec v),
u⃗⋅u⃗=∥u⃗∥2,∥u⃗+v⃗∥2=∥u⃗∥2+2 u⃗⋅v⃗+∥v⃗∥2.\vec u \cdot \vec u = \norm{\vec u}^2, \qquad \norm{\vec u + \vec v}^2 = \norm{\vec u}^2 + 2\,\vec u\cdot\vec v + \norm{\vec v}^2 .

Proof. All follow from the coordinate formula: for instance u⃗⋅(v⃗+w⃗)=x(x′+x′′)+y(y′+y′′)=(xx′+yy′)+(xx′′+yy′′)\vec u \cdot (\vec v + \vec w) = x(x' + x'') + y(y' + y'') = (xx' + yy') + (xx'' + yy''). The last identity expands ∥u⃗+v⃗∥2=(u⃗+v⃗)⋅(u⃗+v⃗)\norm{\vec u + \vec v}^2 = (\vec u + \vec v)\cdot(\vec u + \vec v) using the first three rules — the scalar product behaves exactly like an ordinary product, so the usual algebraic identities apply to vectors. ∎

16.2 Orthogonality

Definition 16.5 (Orthogonal vectors)

Two vectors are orthogonal, written u⃗⊥v⃗\vec u \perp \vec v, if u⃗⋅v⃗=0\vec u \cdot \vec v = 0.

Remark 16.6

For nonzero vectors, formula 2 of Theorem 16.2 shows u⃗⋅v⃗=0  ⟺  cos⁡θ=0  ⟺  \vec u \cdot \vec v = 0 \iff \cos\theta = 0 \iff the directions are perpendicular. The zero vector is orthogonal to everything.

Example 16.7

u⃗ (3,2)\vec u\,(3, 2) and v⃗ (−4,6)\vec v\,(-4, 6): u⃗⋅v⃗=−12+12=0\vec u \cdot \vec v = -12 + 12 = 0: orthogonal. In general, u⃗ (a,b)\vec u\,(a, b) is always orthogonal to v⃗ (−b,a)\vec v\,(-b, a) — a quarter-turn rotation.

16.3 The law of cosines

Theorem 16.8 (Law of cosines)

In a triangle ABCABC with sides a=BCa = BC, b=CAb = CA, c=ABc = AB and angle A^\widehat A at the vertex AA:

a2=b2+c2−2bc cos⁡A^.a^2 = b^2 + c^2 - 2bc\,\cos\widehat A .

Proof. Write BC→=BA→+AC→=AC→−AB→\vect{BC} = \vect{BA} + \vect{AC} = \vect{AC} - \vect{AB} and expand with Proposition 16.4:

a2=∥BC→∥2=∥AC→∥2−2 AC→⋅AB→+∥AB→∥2=b2+c2−2 AB→⋅AC→.a^2 = \norm{\vect{BC}}^2 = \norm{\vect{AC}}^2 - 2\,\vect{AC}\cdot\vect{AB} + \norm{\vect{AB}}^2 = b^2 + c^2 - 2\,\vect{AB}\cdot\vect{AC}.

By the cosine formula, AB→⋅AC→=c b cos⁡A^\vect{AB}\cdot\vect{AC} = c\,b\,\cos\widehat A. ∎

Remark 16.9

When A^\widehat A is a right angle, cos⁡A^=0\cos\widehat A = 0 and the law of cosines reduces to the Pythagorean theorem a2=b2+c2a^2 = b^2 + c^2: it is Pythagoras with a correction term for non-right angles.

The law of cosines: knowing two sides b, c and the angle between them determines the third side a.
The law of cosines: knowing two sides bb, cc and the angle between them determines the third side aa.

Example 16.10

Two sides of a triangle measure b=5b = 5 and c=8c = 8, enclosing an angle A^=π3\widehat A = \frac\pi3. Then

a2=25+64−2×5×8×12=89−40=49,a^2 = 25 + 64 - 2 \times 5 \times 8 \times \frac12 = 89 - 40 = 49,

so a=7a = 7.

16.4 Applications: circles and perpendicular lines

Proposition 16.11 (Circle with a given diameter)

A point MM lies on the circle of diameter [AB][AB] if and only if

MA→⋅MB→=0.\vect{MA} \cdot \vect{MB} = 0 .

