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 `P``Q``R` through the vertex `Q` is the segment `Q``H` perpendicular to the base `P``R`. 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) withu,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 `P``Q` 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 `P``Q` and `R``S` 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 somek,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 eliminatingl; 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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