Question

3.6 Distance from a line (2D/3D geometry). (Section 2.9). A common geometry problem is to determine how far a point S is from a line PQ. The figure to the right shows points P, Q, and S whose position vectors from a point Ao are written in terms of right-handed, orthogonal, unit vectors âx, ây, ây, as FP/Ao = p t Falto = gât F5/40 = szax t syây + száz Denoting C as the point on line PQ closest to S, determine the following (in terms of p, q, 82, Sy, S, and symbols in the table below). Distance between P and Q dQ/P Sy – psy + Distance between P and C dP/C C's position from P FC/P axt dP/C ᏧQ}Ꮲ ay

Not sure how to get distance between p and c AND cs position form P

Answer 1

P/A0 = pan 28/ta , và res/Ao = Syam + syầy + szar het gy ellos traut ty ay Now : s is the shortestes closest to live pe [point c] 2 pp/toc/A). (PS/Ao pelto) 20 as (pp/po malo). ( > 5/70 J/A) ao as ( par-way). (Con-atrape + (sy-bby) ay tszan) zo as p (sorte) -q (sy tylny) = 0; ~ ① Also - Cis on line pa es Etn-an) =-d are as tua ân (1-2) & (ty) ne atay as ty= ay n cz daß 2 (tn-p) = -ap tuz p(16) @ & (ty-o) = a(q) es ty zaq 0 (3 Pute it into equation ④ p(sm-p(1-2)) -q (sy - aq)=0 n psn - p²tp²d - q sy + Qq²20 n a = p² tq sy - psa p2 tq2

put it into equation ① 2 3 =PLI - PIES) soinde = putem a * Psa-934) s clao blew p&a= IPP/Ao ya/Ao | Now distance = rana tay? Distance to plc -/1pplau mstao 1 = ((p-tn7+ ty? tn Aty - O2 - care in + equation c's position rom pus elo pelto pelto How to and = (tn-p) an + ty aj to, ty are in equation 6 4 this but See here yelp is nothing vector pe & pêz ÃoC - AP 24/P = Pc/A PP/AO

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