Proof. If M≠A,BM \neq A, B, the condition says the angle at MM in the triangle AMBAMB is right, and the points seeing a segment under a right angle are exactly those of the circle with that diameter. Here is a direct proof: let OO be the midpoint of [AB][AB] and r=AB2r = \frac{AB}{2}. Then MA→=MO→+OA→\vect{MA} = \vect{MO} + \vect{OA} and MB→=MO→+OB→=MO→−OA→\vect{MB} = \vect{MO} + \vect{OB} = \vect{MO} - \vect{OA}, so

MA→⋅MB→=∥MO→∥2−∥OA→∥2=MO2−r2,\vect{MA}\cdot\vect{MB} = \norm{\vect{MO}}^2 - \norm{\vect{OA}}^2 = MO^2 - r^2,

which vanishes exactly when MO=rMO = r, i.e. when MM is on the circle of center OO and radius rr. ∎

Proposition 16.12 (Normal vector of a line)

The vector n⃗ (a,b)\vec n\,(a, b) is orthogonal to every direction vector of the line d:ax+by+c=0d : ax + by + c = 0; one says n⃗\vec n is a normal vector of dd. Two lines are perpendicular exactly when their normal (or direction) vectors are orthogonal.

Proof. A direction vector of dd is u⃗ (−b,a)\vec u\,(-b, a) (Theorem 15.9), and n⃗⋅u⃗=a(−b)+b(a)=0\vec n \cdot \vec u = a(-b) + b(a) = 0. ∎

Method 16.13 (Using normal vectors)

  • Line through AA with normal n⃗ (a,b)\vec n\,(a,b): write AM→⋅n⃗=0\vect{AM} \cdot \vec n = 0, which expands to ax+by+c=0ax + by + c = 0 with cc determined by AA.
  • Perpendicularity of d:ax+by+c=0d: ax+by+c = 0 and d′:a′x+b′y+c′=0d': a'x+b'y+c' = 0: check aa′+bb′=0aa' + bb' = 0.

Example 16.14

The line through P(2,−1)P(2, -1) perpendicular to d:3x−y+4=0d : 3x - y + 4 = 0: a direction vector of dd is (1,3)(1, 3), which serves as normal vector of the perpendicular. Equation: x+3y+c=0x + 3y + c = 0 with 2−3+c=02 - 3 + c = 0, so c=1c = 1: the line is x+3y+1=0x + 3y + 1 = 0.

Example 16.15 (Equation of a circle)

The circle of center Ω(1,2)\Omega(1, 2) and radius 33 is the set of points M(x,y)M(x,y) with ΩM2=9\Omega M^2 = 9:

(x−1)2+(y−2)2=9.(x - 1)^2 + (y - 2)^2 = 9 .

Conversely, x2+y2−2x−4y−4=0x^2 + y^2 - 2x - 4y - 4 = 0 rewrites, completing the squares (Chapter 10), as (x−1)2+(y−2)2=9(x-1)^2 + (y-2)^2 = 9: the same circle.

16.5 Exercises

Exercise 16.1 ★

Compute u⃗⋅v⃗\vec u \cdot \vec v for

u⃗ (2,5), v⃗ (3,−1);u⃗ (1,−4), v⃗ (8,2);∥u⃗∥=3, ∥v⃗∥=4, θ=π3.\vec u\,(2, 5),\ \vec v\,(3, -1); \qquad \vec u\,(1, -4),\ \vec v\,(8, 2); \qquad \norm{\vec u} = 3,\ \norm{\vec v} = 4,\ \theta = \frac{\pi}{3} .
Solution

Solution of Exercise 16.1.

(2,5)⋅(3,−1)=6−5=1(2,5)\cdot(3,-1) = 6 - 5 = 1.

(1,−4)⋅(8,2)=8−8=0(1,-4)\cdot(8,2) = 8 - 8 = 0 (orthogonal vectors).

u⃗⋅v⃗=3×4×cos⁡π3=12×12=6\vec u\cdot\vec v = 3 \times 4 \times \cos\frac\pi3 = 12 \times \frac12 = 6.

Exercise 16.2 ★

ABCABC is an equilateral triangle of side 66. Compute AB→⋅AC→\vect{AB} \cdot \vect{AC} and AB→⋅BC→\vect{AB} \cdot \vect{BC}.

