As noted before, the algebra permits us to calculate proofs without resorting to coordinate manipulations. As a demonstration, I'll present a very simple calculation, and then a theorem that I didn't know (and I would be very hard-pressed to come up with a classical proof).
Given points P, Q, R, the altitude of the triangle PQR through the vertex Q is the segment QH perpendicular to the base PR. Denoting hgt
.P.Q.R = |Q−H| (we write |•−•| = |‹•−•›| per abusum linguae), it is:
(17) hgt
.P.Q.R = (‹R−P›⊥⋅‹Q−P›)/|R−P|
Proof:
hgt
.P.Q.R = { definition } |Q−H| = { definition of the sine line } |Q−P|·sin.(θ
.‹R−P›.‹Q−P›) = { (16) with u, v := ‹R−P›, ‹Q−P› } (‹R−P›⊥⋅‹Q−P›)/|R−P|
The line through point P with direction vector v is the set of points P⇒v that, in the words of Euclid, "lies equally on the points of itself":
(18.0) Q ∈ P⇒v ≡ |v|⋅‹Q − P› = |Q − P|⋅v
The following are theorems:
(18.1) P ∈ P⇒v (18.2) 〈∀ k : k ≠ 0 : P⇒(k⋅v) = P⇒v 〉 (18.2) P⇒v = 〈k ∈ R | P → k⋅v〉
The parametric line through points P and Q is defined as the real function:
(18) ray
.P.Q.k = P → k·‹Q−P›
From (18.2), it is immediate that ray
.P.Q maps R to the line P⇒‹Q−P›.
As a special case, the midpoint of the segment PQ is:
(19) ray
.P.Q.½ = P → ½·‹Q−P›
Given points P, Q, R and S, the intersection of the lines containing the segments PQ and RS is:
(20)meet
.P.Q.R.S =ray
.P.Q.((‹S−R›⊥⋅‹R−P›)/(‹S−R›⊥⋅‹Q−P›))
Proof: At the intersection point
meet
.P.Q.R.S, for some k, l ∈ R:ray
.P.Q.k =ray
.R.S.l = { (18) } P → k·‹Q−P› = R → l·‹S−R› = { (6) } ‹(P → k·‹Q−P›) − (R → l·‹S−R›)› = 0 = { (9) } ‹P−R› + k·‹Q−P› − l·‹S−R› = 0 = { heading towards eliminating l; inner product } ‹S−R›⊥⋅(‹P−R› + k·‹Q−P› − l·‹S−R›) = 0 = { linearity of inner product } ‹S−R›⊥⋅‹P−R› + k·‹S−R›⊥⋅‹Q−P› − l·‹S−R›⊥⋅‹S−R› = 0 = { (14) } ‹S−R›⊥⋅‹P−R› + k·‹S−R›⊥⋅‹Q−P› = 0 = { (5); algebra } k = (‹S−R›⊥⋅‹R−P›)/(‹S−R›⊥⋅‹Q−P›)In Part III, the theorem.
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