604 Geometry and TrigonometryHence the solution to the equation is−→x = m
−→a ·−→b−→
b +1
−→a ·−→b−→a ×−→c.(C. Co ̧sni ̧ta, I. Sager, I. Matei, I. Dragot ̆ a, ̆ Culegere de probleme de Geometrie
Analitica ̆(Collection of Problems in Analytical Geometry), Editura Didactic ̆a ̧si Ped-
agogica, Bucharest, 1963) ̆
576.The vectors−→
b−−→a and−→c−−→a belong to the plane under discussion, so the vector
(−→
b −−→a)×(−→c −−→a)is perpendicular to this plane. Multiplying out, we obtain(−→
b −−→a)×(−→c −−→a)=−→
b ×−→c−−→a ×−→c −−→
b ×−→a
=−→
b ×−→c +−→c ×−→a +−→a ×−→
b.Hencethe conclusion.
577.The hypothesis implies that(−→a ×−→
b)−(−→
b ×−→c)=−→
0.
It follows that−→
b ×(−→a +−→c)=−→
0 , hence−→
b =λ(−→a +−→c), whereλis a scalar.
Analogously, we deduce−→c ×(−→a +
−→
b)=−→
0 , and substituting the formula we found
for−→
b, we obtain
−→c ×(−→a +λ−→a +λ−→c)=−→ 0.Hence( 1 +λ)−→c×−→a =−→
0. It follows thatλ=−1 and so−→
b =−−→a−−→c. Therefore,
−→a +−→b +−→c =−→0.
(C. Co ̧sni ̧t ̆a, I. Sager, I. Matei, I. Dragot ̆a,Culegere de probleme de Geometrie
Analitica ̆(Collection of Problems in Analytical Geometry), Editura Didactic ̆a ̧si Ped-
agogica, ̆Bucharest, 1963)
578.Differentiating the equation from the statement, we obtain
−→u′×−→u′+−→u×−→u′′=−→u ×−→u′′=−→v′.It follows that the vectors−→uand−→v′are perpendicular. But the original equation shows
that−→u and−→v are also perpendicular, which means that−→u stays parallel to−→v ×−→
v′.
Then we can write−→u =f−→v ×−→v′for some scalar functionf=f(t). The left-hand
side of the original equation is therefore equal tof(−→v ×−→v′)×[f′−→v ×−→
v′+f−→v′×−→v′+f−→v ×−→v′′]