Solution

Solution of Exercise 16.2.

AB→⋅AC→=6×6×cos⁡60∘=18\vect{AB}\cdot\vect{AC} = 6 \times 6 \times \cos 60^\circ = 18.

The vectors AB→\vect{AB} and BC→\vect{BC} make an angle of 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ (place them tail to tail at BB), so AB→⋅BC→=36cos⁡120∘=−18\vect{AB}\cdot\vect{BC} = 36 \cos 120^\circ = -18.

Exercise 16.3 ★

For which value of tt are u⃗ (3,t)\vec u\,(3, t) and v⃗ (t−8,1)\vec v\,(t - 8, 1) orthogonal? And for which value(s) of tt is u⃗ (t,2)\vec u\,(t, 2) orthogonal to itself minus v⃗ (4,2)\vec v\,(4, 2), i.e. u⃗⊥(u⃗−v⃗)\vec u \perp (\vec u - \vec v)?

Solution

Solution of Exercise 16.3.

u⃗⋅v⃗=3(t−8)+t=4t−24=0\vec u \cdot \vec v = 3(t - 8) + t = 4t - 24 = 0 gives t=6t = 6.

u⃗−v⃗=(t−4, 0)\vec u - \vec v = (t - 4,\ 0), so u⃗⋅(u⃗−v⃗)=t(t−4)+0=t2−4t\vec u \cdot (\vec u - \vec v) = t(t-4) + 0 = t^2 - 4t, which vanishes for t=0t = 0 or t=4t = 4.

Exercise 16.4 ★

Compute the angle θ\theta between u⃗ (1,2)\vec u\,(1, 2) and v⃗ (3,1)\vec v\,(3, 1) (give cos⁡θ\cos\theta exactly, then θ\theta to the nearest degree).

Solution

Solution of Exercise 16.4.

u⃗⋅v⃗=3+2=5\vec u \cdot \vec v = 3 + 2 = 5, ∥u⃗∥=5\norm{\vec u} = \sqrt5, ∥v⃗∥=10\norm{\vec v} = \sqrt{10}, so

cos⁡θ=55 10=552=22,\cos\theta = \frac{5}{\sqrt5\,\sqrt{10}} = \frac{5}{5\sqrt2} = \frac{\sqrt2}{2},

hence θ=45∘\theta = 45^\circ exactly.

Exercise 16.5 ★★

A triangle has sides b=4b = 4, c=6c = 6 and angle A^=2π3\widehat A = \frac{2\pi}{3} between them. Compute the third side aa. Then, in a triangle with sides a=7a = 7, b=5b = 5, c=4c = 4, compute cos⁡A^\cos\widehat A and decide whether A^\widehat A is acute or obtuse.

Solution

Solution of Exercise 16.5.

a2=16+36−2×4×6×cos⁡2π3=52−48×(−12)=76a^2 = 16 + 36 - 2 \times 4 \times 6 \times \cos\frac{2\pi}{3} = 52 - 48 \times \left(-\frac12\right) = 76, so a=76=219≈8.7a = \sqrt{76} = 2\sqrt{19} \approx 8.7.

For the second triangle, the law of cosines solved for the angle gives

cos⁡A^=b2+c2−a22bc=25+16−492×5×4=−840=−15,\cos\widehat A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{25 + 16 - 49}{2 \times 5 \times 4} = -\frac{8}{40} = -\frac15,

negative: A^\widehat A is obtuse.

Exercise 16.6 ★★

Let A(1,1)A(1, 1), B(5,−1)B(5, -1) and C(3,5)C(3, 5). Compute AB→⋅AC→\vect{AB} \cdot \vect{AC}; is the angle A^\widehat A right? Compute the three side lengths and check your answer with the law of cosines.

Solution

Solution of Exercise 16.6.

AB→ (4,−2)\vect{AB}\,(4, -2) and AC→ (2,4)\vect{AC}\,(2, 4): AB→⋅AC→=8−8=0\vect{AB}\cdot\vect{AC} = 8 - 8 = 0, so the angle at AA is right. Sides: AB=16+4=20AB = \sqrt{16 + 4} = \sqrt{20}, AC=4+16=20AC = \sqrt{4 + 16} = \sqrt{20}, and BC→ (−2,6)\vect{BC}\,(-2, 6) gives BC=40BC = \sqrt{40}. Consistency: BC2=40=20+20=AB2+AC2BC^2 = 40 = 20 + 20 = AB^2 + AC^2, the law of cosines with cos⁡A^=0\cos\widehat A = 0 (Pythagoras).

Exercise 16.7 ★★

Give an equation of the circle of center Ω(−2,3)\Omega(-2, 3) and radius 55. Then determine the center and radius of the circle

x2+y2−6x+2y−6=0.x^2 + y^2 - 6x + 2y - 6 = 0 .
Solution

Solution of Exercise 16.7.

Center–radius form: (x+2)2+(y−3)2=25(x + 2)^2 + (y - 3)^2 = 25.

For the second circle, complete the squares:

x2−6x+y2+2y=6  ⟺  (x−3)2−9+(y+1)2−1=6  ⟺  (x−3)2+(y+1)2=16:x^2 - 6x + y^2 + 2y = 6 \iff (x - 3)^2 - 9 + (y + 1)^2 - 1 = 6 \iff (x-3)^2 + (y+1)^2 = 16 :

center (3,−1)(3, -1), radius 44.

Exercise 16.8 ★★

Let A(−3,0)A(-3, 0) and B(0,2)B(0, 2). Using Proposition 16.11, find an equation of the circle of diameter [AB][AB], and check that the origin O(0,0)O(0,0) lies on it.

Solution

Solution of Exercise 16.8.

M(x,y)M(x,y) is on the circle when MA→⋅MB→=0\vect{MA}\cdot\vect{MB} = 0 with MA→ (−3−x, −y)\vect{MA}\,(-3 - x,\ -y) and MB→ (−x, 2−y)\vect{MB}\,(-x,\ 2 - y):

(−3−x)(−x)+(−y)(2−y)=x2+3x+y2−2y=0.(-3 - x)(-x) + (-y)(2 - y) = x^2 + 3x + y^2 - 2y = 0 .

For the origin: 0+0+0−0=00 + 0 + 0 - 0 = 0, so OO lies on the circle — geometrically, OA→ (−3,0)\vect{OA}\,(-3,0) and OB→ (0,2)\vect{OB}\,(0,2) are orthogonal, so OO sees [AB][AB] under a right angle. (Completing the squares: center (−32,1)\left(-\frac32, 1\right), radius 132\frac{\sqrt{13}}{2}.)

Exercise 16.9 ★★

Find an equation of the line through P(1,3)P(1, 3) perpendicular to d:2x+y−7=0d : 2x + y - 7 = 0, and the coordinates of the foot HH of the perpendicular (the intersection of the two lines). Deduce the distance from PP to dd.

Solution

Solution of Exercise 16.9.

The perpendicular through PP has the direction vector of dd, namely (−1,2)(-1, 2), as normal vector: equation −x+2y+c=0-x + 2y + c = 0; through P(1,3)P(1,3): −1+6+c=0-1 + 6 + c = 0, c=−5c = -5, so x−2y+5=0x - 2y + 5 = 0.

Intersection with dd: from 2x+y=72x + y = 7, y=7−2xy = 7 - 2x; substituting, x−2(7−2x)+5=5x−9=0x - 2(7 - 2x) + 5 = 5x - 9 = 0, so x=95x = \frac95, y=175y = \frac{17}5: H(95,175)H\left(\frac95, \frac{17}5\right). Then PH→(45,25)\vect{PH}\left(\frac45, \frac25\right) and

PH=1625+425=205=255≈0.89.PH = \sqrt{\frac{16}{25} + \frac{4}{25}} = \frac{\sqrt{20}}{5} = \frac{2\sqrt5}{5} \approx 0.89 .

Exercise 16.10 ★★

Using ∥u⃗+v⃗∥2=∥u⃗∥2+2u⃗⋅v⃗+∥v⃗∥2\norm{\vec u + \vec v}^2 = \norm{\vec u}^2 + 2\vec u\cdot\vec v + \norm{\vec v}^2 and its analogue for u⃗−v⃗\vec u - \vec v, prove the parallelogram identity:

∥u⃗+v⃗∥2+∥u⃗−v⃗∥2=2∥u⃗∥2+2∥v⃗∥2,\norm{\vec u + \vec v}^2 + \norm{\vec u - \vec v}^2 = 2\norm{\vec u}^2 + 2\norm{\vec v}^2 ,

and interpret it in a parallelogram (diagonals versus sides).

Solution

Solution of Exercise 16.10.

Adding

∥u⃗+v⃗∥2=∥u⃗∥2+2u⃗⋅v⃗+∥v⃗∥2and∥u⃗−v⃗∥2=∥u⃗∥2−2u⃗⋅v⃗+∥v⃗∥2,\norm{\vec u + \vec v}^2 = \norm{\vec u}^2 + 2\vec u\cdot\vec v + \norm{\vec v}^2 \quad\text{and}\quad \norm{\vec u - \vec v}^2 = \norm{\vec u}^2 - 2\vec u\cdot\vec v + \norm{\vec v}^2,

the cross terms cancel and ∥u⃗+v⃗∥2+∥u⃗−v⃗∥2=2∥u⃗∥2+2∥v⃗∥2\norm{\vec u + \vec v}^2 + \norm{\vec u - \vec v}^2 = 2\norm{\vec u}^2 + 2\norm{\vec v}^2. In a parallelogram with sides u⃗\vec u and v⃗\vec v, the diagonals are u⃗+v⃗\vec u + \vec v and u⃗−v⃗\vec u - \vec v: the sum of the squares of the two diagonals equals the sum of the squares of the four sides.

Exercise 16.11 ★★★

Let AA and BB be two points with AB=4AB = 4, and II the midpoint of [AB][AB].

  1. Show that for every point MM: MA→⋅MB→=MI2−4\vect{MA}\cdot\vect{MB} = MI^2 - 4. (Insert II: MA→=MI→+IA→\vect{MA} = \vect{MI} + \vect{IA}.)
  2. Deduce the set of points MM such that MA→⋅MB→=12\vect{MA}\cdot\vect{MB} = 12.
Solution

Solution of Exercise 16.11.

1. Since II is the midpoint, IB→=−IA→\vect{IB} = -\vect{IA} and IA=2IA = 2. Then

MA→⋅MB→=(MI→+IA→)⋅(MI→−IA→)=∥MI→∥2−∥IA→∥2=MI2−4.\vect{MA}\cdot\vect{MB} = (\vect{MI} + \vect{IA})\cdot(\vect{MI} - \vect{IA}) = \norm{\vect{MI}}^2 - \norm{\vect{IA}}^2 = MI^2 - 4 .

2. The condition MA→⋅MB→=12\vect{MA}\cdot\vect{MB} = 12 becomes MI2=16MI^2 = 16, i.e. MI=4MI = 4: the set is the circle of center II and radius 44.

16.6 Problem: Two gifts of the scalar product

Problem 16.1

Weekend problem — the addition formulas fall out of one dot product, the distance to a line out of another, and the three altitudes meet as a bonus

The scalar product looks like bookkeeping — multiply, add — yet it hands over, almost free of charge, two treasures that resisted centuries of angle-chasing: the addition formulas of trigonometry (compute cos⁡15∘\cos 15^\circ exactly!) and the distance from a point to a line in one clean fraction. As a parting gift it proves that the three altitudes of every triangle pass through one point. All four faces of Theorem 16.2 will report for duty.

Part I — Fluency.

  1. For u⃗(3,4)\vec u(3, 4) and v⃗(−1,2)\vec v(-1, 2): compute u⃗⋅v⃗\vec u \cdot \vec v, the two norms, and the angle between the vectors (to the tenth of a degree).
  2. Find kk so that (2,k)(2, k) and (3,−4)(3, -4) are orthogonal (Definition 16.5).
  3. Physics face: a sled is pulled 1010 m by a 5050 N force at 60∘60^\circ to the ground. Compute the work W=F⃗⋅d⃗W = \vec F \cdot \vec d.
  4. Law of cosines (Theorem 16.8): two sides of 55 and 77 enclose an angle of 60∘60^\circ; find the third side exactly.
  5. Reverse: a triangle has sides 44, 66, 88. Compute the cosine of the angle opposite the side 88, classify the angle, and give it to the tenth of a degree.

Part II — First gift: the addition formulas. Let u⃗=(cos⁡a,sin⁡a)\vec u = (\cos a, \sin a) and v⃗=(cos⁡b,sin⁡b)\vec v = (\cos b, \sin b): two unit vectors at angles aa and bb.

  1. Compute u⃗⋅v⃗\vec u \cdot \vec v twice — once with coordinates, once with norms and the angle between the vectors — and conclude the mother formula:

    cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b.\cos(a - b) = \cos a \cos b + \sin a \sin b .
  2. Substitute −b-b for bb (parities of sine and cosine!) to obtain cos⁡(a+b)\cos(a + b).
  3. Obtain sin⁡(a+b)\sin(a + b) from the complementary identity sin⁡x=cos⁡(π2−x)\sin x = \cos\left(\frac\pi2 - x\right) applied to x=a+bx = a + b.
  4. Exact values at last: compute cos⁡15∘\cos 15^\circ (as cos⁡(45∘−30∘)\cos(45^\circ - 30^\circ)) and sin⁡75∘\sin 75^\circ. Check both numerically.
  5. Set b=ab = a to derive the double-angle formulas: cos⁡2a\cos 2a (in its three forms) and sin⁡2a\sin 2a.
  6. From cos⁡2a=1−2sin⁡2a\cos 2a = 1 - 2\sin^2 a, compute sin⁡215∘\sin^2 15^\circ, then sin⁡15∘\sin 15^\circ as a nested radical — and verify that its square agrees with the value 6−24\frac{\sqrt6 - \sqrt2}{4} that question 9’s method gives.

Part III — Second gift: the distance to a line.

  1. Let dd be the line ax+by+c=0ax + by + c = 0, with normal vector n⃗(a,b)\vec n(a, b) (Proposition 16.12), and P(x0,y0)P(x_0, y_0) a point. Taking any point AA on dd, explain why the distance from PP to dd is ∣AP→⋅n⃗∣∥n⃗∥\dfrac{\abs{\vect{AP} \cdot \vec n}}{\norm{\vec n}}, and deduce the formula

    dist⁡(P,d)=∣ax0+by0+c∣a2+b2.\operatorname{dist}(P, d) = \frac{\abs{a x_0 + b y_0 + c}}{\sqrt{a^2 + b^2}} .

    Compute the distance from P(5,6)P(5, 6) to 3x+4y−12=03x + 4y - 12 = 0.

  2. Write the equation of the circle centered at (5,6)(5, 6) tangent to that line.
  3. Cross-check the formula: compute the distance from the origin to 3x+4y−12=03x + 4y - 12 = 0, then verify it by computing the foot of the perpendicular from the origin explicitly.
  4. Triangle A(0,0)A(0,0), B(6,0)B(6,0), C(2,5)C(2,5): compute its area using the base [AB][AB]; then compute the distance from BB to the line (AC)(AC) and check it reproduces the same area.
  5. The circle of center (2,1)(2, 1) and radius 22, against the line x+y−6=0x + y - 6 = 0: compute the center’s distance to the line and classify their position (secant, tangent, disjoint).

Part IV — The bonus: altitudes concur.

  1. Generalize Exercise 16.11: for any two points AA, BB with midpoint II, prove

    MA→⋅MB→=MI2−AB24for every point M.\vect{MA} \cdot \vect{MB} = MI^2 - \frac{AB^2}{4} \quad\text{for every point } M .
  2. Deduce Proposition 16.11 in one line: the set of points MM with MA→⋅MB→=0\vect{MA} \cdot \vect{MB} = 0 is the circle of diameter [AB][AB].

  3. The altitudes: let HH be the intersection of the altitudes from AA and from BB in a triangle ABCABC (so HA→⋅BC→=0\vect{HA} \cdot \vect{BC} = 0 and HB→⋅CA→=0\vect{HB} \cdot \vect{CA} = 0). Prove the identity

    HA→⋅BC→+HB→⋅CA→+HC→⋅AB→=0\vect{HA} \cdot \vect{BC} + \vect{HB} \cdot \vect{CA} + \vect{HC} \cdot \vect{AB} = 0

    — expand everything from HH with Chasles — and conclude that HH also lies on the altitude from CC: the three altitudes are concurrent (at the orthocenter).

  4. Finale: four faces, three trophies. Which face of Theorem 16.2 carried the addition formulas, which the distance formula, which the orthocenter — and why is “compute the same quantity twice” the motto of the whole chapter?
Solution

Solution of Problem 16.1.

1. u⃗⋅v⃗=−3+8=5\vec u \cdot \vec v = -3 + 8 = 5; ∥u⃗∥=5\norm{\vec u} = 5, ∥v⃗∥=5\norm{\vec v} = \sqrt5; cos⁡θ=555=15\cos\theta = \frac{5}{5\sqrt5} = \frac{1}{\sqrt5}: θ≈63.4∘\theta \approx 63.4^\circ.

2. 6−4k=06 - 4k = 0: k=32k = \frac32.

3. W=50×10×cos⁡60∘=250W = 50 \times 10 \times \cos 60^\circ = 250 joules: only the component along the motion works.

4. c2=25+49−2×35×12=39c^2 = 25 + 49 - 2 \times 35 \times \frac12 = 39: c=39c = \sqrt{39}.

5. cos⁡γ=16+36−642×4×6=−14\cos\gamma = \frac{16 + 36 - 64}{2 \times 4 \times 6} = -\frac14: negative, the angle is obtuse: γ≈104.5∘\gamma \approx 104.5^\circ.

6. Coordinates: u⃗⋅v⃗=cos⁡acos⁡b+sin⁡asin⁡b\vec u \cdot \vec v = \cos a \cos b + \sin a \sin b. Geometry: two unit vectors, u⃗⋅v⃗=1×1×cos⁡(a−b)\vec u \cdot \vec v = 1 \times 1 \times \cos(a - b). Equating the two computations of the same number: cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos(a - b) = \cos a \cos b + \sin a \sin b.

7. cos⁡(a+b)=cos⁡(a−(−b))=cos⁡acos⁡(−b)+sin⁡asin⁡(−b)=cos⁡acos⁡b−sin⁡asin⁡b\cos(a + b) = \cos(a - (-b)) = \cos a \cos(-b) + \sin a \sin(-b) = \cos a \cos b - \sin a \sin b.

8. sin⁡(a+b)=cos⁡(π2−a−b)=cos⁡((π2−a)−b)=cos⁡(π2−a)cos⁡b+sin⁡(π2−a)sin⁡b=sin⁡acos⁡b+cos⁡asin⁡b\sin(a + b) = \cos\left(\frac\pi2 - a - b\right) = \cos\left(\left(\frac\pi2 - a\right) - b\right) = \cos\left(\frac\pi2 - a\right)\cos b + \sin\left(\frac\pi2 - a\right)\sin b = \sin a \cos b + \cos a \sin b.

9. cos⁡15∘=cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘=22⋅32+22⋅12=6+24≈0.966\cos 15^\circ = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt2}{2} \cdot \frac{\sqrt3}{2} + \frac{\sqrt2}{2} \cdot \frac12 = \frac{\sqrt6 + \sqrt2}{4} \approx 0.966. And sin⁡75∘=cos⁡15∘\sin 75^\circ = \cos 15^\circ: same value.

10. b=ab = a in questions 7 and 8: cos⁡2a=cos⁡2a−sin⁡2a=2cos⁡2a−1=1−2sin⁡2a\cos 2a = \cos^2 a - \sin^2 a = 2\cos^2 a - 1 = 1 - 2\sin^2 a (using cos⁡2+sin⁡2=1\cos^2 + \sin^2 = 1), and sin⁡2a=2sin⁡acos⁡a\sin 2a = 2 \sin a \cos a.

11. sin⁡215∘=1−cos⁡30∘2=2−34\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{2 - \sqrt3}{4}, so sin⁡15∘=2−32\sin 15^\circ = \frac{\sqrt{2 - \sqrt3}}{2}. Squaring 6−24\frac{\sqrt6 - \sqrt2}{4}: 6−212+216=8−4316=2−34\frac{6 - 2\sqrt{12} + 2}{16} = \frac{8 - 4\sqrt3}{16} = \frac{2 - \sqrt3}{4}: the same number in two costumes.

12. AP→\vect{AP} decomposes into a part along dd (invisible to n⃗\vec n) and a part perpendicular to dd, whose length is the distance sought; dotting with the unit normal n⃗∥n⃗∥\frac{\vec n}{\norm{\vec n}} kills the first part and measures the second. With AA on dd: AP→⋅n⃗=ax0+by0−(axA+byA)=ax0+by0+c\vect{AP} \cdot \vec n = a x_0 + b y_0 - (a x_A + b y_A) = a x_0 + b y_0 + c, whence the formula. Example: ∣15+24−12∣5=275=5.4\frac{\abs{15 + 24 - 12}}{5} = \frac{27}{5} = 5.4.

13. Radius =5.4= 5.4: (x−5)2+(y−6)2=29.16(x - 5)^2 + (y - 6)^2 = 29.16.

14. Distance =∣−12∣5=2.4= \frac{\abs{-12}}{5} = 2.4. Foot: the perpendicular through the origin is (3t,4t)(3t, 4t); on the line: 9t+16t=129t + 16t = 12, t=1225t = \frac{12}{25}: foot (3625,4825)\left(\frac{36}{25}, \frac{48}{25}\right), at distance 5t=6025=2.45t = \frac{60}{25} = 2.4. The formula and the honest computation agree.

15. Base AB=6AB = 6, height == distance from CC to (AB)(AB) (the axis y=0y = 0): 55; area 1515. Line (AC)(AC): 5x−2y=05x - 2y = 0, and dist⁡(B,(AC))=∣30∣29\operatorname{dist}(B, (AC)) = \frac{\abs{30}}{\sqrt{29}}; then 12×AC×3029=12×29×3029=15\frac12 \times AC \times \frac{30}{\sqrt{29}} = \frac12 \times \sqrt{29} \times \frac{30}{\sqrt{29}} = 15: same area, either altitude.

16. dist⁡=∣2+1−6∣2=32≈2.12>2\operatorname{dist} = \frac{\abs{2 + 1 - 6}} {\sqrt2} = \frac{3}{\sqrt2} \approx 2.12 > 2: the line passes clear of the circle — disjoint.

17. Insert II: MA→⋅MB→=(MI→+IA→)⋅(MI→+IB→)=MI2+MI→⋅(IA→+IB→)+IA→⋅IB→\vect{MA} \cdot \vect{MB} = (\vect{MI} + \vect{IA}) \cdot (\vect{MI} + \vect{IB}) = MI^2 + \vect{MI} \cdot (\vect{IA} + \vect{IB}) + \vect{IA} \cdot \vect{IB}. The middle term dies (IA→+IB→=0⃗\vect{IA} + \vect{IB} = \vec 0), and IA→⋅IB→=−IA2=−AB24\vect{IA} \cdot \vect{IB} = -IA^2 = -\frac{AB^2}{4} (opposite vectors of length AB2\frac{AB}2).

18. MA→⋅MB→=0  ⟺  MI2=AB24  ⟺  MI=AB2\vect{MA} \cdot \vect{MB} = 0 \iff MI^2 = \frac{AB^2}{4} \iff MI = \frac{AB}{2}: the circle of center II and radius AB2\frac{AB}2 — the circle with diameter [AB][AB].

19. Expanding each pair from HH (BC→=HC→−HB→\vect{BC} = \vect{HC} - \vect{HB}, etc.):

HA→⋅HC→−HA→⋅HB→+HB→⋅HA→−HB→⋅HC→+HC→⋅HB→−HC→⋅HA→=0:\vect{HA} \cdot \vect{HC} - \vect{HA} \cdot \vect{HB} + \vect{HB} \cdot \vect{HA} - \vect{HB} \cdot \vect{HC} + \vect{HC} \cdot \vect{HB} - \vect{HC} \cdot \vect{HA} = 0 :

everything cancels in pairs — the identity holds for any four points. For our HH, the first two dot products vanish by hypothesis, so HC→⋅AB→=0\vect{HC} \cdot \vect{AB} = 0: HH lies on the altitude from CC. Three altitudes, one point.

20. The addition formulas came from computing one dot product with the coordinate face and the norms-and-angle face; the distance formula from the projection face; the orthocenter from the bilinearity (Chasles-and-expand) face. Compute the same quantity twice, equate, and a theorem falls out — the scalar product’s whole trade.